### Fluid mechanics ( 2019 2020)

• 1. ACE 20- FLUID MECHANICS/HYDRAULICS By: Yuri G. Melliza COURSE DESCRIPTION This class provides students with an introduction to principal concepts and methods of fluid mechanics. Topics covered in the course include pressure, hydrostatics, and buoyancy; open systems and control volume analysis; mass conservation and momentum conservation for moving the models necessary to study, analyze, and design fluid systems through the application of these concepts, and to develop the problem-solving skills essential to good engineering practice of fluid mechanics in practical applications. Students will have the opportunity to demonstrate a familiarity and ability to work on fluid mechanics. These outcomes will be demonstrated through an assessment of quizzes, assignments and major examinations. GENERAL OBJECTIVES By the end of this course, you should expect to be able to calculate: • fundamental fluid properties for different fluids and flows • forces on objects submerged in both static and flowing fluids • pressures in both static and flowing fluids, and the velocities associated with different flows • forces in complicated momentum balance problems • energy loss and the flow rates associated with different flow networks in channels and pipes • dimensionless numbers important for design of experiments and practical engineering work • numerical solutions for simple fluid flow problems • properties of a boundary layer, both turbulent and laminar problems with the application of Bernoulli’s Principle Fluid Mechanics - it is a science that deals with the mechanics of fluids (liquid or gas) and is based on the same fundamental principles employed in the mechanics of solids. STUDENT COURSE OUTCOMES This is an introductory course on fluid motion, the forces that fluids exert, and the forces that are exerted on them. The study of fluid mechanics has numerous engineering applications. Fluids interact with structures such as high-rise buildings, dams, and bridges and the static and dynamic loads imposed by the fluids must be considered in the design and construction of these structures. Cars, aircraft, and ships all move through fluids, and frictional (fluid drag) forces represent a major energy sink. Water is an important resource to California and fluid mechanical problems abound in the complex system of dams, aqueducts, treatment plants, pipes, and valves used to deliver water to urban and agricultural consumers. Finally, the motions of contaminants in water and air are governed by the mechanics of fluid flow. After this course, the students are expected to: Branches of Fluid Mechanics 1. Fluid Static - deals with the forces exerted by or upon the fluid at rest. 2. Kinematics - deals with velocities and accelerations without considering forces and energy. 3. Hydrodynamics - deals with the relation of velocities and accelerations and the forces exerted by or upon the fluid in motion. Fluid - is a substance capable of flowing and having particles which easily moved and changed their relative position without the separation of mass. Distinction between Liquid and Gas Liquid 1. It has a free surface 2. A given quantity of liquid occupies a given volume of a container 3. It is incompressible Gas 1. It has no free surface 2. A given quantity of gas occupies all portions of the container regardless of its shape and size. 3. It is compressible Free Surface - a surface in which all pressures can be removed except its own vapor pressure. Vapor - is a gas in which its pressure and temperature are such that very near to its liquid state. PROPERTIES OF FLUIDS 1. Density (): It is the mass on a unit volume. 3 m kg V m ρ= 2. Specific Volume (): It is the volume per unit mass, or the reciprocal of its density. kg m m V 3 υ= 3. Specific Weight or Weight Density (): It is the weight on a unit volume. 3 3 m K N 1 0 0 0 g ρ V 1 0 0 0 mg γ m K N V W γ = = =
• 2. 4. Specific Gravity or Relative Density (S): For Liquids It is the ratio of its density to density of water at standard temperature and pressure. w L w L L γ γ ρ ρ S = = For Gases It is the ratio of its density to the density of either air or hydrogen at some specified temperature and pressure. AH G G ρ ρ S = For Gases 3 m Kg RT P = ρ Where: P - Absolute pressure in KPaa R -Gas constant in KJ/kg-K T -Absolute temperature in K m ole kg/kg in weight molecular - M K - kg KJ M 3143 . 8 R  = where: w = 1000 kg/m3 (Standard condition) w = 9.81 KN/m3 (Standard condition) AH = is the density of either air or hydrogen at some value of P and T. 5. Temperature: It is the measure of the intensity of heat of a fluid. Celsius and Fahrenheit Scale 3 2 ) C ( 8 . 1 F 8 . 1 3 2 F C +  =  −  =  Absolute Scale RANKINE: 460 F R +  =  KELVIN: 273 C K +  =  6. Pressure (P): It is the normal component of a force per unit area. Pascal) (Kilo KPa 1 m KN 1 KPa o r m KN A F P 2 2 = = If a force dF acts on an infinitesimal area dA, the intensity of pressure is: KPa or m KN A dF P 2 = where: F - force, KN A - area , m2
• 3. 3 1 3 1 3 1 3 3 1 1 3 3 1 1 P P 1 equation to 2 Equation 2 eq. sin A A ; A A sin Figure From 1 eq. sin A P A P 0 sin A P A P 0 Fx = →  = =  →  = =  − =  3 2 1 3 2 3 2 3 2 3 3 2 2 3 3 2 2 P P P therefore P P 3 equation to 4 Equation eq.4 cos A A A A cos eq.3 cos A P A P cos A P A P 0 Fy = = = →  = =  →  =  − =  PASCAL`S LAW At any point in a homogeneous fluid at rest the pressures are the same in all directions. Atmospheric Pressure (Pa): It is the absolute pressure exerted by the atmosphere. At sea level Pa = 101.325 KPa = 0.101325 MPa = 1.01325 Bar = 760 mm Hg = 10.33 m of H2O = 1.033 kg/cm2 = 14.7 psi = 29.921 in Hg = 33.878 ft of H2O Absolute and Gage Pressure Absolute Pressure - is the pressure measured referred to absolute zero and using absolute zero as the base. Gage Pressure - is the pressure measured referred to the existing atmospheric pressure and using atmospheric pressure as the base. Pgage - if above atmospheric Pvac- vacuum or negative gage pressure, if below atmospheric vacuum abs Gage a abs P Pa P P P P − = + =
• 4. 7. Viscosity: It is the property of a fluid that determines the amount of its resistance to shearing stress. Assumptions: 1. Fluid particles in contact with the moving surface moves with the same velocity of that surface. 2. The rate of change of velocity dv/dx is constant in the direction perpendicular to the direction of motion. 3. The shearing is directly proportional to the rate of change of velocity. let S - shearing stress in Pa or N/m2 2 m sec - N o r sec - Pa v x S μ x v μ S x v d x d v b u t d x d v μ S d x d v S = = = =  Where  - Absolute or dynamic viscosity in Pa-sec or N-sec/m2 S - shearing stress in Pascal (Pa) x - distance apart in meters v - velocity in m/sec 8. Kinematics Viscosity: It is the ratio of the absolute or dynamic viscosity to the mass density. Conversion 1 Pa-sec = 1 N-sec/m2 1 Pa-sec = 10 Poise 1 Pa-sec = 0.020885 lb-sec/ft2 1 Pa-sec = 0.10197 kg-sec/m2 1 Pa-sec = 1000 Centipoise 1 m2 /sec = 10,000 stokes 1 m2 /sec = 10.764 ft2 /sec 1 m2 /sec = 1,000,000 Centistokes Stoke = 1 cm2 /sec 1 kg/m3 = 0.062428 llb/ft3 1 lb/ft3 = 16.018 kg/m3 9. Elasticity: If a pressure is applied to a fluid, it contracts; if the pressure is released, it expands, the elasticity of a fluid is related to the amount of deformation(expansion or contraction) for a given pressure change. Quantitatively, the degree of elasticity is equal to: Where negative sign is used because dV/V is negative for a positive dP. P P sec m ρ μ ν 2 = K Pa V d V d P Ev − = ρ ρ d V dV - KPa ρ ρ d dP Ev = =
• 5. where: Ev - bulk modulus of elasticity dV - is the incremental volume change, m3 V - is the original volume, m3 dP - is the incremental pressure change in Kpa 10. Surface Tension The attractive force exerted upon the surface molecules of a liquid by the molecules beneath that tends to draw the surface molecules into the bulk of the liquid and makes the liquid assume the shape having the least surface area. Surface tension is the elastic tendency of a fluid surface which makes it acquire the least surface area possible. Surface tension allows insects (e.g. water spiders), usually denser than water, to float and stride on a water surface. At liquid–air interfaces, surface tension results from the greater attraction of liquid molecules to each other (due to cohesion) than to the molecules in the air (due to adhesion). The net effect is an inward force at its surface that causes the liquid to behave as if its surface were covered with a stretched elastic membrane. Thus, the surface becomes under tension from the imbalanced forces, which is probably where the term "surface tension" came from. Because of the relatively high attraction of water molecules for each other through a web of hydrogen bonds, water has a higher surface tension (72.8 millinewtons per meter at 20 °C) compared to that of most other liquids. Surface tension is an important factor in the phenomenon of capillarity. Surface tension has the dimension of force per unit length, or of energy per unit area. The two are equivalent, but when referring to energy per unit of area, it is common to use the term surface energy, which is a more general term in the sense that it applies also to solids. Floating objects Cross-section of a needle floating on the surface of water. Fw is the weight and Fs are surface tension resultant forces. When an object is placed on a liquid, its weight Fw depresses the surface, and if surface tension and downward force becomes equal than is balanced by the surface tension forces on either side Fs, which are each parallel to the water's surface at the points where it contacts the object. Notice that small movement in the body may cause the object to sink. As the angle of contact decreases surface tension decreases the horizontal components of the two Fs arrows point in opposite directions, so they cancel each other, but the vertical components point in the same direction and therefore add up to balance Fw. The object's surface must not be wettable for this to happen, and its weight must be low enough for the surface tension to support it. Examples of Surface Tension Drops of water. When using a water dropper, the water does not flow in a continuous stream, but rather in a series of drops. The shape of the drops is caused by the surface tension of the water. The only reason the drop of water isn't completely spherical is because of the force of gravity pulling down on it. In the absence of gravity, the drop would minimize the surface area in order to minimize tension, which would result in a perfectly spherical shape. Insects walking on water. Several insects are able to walk on water, such as the water strider. Their legs are formed to distribute their weight, causing the surface of the liquid to become depressed, minimizing the potential energy to create a balance of forces so that the strider can move across the surface of the water without breaking through the surface. This is similar in concept to wearing snowshoes to walk across deep snowdrifts without your feet sinking. Needle (or paper clip) floating on water. Even though the density of these objects is greater than water, the surface tension along the depression is enough to counteract the force of gravity pulling down on the metal object. Click on the picture to the right, then click "Next," to view a force diagram of this situation or try out the Floating Needle trick for yourself. Capillarity Capillary Action Capillary action is the result of adhesion and surface tension. Adhesion of water to the walls of a vessel will cause an upward force on the liquid at the edges and result in a meniscus which turns upward. The surface tension acts to hold the surface intact, so instead of just the edges moving upward, the whole liquid surface is dragged upward.
• 6. Where:  - surface tension, N/m  - specific weight of liquid, N/m3 r – radius, m h – capillary rise, m Table 1.Surface Tension of Water C  0 0.0756 10 0.0742 20 0.0728 30 0.0712 40 0.0696 60 0.0662 80 0.0626 100 0.0589   r h   r γ θ cos σ 2 h =
• 7. IDEAL GAS PRINCIPLES IDEAL OR PERFECT GAS 1. CHARACTERISTIC EQUATION PV = mRT→ 1 P = RT→ 2 where: P - absolute pressure in KPa V - volume in m3 m -mass in kg R -Gas constant in KJ/kg-K T - absolute temperature in K  - specific volumein m3 /kg  - density in kg/m3 2. GAS CONSTANT 3. BOYLE`S LAW ( T = C ): Robert Boyle (1627-1691) If the temperature of a certain quantity of gas is held constant, the volume V is inversely proportional to the pressure P, during a quasi-static change of state. P 1 C V or P 1 α V = 1. CHARLE`S LAW ( P = C and V = C):Jacques Charles (1746-1823) and Joseph Louis Gay- Lussac (1778-1850) A) At constant pressure (P=C), the volume V of a certain quantity of gas is directly proportional to the absolute temperature T, during a quasi static change of state. V  T or V = CT B) At constant volume (V = C), the pressure P of a certain quantity of gas Is directly proportional to the absolute temperature T, during a quasi-static change of state. P  T or P = CT 5. AVOGADRO`S LAW:Amadeo Avogadro (1776-1856) All gases at the same temperature and pressure, under the action of a given value of g, have the same number of molecules per unit of volume. From which it follows that the  M. 1 2 2 1 2 1 R R M M γ γ = = 6 T V P T V P 5 C T PV 4 P RT υ 3 RT P ρ 2 2 2 1 1 1 → = → = → = → = mol mol kg kg in gas a of weight molecular M Constant) Gas (Universal K - kg KJ 3143 . 8 R 7 K - kg KJ M R R −  = →  = 8 V P V P C PV 2 2 1 1 → = = 9 T V T V C T V 2 2 1 1 → = = 1 0 T P T P C T P 2 2 1 1 → = =
• 8. Cv Cp k 1 k R Cv 1 k Rk Cp R Cv Cp = − = − = + = 6. SPECIFIC HEATS 7. ENTROPY CHANGE (S) Entropy is that property of a substance that determines the amount of randomness and disorder of a substance. If during a process, an amount of heat is taken and is by divided by the absolute temperature at which it is taken, the result is called the ENTROPY CHANGE.  = = T d Q S Δ T d Q d S 8. Isentropic Process: It is an internally reversible adiabatic process of an ideal or perfect gas in which S = C or PVk = C. P, V, & T relationship: 1 k 2 1 k 1 k 1 2 1 2 k 2 2 k 1 1 V V P P T T V P V P − −         =         = = 9. Polytropic Process: It is an internally reversible process of an ideal or perfect gas in which PVn = C. Where n stands for any constant (but not equal to infinity). P, V, & T relationship: 1 n 2 1 n 1 n 1 2 1 2 n 2 2 n 1 1 V V P P T T V P V P − −         =         = = 1. Total moles of a mixture n = ni 2. Mole Fraction yi = ni/n 3. Total mass of a mixture m = mi 4. Mass Fraction xi = mi/m 5. Equation of State A. Mass Basis a. For the mixture PV = mRT b. For the components PiVi = miRiTi B. Mole Basis a. For the mixture PV = nRT b. For the components PiVi = niRTi 6. Amagat's Law: The total volume V of a mixture is equal to the sum of the volume occupied by each component at The mixture pressure P, and temperature T. V = Vi n1 m1 V1 n2 m2 V2 n3 m3 V3 P,T
• 9. 1 n1 P1 2 n2 P2 3 n3 P3 mixture n P               + + = P T R T R PV T R PV T R PV T R PV 3 2 1               + + = P T R T R V P T R V P T R V P T R PV 3 2 1 P = P1 = P2 = P3 T = T1 = T2 = T3 n = n1 + n2 + n3 V = V1 + V2 + V3 V = V1 + V2 + V3 V = V 7. Dalton's Law The total pressure of a mixture P is equal to the sum of the partial pressure that each gas would exert at the mixture volume V and temperature T. P = Pi T = T1 = T2 = T3 V = V1 = V2 = V3 n = n1 + n2 + n3 P = P1 + P2 + P3 8. Molecular Weight of a Mixture 9. Gas Constant 10. Specific Heat 11. Gravimetric and Volumetric Analysis Gravimetric Analysis gives the mass fractions of the components in the mixture. Volumetric Analysis gives the volumetric or molal fractions of the components in the mixture. V V y i i = P P y i i = mol i i k g k g R 3 1 4 3 . 8 R R M M y M = = =  K - k g K J R 3 1 4 3 . 8 M R R R x R i i  = = =  1 k R C 1 k Rk C K - k g K J R C C K - k g K J C x Cv K - k g K J C x C V P V P V i i P i p i − = − =  + =  =  =     = = = i i i i i i i i i i i i M x M x y M M y M y M y x
• 10. dh γ dP − = VARIATION OF PRESSURE WITH ELEVATION By applying equations for equilibrium h Δ γ - P Δ constant is ρ density or γ weight specific the If dh γ - dP Equation General ) dh ( γ dP ) dh ( A γ dPA ) dh ( A V V γ W But W dPA 0 W PA dPA PA 0 W PA A ) dP P ( 0 Fy = = = = = = = = − − + = − − + =  Note: Negative sign is used because pressure decreases with increasing elevation and increases with decreasing elevation. h is positive if measured upward h is negative if measured downward Where: P – Pressure in KPa  - specific weight in KN/m3 h – elevation in meters CONDITIONS OF PRESSURE VARIATION WITH ELEVATION General Formula a. Constant Density downward measured if - h - upward measured if h Note Elevation P P h Pressure Final h P P h ) h h ( ) h h ( P P dh dP 2 1 1 2 1 2 1 2 1 2 2 1 2 1 − + →  − = →  − = = − −  − = −  − =   b. For Isothermal Condition (T = C)
• 11. c. Isentropic Condition ( ) Pressure Fin al C k 1 k h P P Elevation k 1 k P P C h C ) h ( k 1 k P P ) h h ( C 1 k 1 1 P k / 1 k 1 k 1 1 k k 2 k 1 k 1 k 1 k 2 k / 1 k / 1 k 1 k 1 k 1 k 2 1 2 k / 1 2 1 k 1 1 → − − = → −       − − = − − = − − − =             −       − − − − − − − d. Temperature decreases at a Standard Lapse Rate (6.5K per Kilometer) Elev atio n 0 0 6 5 . 0 P P 1 T h Pessu re Fin al T ) h 0 0 6 5 . 0 T ( P P T ) h 0 0 6 5 . 0 T ( P P T ) h 0 0 6 5 . 0 T ( ln R 5 . 6 g P P ln ) h 0 0 6 5 . 0 T ( d h ) 0 0 6 5 . 0 ( R 1 0 0 0 ) 0 0 6 5 . 0 ( g P d P co n stan t is T ) h 0 0 6 5 . 0 T ( d h R 1 0 0 0 g P d P ) h 0 0 6 5 . 0 T ( d h R 1 0 0 0 g P d P ) h 0 0 6 5 . 0 T ( R 1 0 0 0 ) d h ( Pg d P ) h 0 0 6 5 . 0 T ( R 1 0 0 0 Pg 1 0 0 0 g ) h 0 0 6 5 . 0 T ( R P RT P ) h 0 0 6 5 . 0 T ( T g R 5 . 6 1 2 s R 5 . 6 g s s 1 2 R 5 . 6 g s s 1 2 s s 1 2 2 1 s 2 1 s 2 1 s 2 1 s s s s s →                   − = →       − =       − =       − =         − − − − =         − − =         − − = − − = − =  =  − = =  − =       − = − = − = − = = = 2 1 2 1 dh RT 1000 g P dP dh RT 1000 g P dP dh RT 1000 Pg dP dh γ dP RT 1000 Pg γ RT P ρ Pressure Final e P P Elevation P P ln g RT 1000 h RT 1000 gh P P ln h h h ) h h ( RT 1000 g P P ln RT 1000 gh 1 2 2 1 2 1 1 2 1 2 1 2 → = →       = = = − − − =   − = − = − = − = − = = = = =  = = − − 2 1 k / 1 2 1 k / 1 k / 1 k / 1 k / 1 k / 1 k / 1 k / 1 k / 1 k / 1 k 2 2 k 1 1 k k d h C 1 d P P d h C 1 d P P d h C 1 P d P d h C P d P d h γ d P C P γ γ P γ P C C γ P γ ρ ρ 1 υ C υ P
• 12. 3 Hg 3 Hg water Hg water Hg Hg 3 water m KN 13 2.9 255 ) 81 . 9 ( 55 . 13 m kg 55 0 , 13 ) 10 00 ( 55 . 13 S m kg 10 00 = =  = =    =   = =  Pressure Variation in the Atmosphere TROPOSPHERE: It is the layer in the atmosphere between sea level and 10.769 km and the temperature decreases linearly with increasing elevation at a lapse rate of 6.5K/km. meters 0 .0 0 6 5 P P 1 T h T 0 .0 0 6 5 h T P P g 6.5R s s 6.5R g s s s                   − =       − = STRATOSPHERE: It is the layer that begins at the top of the troposphere and extends to an elevation of 32.3 km, and in this layer the temperature is constant at -55C. meters P P ln g 1000RT h e 1 P P s 1000RT gh s = = where: P - pressure at elevation h, KPa Ps - Pressure at sea level, KPa h - elevation, meters Ts - temperature at sea level in K FROM NASA 1. Atmospheric pressure decreases at the rate of 83.312 mm Hg per 1000 meters rise in elevation. 2. Atmospheric temperature decreases at the rate of 6.5K per 1000 meters rise in elevation. Example 1 The specific weight of water at ordinary pressure and temperature is 9.81 KN/m3 . The specific gravity of mercury is 13.55. Compute the density of water, and the specific weight and density of mercury. Example 2 A. Calculate the density, specific weight and specific volume of oxygen (M = 32 ; k = 1.395) at 38C and 104 KPa. B. What would be the temperature and pressure of the gas if it where compressed isentropically to 40 percent of its original volume. C. If the process described in (B) had been isothermal, what would be the temperature and pressure have been? (For O2: M = 32 ; k = 1.395) Example 3
• 13. 2 6 . 1 1 0 0 0 5 . 1 2 6 0 S m KN 4 . 1 2 1 0 0 0 ) 8 1 . 9 ( 5 . 1 2 6 0 1 0 0 0 g m k g 5 . 1 2 6 0 9 5 2 . 0 1 2 0 0 V m KN 7 7 2 . 1 1 1 0 0 0 ) 8 1 . 9 ( 2 0 0 , 1 1 0 0 0 mg W water 3 3 = =   = = =  =  = = =  = = = Slug 1 kg 59 . 14 Slugs 73 m kg 500 a F m N ma F = = = = = A vessel contains 85L of water at 10C ( = 999.7 kg/m3 ) and atmospheric pressure. If it is heated to 70C ( = 977.8 kg/m3 ), what will be the percentage change in its volume? What mas of water must be removed, to maintain the volume at the original volume? Example 4 A reservoir of glycerine has a mass of 1,200 kg and a volume of 0.952 m3 . Find the glycerin’s weight, density, specific weight and specific gravity. Example 5 A body requires a force of 100 N to accelerate it at a rate of 0.20 m/sec2 . Determine the mass of the body in kgs and in slugs. Example 6 Avertical cylindrical tank with a diameter of 12 m and a depth of 4 m is filled to the top with water at 20C ( = 9.79 KN/m3 ). If the water is heated to 50C ( = 9.69 KN/m3 ), how much water will spill over? (4.7 m3 ) Example No. 7 A mercury barometer at the ground floor of Kingston Tower in Chicago reads 735 mm Hg. At the same time another barometer at the top of the tower reads 590 mmHg. Assuming the air to be constant at 1.21 kg/m3 , what is the approximate height of the tower using g = 9.7 m/sec2 . h • 1 • 2 meters 1 6 5 8 .0 8 γ P P h ) h ( γ P P h h ; 0 h ) h h ( γ P P d h γ d P m K N 1 0 x 1 1 .6 4 1 0 0 0 ) 7 . 9 ( 2 1 . 1 1 0 0 0 g ρ γ K Pa 7 . 7 8 7 6 0 ) 3 2 5 . 1 0 1 ( 5 9 0 P K Pa 9 8 7 6 0 ) 3 2 5 . 1 0 1 ( 7 3 5 P 1 2 1 2 2 1 1 2 1 2 3 3 - 2 1 = − − = − = − = = − − = − − = = = = = = = = Example No. 8 A tank contains a mixture of 20 kg of nitrogen and 20 kg of carbon monoxide. The total tank volume is 20 m3 . Determine the density, specific volume and specific weight of the mixture if local g = 9.81 m/sec2 k g m 5 . 0 ρ 1 υ m k g 2 2 0 4 0 V m ρ k g 4 0 2 0 2 0 m m m m 3 3 2 1 = = = = = = + = + = Example No. 9 A pressure gage registers 345 KPag in a region where the barometric pressure is 736.7 mm hg. Find the absolute in Bars.
• 14. sec m 10 x 66 . 2 ν 972 . 864 023 . 0 ρ μ ν m kg 972 . 864 ft lb 1 m kg 018 . 16 x ft lb 54 ρ sec - Pa 023 . 0 ft sec - lb 0.020885 sec - Pa 1 x ft sec lb 10 x 8 . 4 μ 2 5 - 3 3 3 3 2 2 4 - = = = = = = − = ft 10.764 m sec - Pa 10 x 002 .1 μ ft sec - lb 10 x 09 .2 sec - Pa ft sec - lb 0.020885 sec - Pa 10 x 002 .1 μ 2 2 2 3 2 5- 2 3 = = − − Bars 4 .4 3 2 P K Pa 1 0 0 Bar x K Paa 4 4 3 .2 3 4 5 7 6 0 ) 3 2 5 . 1 0 1 ( 7 . 7 3 6 P P P P abs abs gage a abs = = +       = + = Example No. 10 The level of water in an enclosed water tank is 40 m above ground level. If the pressure of the air space above the water is 120 KPa, what is the pressure at ground level in KPa. Air • 1 • 2 40 m Example No. 11 A liquid has an absolute viscosity of 4.8 x 10-4 lb-sec/ft2 . It weighs 54 lb/ft3 . What are its absolute and kinematic viscosities in SI. Example No. 12 The absolute viscosity of water at 20C is 1.002 x 10-3 Pa-sec and the density is 998 kg/m3 . What are its absolute and kinematic viscosity in English units. Example No. 13 In a fluid the velocity measured at a distance of 75 mm from the boundary is 1.125 ./sec. The fluid has an absolute viscosity of 0.048 Pa-sec and a relative density of 0.913. What is the velocity gradient and shear stress at the boundary assuming a linear velocity distribution? Determine also its kinematic viscosity. K Pa 5 1 2 .4 ) 4 0 ( 8 1 . 9 1 2 0 P m 4 0 h 0 h ) h h ( γ P P ) h h ( γ P P d h γ d P n in teg ratio By d h γ d P 2 2 1 1 2 1 2 1 2 1 2 2 1 2 1 = − − = − = = − − = − − = − − = − =   s ec m 1 0 x 2 6 . 5 9 1 3 0 4 8 . 0 m k g 9 1 3 ) 1 0 0 0 ( 9 1 3 . 0 P a 7 2 0 . 0 ) 1 5 ( 0 4 8 . 0 S s ec 1 1 5 0 7 5 . 0 1 2 5 . 1 d x d v d x d v S 2 5 - 3 = =  = =    =  = = = =  =
• 15. 2 2 2 2 2 2 18 8 2 2 2 2 2 2 m n N 1 0 . 6 1 O 7 5 . 3 O 9 H 8 CO N 1 0 . 6 1 O 2 5 . 1 6 H C 7 5 . 3 d 9 c 8 b 1 2 .5 a 0 .2 5 m) e(n d 0 .5 m c n b 0 .2 5 m n a wh ere e)a(3 .7 6 )N (1 d O O cH b CO e)a(3 .7 6 )N (1 e)aO (1 H C is e" " air ex cess with Cn Hm o f eq u atio n co mb u stio n th e , co mb u stio n al th eo retic Fro m + + + → + + = = = = + = = = + = + + + + → + + + + 3 3 mol m KN 0115 . 0 1000 ) 81 . 9 ( 17 . 1 1000 g m kg 17 . 1 ) 298 ( 290 . 0 101 RT P K kg KJ 290 . 0 M 3143 . 8 R kg kg 65 . 28 ) 28 ( 7465 . 0 ) 32 ( 0458 . 0 ) 18 ( 11 . 0 ) 44 ( 0977 . 0 yiMi M = =  =  = = =  − = = = + + + =  = Example no. 14 A hydrocarbon fuel C8H18 (Octane) is burned in an engine with 30% excess air required for combustion. For a pressure P = 101 KPa and temperature T = 298 K of the products of combustion, Determine a. The density in kg/m3 b. The specific weight in KN/m3 c. The molecular weight M (kg/kgmol) and gas constant R (KJ/kg-K) Fuel: CnHm n = 8 ; m = 18 Mixture: Products of combustion Gases ni(kgmol) Mi mi(kg) xi(%) yi(%) CO2 8 44 352 15 9.77 H2O 9 18 162 6.9 11 O2 3.75 32 120 5.1 4.58 N2 61.1 28 1,710.80 73 74.65 81.85 2,344.80 100 100 MANOMETERS Manometer is an instrument used in measuring gage pressure in length of some liquid column. 1. Open Type Manometer : It has an atmospheric surface and is capable in measuring gage pressure. 2. Differential Type Manometer : It has no atmospheric surface and is capable in measuring differences of pressure.
• 16. Determination of S using a U - tube Specific gravity or relative density of an unknown liquid can be determine using a U - tube with both ends open to the atmosphere, and applying the Variation of Pressure with Elevation principle, provided that one liquid of known specific gravity is available. Problem No. 1 A mercury manometer shown is connected to a pipeline carrying water ( = 9.81 KN/m3 ). If the elevation at point B is 3 m above A and the mercury reading y = 1200 mm, what is the pressure in the pipe in KPa. ( of mercury = 133.4 KN/m3 ) 0 1 1 2 P1 = PB P2 = PA KPa 189.51 P 0 P P ) 81 . 9 ( 3 ) 2 . 1 ( 4 . 133 P A 0 A 0 = = = + + Problem no. 2 In the figure shown, fluid A is water ( = 9.81 KN/m3 , fluid B is mercury ( = 133.42 KN/m3 ). Z = 450 mm and Y = 900 mm. Compute the pressure difference (Pn – Pm) in KPa.
• 17. θ θ mm 29.5 meters 0295 . 0 h 0 ) 81 . 9 )( 80 . 0 ( 45 . 0 ) 81 . 9 )( 56 . 13 ( h ) 81 . 9 ( 76 . 0 0 = = = − − + 1 2 3 4 3 5 Pn = P1 Pm = P5 K Pa 5 . 6 4 P P K Pa ) 8 1 . 9 ( z ) 4 2 . 1 3 3 ( z ) 8 1 . 9 ( y P P ) 8 1 . 9 ( x ) 8 1 . 9 ( z ) 4 2 . 1 3 3 ( z ) 8 1 . 9 ( x ) 8 1 . 9 ( y P P ) 8 1 . 9 ( x ) 8 1 . 9 ( z ) 4 2 . 1 3 3 ( z 8 1 . 9 ) x y ( P P Pm ) 8 1 . 9 ( x ) 8 1 . 9 ( z ) 4 2 . 1 3 3 ( z 8 1 . 9 ) x y ( Pn m n m n m n m n − = − + − − = − + + − − − = − + + − + − = − = − − + + + Problem no. 3 A U tube with both ends open to the atmosphere contains mercury in the lower portion. In one leg, water stands 760 mm above the surface of mercury, in the other leg , oil (S = 0.80) stands 450 mm above the surface of the mercury. What is the difference in elevation between the surfaces of mercury in contact with oil and water columns? HYDROSTATIC FORCES ACTING ON PLANE SURFACES Fig. (a) Fig. (b) C.G. - center of gravity C.P. - center of pressure General Equation:
• 18. Therefore, the total hydrostatic force exerted by the fluid on any plane surfaces submerged in a homogeneous fluid at rest, is equal to the product of the surface area A and the pressure at its centroids h  . LOCATION OF THE CENTER OF PRESSURE y A Ss Ss Ig e y A Ig e y y A Ig e y e y y y A y A Ig y p 2 p = = = + = + + = + = where: e - is the perpendicular distance between C.G. andC.P. Ig - moment of inertia with respect to the axis at its centroids and lying on its plane Ss- statical moment of inertia with respect to the axis SS not lying on its plane HYDROSTATIC FORCE ACTING ON CURVED SURFACES C'C = B'B = L where L - length of the curved surface AB perpendicular to the paper KN F F F 2 v 2 h + = KN A h γ Fh = A = BC x L m2 where: A - area of the vertical projection of the curved surface AB KN V FV γ = V = AABCDEA (L) m3 Fh - horizontal component of F in KN that passes through the center of pressure of the vertical projection of the curved surface AB Fv - vertical component of F, KN F - total hydrostatic force acting on the curved surface AB Note: If the liquid is underneath the surface the Force acts upward. HOOP TENSION 1m D H P = h A h γ F =
• 19. Considering a semi-circular segment of 1 m length F = 0 F = 2T → 1 2 F T = → 2 On the vertical projection: D 1 m A F P = where A = 1D F = P(1D) let : S - tensile or hoop stress in KPa t A T S = where At = 1t (area subjected to tensile stress) ) t 1 ( 2 ) D 1 ( P ) t 1 ( 2 F S = = t 2 PD S = T T F t 1 m T T F
• 20. ARCHIMEDES PRINCIPLE LAWS OF BUOYANCY:Any body partly or wholly submerged in a liquid is subjected to a Buoyant or Upward force, which is equal to the weight of the liquid displaced. If the buoyant force is less than to the weight of the body, the body sinks, and if the buoyant force is greater than the weight of the body the body floats. A. W = BF W = BVBKN; W = BVB kg BF = VsKN; BF = Vs kg B.
• 21. W = BF - T W = BVBKN; W = BVB kg BF = VsKN; BF = Vs kg C. W = BF + T W = BVBKN; W = BVB kg BF = VsKN ; BF = Vs kg D. W = BF - T W = BVBKN ; W = BVB kg BF = VsKN ; BF = Vs kg VB = Vs E. W = BF + T W = BVBKN ; W = BVB kg BF = VsKN ; BF = Vs kg VB = Vs where: W - weight of the body; KN, kg BF - buoyant force, KN, kg
• 22. VB - volume of body, m3 Vs - volume submerged, m3 B - specific gravity of the body, KN/m3  - specific gravity of the liquid, KN/m3 B - density of the body, kg/m3  - density of the liquid, kg/m3 SPECIFIC GRAVITY OF SOLIDS HEAVIER THAN WATER 2 1 1 B W W W S − = SPECIFIC GRAVITY OF AN UNKNOWN LIQUIDS 2 1 3 1 B W W W W S − = − Where W1 – weight of the body in air W2 - weight of the body in water W3 - weight of the body in an unknown liquid SAMPLE PROBLEMS Problem No. 1 (Manometer) A mercury manometer shown is connected to a pipeline carrying water ( = 9.81 KN/m3 ). If the elevation at point B is 3 m above A and the mercury reading y = 1200 mm, what is the pressure in the pipe in KPa. ( of mercury = 133.4 KN/m3 ) 0 1 1 2 P1 = PB P2 = PA KPa 189.51 P 0 P P ) 81 . 9 ( 3 ) 2 . 1 ( 4 . 133 P A 0 A 0 = = = + + Problem no. 2 (Manometer) In the figure shown, fluid A is water ( = 9.81 KN/m3 , fluid B is mercury ( = 133.42 KN/m3 ). Z = 450 mm and Y = 900 mm. Compute the pressure difference (Pn – Pm) in KPa. 1 2 3 4 3 5 Pn = P1 Pm = P5
• 23. KPa 5 . 64 P P KPa ) 81 . 9 ( z ) 42 . 133 ( z ) 81 . 9 ( y P P ) 81 . 9 ( x ) 81 . 9 ( z ) 42 . 133 ( z ) 81 . 9 ( x ) 81 . 9 ( y P P ) 81 . 9 ( x ) 81 . 9 ( z ) 42 . 133 ( z 81 . 9 ) x y ( P P Pm ) 81 . 9 ( x ) 81 . 9 ( z ) 42 . 133 ( z 81 . 9 ) x y ( Pn m n m n m n m n − = − + − − = − + + − − − = − + + − + − = − = − − + + + Problem No. 3 (Variation of Pressure) An airplane barometer reads 762 mm Hg at sea level and 736 mm Hg at some unknown elevation. If the temperature at sea level is 15 C, what is the elevation assuming: a. air at constant density b. air under isothermal condition c. air temperature decreases linearly with elevation at the rate of 6.5 K per kilometer. Given: P1 = 762 mm Hg = 101.6 KPa P2 = 736 mm Hg = 98.23 KPa T1 = 15 + 273 = 288 K a. Constant density meters 4 . 279 P P h ) h h ( P P m KN 01206 . 0 1000 ) 81 . 9 ( 23 . 1 m kg 23 . 1 ) 288 ( 287 . 0 6 . 101 RT P 2 1 1 2 1 2 3 3 =  − = −  − = − = =  = = =  b. Constant Temperature c. Air temperature decreases linearly with elevation at the rate of 6.5 K per kilometer meters 0.0065 P P 1 T h T 0.0065h T P P g 6.5R 1 2 1 6.5R g 1 1 1 2                   − =       − = h = 283.04 meters Problem No. 4 (Variation of Pressure) The bottom of a river is 12 m below the water surface. Underneath which is a silt having a specific gravity of 1.75 and a thickness t. The pressure at the bottom of silt is 450 KPa. Determine the thickness of the silt. M 355 . 19 ) 81 . 9 ( 75 . 1 12 ) 81 . 9 ( 450 t 450 t 1.75(9.81) 9.81(12) 0 2 to 0 at = − = = + + Problem No. 5 (Hydrostatic forces) A rectangular gate 1.6 m wide and 2 m high has water on one side and is inclined at 45with the horizontal. Water is 1.5 m above the top of the gate. a) Compute the total force acting on the gate and its location b) If the gate is hinged at the top(B)what force P is needed at the bottom(A) to open the gate. meters 2 . 284 h P P ln g RT 1000 h 2 1 =       =
• 24. KN 3 . 38 P 2 ) 10 . 1 ( 3 . 69 P 0 ) 10 . 1 ( F P 2 A at M 45 sin 1 5 . 1 h = = = −  + =  m 109 . 1 y 45 sin 5 . 1 23 . 3 45 sin 5 . 1 y y m 23 . 3 y ) 12 . 3 ( 2 . 3 ) 12 . 3 ( 2 . 3 0666 . 1 y A y A I y m 0666 . 1 12 ) 2 ( 6 . 1 12 bh I rectangle a For m 12 . 3 45 sin 207 . 2 y y h 45 sin Figure From KN 3 . 69 F 2) )(3. 9.81(2.207 F m 2 . 3 1.6(2) A m 207 . 2 h 45 sin 1 5 . 1 h ' p p p' p 2 2 g p 4 3 3 g 2 = − = − = = + = + = = = = = = = = = = = =  + = Problem No. 6 (Forces of Curved Surfaces) The curve surface represented by AB in the figure is the surface of a quadrant of a circular cylinder 3 m long. Determine the horizontal and vertical component of the total hydrostatic force on the surface if the liquid is gasoline with S = 0.72. L - length L = 3 m R = 2 m (radius) ( )( ) ( ) ( ) ( ) K N 4 1 . 2 5 7 F F F K N 7 1 . 1 9 3 ) 4 2 5 . 2 7 ( 0 6 3 2 . 7 F m 4 2 5 . 2 7 3 ) 3 ( 2 2 π 4 1 V V γ F K N 5 2 . 1 6 9 F m K N 0 6 3 2 . 7 ) 8 1 0 . 9 ( 7 2 . 0 γ m 4 1 3 h m 6 ) 3 ( 2 A A h γ F 2 v 2 h v 3 2 v h 3 2 h = + = = = =       + = = = = = = + = = = = Problem No. 7 (Buoyancy) Two spheres each 1.2 m in diameter weighs 4 KN and 12 KN, respectively. They are connected with a short rope and placed in water. What portion of the lighter sphere protrudes from the water. B A F P yp’ 2 m Liquid surface 2 m 3 m m 4 h = 2 m L = 3 m Fv Fh
• 25. 12 bh Ig 3 = m 453 . 3 ) 22 . 3 ( 9 ) 22 . 3 ( 9 75 . 6 y m 22 . 3 60 sin h y m 75 . 6 12 ) 3 ( 3 I y A y A I y KN 06 . 197 (2.79)(9) 0.80(9.81) F m 79 . 2 60 sin 5 . 1 5 . 1 h 2 p 3 g 2 g p = + = =  = = = + = = = =  + = 3 3 1 F 1 F 3 1 S 3 1 S 2 1 2 1 3 2 1 m 1 7 5 . 0 7 3 . 0 ) 6 0 . 0 ( π 3 4 V su rface liq u id ab o v e p ro tru d es v o lu me V m 7 3 . 0 V ) 6 0 . 0 ( π 3 4 8 1 . 9 V 8 1 . 9 1 2 4 BF BF W W R π 3 4 Sp h ere o f V o lu me m 6 0 . 0 2 1 .2 R K N ; 1 2 W K N ; 4 W = −       = = =       + = + + = + = = = = = Problem No. 8 (Buoyancy) A block of material of an unknown volume is submerged in water and weighs 500 N. The same block weighs 650 N when weigh in air. Determine the volume of the material in m3 . 3 3 2 1 1 m 0 1 5 3 . 0 4 3 3 0 2 6 . 6 6 V V m ρ k g 2 6 . 6 6 8 1 . 9 6 5 0 m mg W m k g 4 3 3 0 ) 1 0 0 0 ( 3 3 . 4 ρ 3 3 . 4 5 0 0 6 5 0 6 5 0 W W W S = = = = = = = = = − = − = Problem No. 9 (Forces on Plane surface) A square gate (3 m x 3 m) is hinged at the top (at A) and rest on a smooth floor making an angle of 60 with the horizontal. Oil (S = 0.80) stands on the left side of the gate to a height of 1.5 m above point A. Determine the amount of the hydrostatic force exerted by the oil on the gate and its location of the center of pressure. Ig for a square: SOLUTION Problem No. 10 (Hoop Tension) A 1.2 m diameter steel pipe, 6 mm thick, carries oil with S = 0.822 under a head of 122 m of oil. Compute a. The stress in the steel in KPa b. The thickness of the steel pipe required to carry a pressure of 1724 KPa with an allowable stress of 124 MPa
• 26. mm 8 .3 m 0 0 8 3 . 0 ) 2 ( 0 0 0 , 1 2 4 1 7 2 4 (1 .2 ) t K Pa 3 7 9 , 9 8 ) 0 0 6 . 0 ( 2 ) 2 . 1 ( 8 . 9 8 3 S K Pa 8 . 9 8 3 ) 1 2 2 )( 8 1 . 9 ( 8 2 2 . 0 P t 2 PD S = = = = = = = = SAMPLE OF MANOMETER PROBLEMS 1. The closed tank in the figure is filled with water. The pressure gage on the tank reads 48 KPa. Determine a. The height h in mm in the open water column b. The gage pressure acting on the bottom of the tank surface AB c. The absolute pressure of the air in the top of the tank if the local atmospheric pressure is 101 KPa absolute. 2. The mercury manometer in the figure indicates a differential reading of 30 m when the pressure in pipe A is 30 mm Hg vacuum. Determine the pressure in pipe B. 3. In the figure pipe A contains carbon tetrachloride (S = 1.60) and the closed storage tank B contains a salt brine (S = 1.15). Determine the air pressure in tank B in KPa if the gage pressure in pipe A is 1.75 kg/cm2 . 4. A U tube mercury manometer is connected to a closed pressurized tank as shown. If the air pressure is 14 KPa, determine the differential reading h. The specific weight of the air is negligible. 5. A closed tank contains compressed air and oil (S = 0.90) as shown in the figure.a U – tube manometer using mercury (S = 13.6) is connected to the tank as shown. For column heights h1 = 90 cm; h2 = 15 cm and h3 = 22 cm, determine the pressure reading of the gage.
• 27. 6. The pressure of gas in a pipeline is measured with a mercury manometer having one limb open to the atmosphere. If the difference in the height of mercury in the limbs is 562 mm, calculate the absolute gas pressure. The barometer reads 761 mm Hg, the acceleration due to gravity is 9.79 m/sec2 and SHg = 13.64. 7. A turbine is supplied with steam at a gauge pressure of 1.4 MPa, after expansion in the turbine the steam flows into a condenser which is maintained at a vacuum of 710 mm Hg. The barometric pressure is 772mm Hg. Express the inlet and exhaust pressure in kg/cm2 . Take the S of mercury is 13.6. 8. The pressure of steam flowing in a pipe line is measured with a mercury manometer. Some steam condenses in to water. /estimate the steam pressure in KPa. Take the density of mercury as 13,600 kg/m3 , the barometer reading as 76.1 cm Hg and g = 9.806 m/sec2 . 9. A manometer is attached to a tank containing three different fluids, as ashow in the figure below. What will be the difference in elevation h of th mercury column in the manometer. Quiz no. _______ 1. The pressure on top of a mountain is 90 KPa. If the pressure and temperature ate sea level are 101 KPa and 288 K, respectively, Determine the height of the mountain and the temperature on top, Assuming a. Constant density b. Isothermal condition c. Isentropic condition d. Standard atmospheric lapse rate prevails 2. A manometer is attached to a tank containing different fluids, as shown in the figure below. What will be the difference in elevation of the mercury column in the manometer.
• 28. 1. A differential manometer is shown in the figure below. Calculate the pressure difference (PA – PB) in kg/cm2 . (1 inch = 25.4 mm; 1 m = 1000 mm). 2. On top of a mountain the pressure is 85 KPa. If the pressure and temperature ate sea level are 101.33 KPa and 21C respectively, Determine the altitude and temperature on top of the mountain, assuming Constant density b. Isothermal condition c. Isentropic condition d. Standard atmospheric lapse rate prevails FLUID MECHANICS/QUIZ NO. 1 (December 10, 2016) Group 3 1 Lampago, Mikel N. 2 Leonor Jr., Eduardo E. 3 Nayon, Carl Christian J. 4 Ompoc, Horace B. Opon, Rashiel Jan M. 6 Repollo, Ryan Z. 1. For the configuration shown below, calculate the weight of the piston if the gage pressure reading is 70 KPa. 2. On a certain day a PAL plane is flying at an altitude of 3000 m. If the atmospheric pressure and temperature at sea level are 101 KPa and 288 K, respectively. Determine the pressure and temperature in the plane, assuming
• 29. a. Constant density from sea level to elevation 3000 m b. Isothermal condition c. Isentropic condition d. Standard atmospheric lapse rate prevails FLUID MECHANICS/QUIZ NO. 1 (December 10, 2016) Group 4 1 Somodlayon, Michael John T. 2 Tabamo, Brael Y. 3 Ubaub, Matt Vann Ernee V. 4 Vasallo Jr., Edward A. 5 Zalsos, Edzel A. 1. A mountain is 2000 meters above sea level. If the pressure and temperature ate sea level are 101 KPa and 288 K, respectively, Determine the temperature and pressure on top of the mountain, assuming A .Constant density from sea level to the top of the mountain b. Isothermal condition c. Isentropic condition d. Standard atmospheric lapse rate prevails 2. If the atmospheric pressure is 101.03 KPa and the absolute pressure at the bottom of the tank is 231.3 KPa. What is the specific gravity of olive oil? (S of SAE oil = 0.89 ; S of Hg = 13.6) 36 . 1 S m KN 34 . 13 3 . 231 ) 4 . 0 )( 81 . 9 ( 6 . 13 ) 9 . 2 ( ) 81 . 9 ( 5 . 2 5 . 1 ) 81 . 9 )( 89 . 0 ( 3 . 101 oil 3 oil oil = =  = +  + + + SAMPLE PROBLEMS 1. A vertical piston cylinder device contains a gas at an unknown pressure. If the outside pressure is 100 kPa, determine (a) the pressure of the gas if the piston has an area of 0.2 m2 and a mass of 20 kg. Assume g = 9.81 m/s2. (b) What-if-Scenario: How would the answer in (a) changes if the orientation of the device is changed upside down. Answers: (a) 101 kPa, (b) 99 kPa. 2. A piston with a diameter of 50 cm and a thickness of 5 cm is made of a composite material with a density of 4000 kg/m3 . (a) If the outside pressure is 101 KPa, determine the pressure inside the piston-cylinder assembly if the cylinder contains air. (b) How would the answer change if the piston diameter was 100 cm instead? Answers: (a) 104.8 KPa 3. Water flows through a variable-area pipe with a mass flow rate of 10,000 kg/min. Determine the minimum diameter of the pipe if the flow velocity is not to exceed 5 m/s. Assume density of water to be 1000 kg/m3 . Answers: 0.206 m 4. A bucket of concrete with a mass of 5000 kg is raised without any acceleration by a crane through a height of 20 m. (a) Determine the work transferred into the bucket. (b) What happens to the energy as it is transferred to the bucket? (c) Also determine the power delivered to the bucket if it is raised at a constant speed of 1 m/s. Answers: (a) 981 kJ, (c) 49.05 kW 3. A building in Makati is 84.5 m high above the street level. The required static pressure of the water line at the top of the building is 2.5 kg/cm2 . What must be the pressure in KPa in the main water located 4.75 m below the street level. P1 = 245.166 KPa KPa 71 . 1120 ) 81 . 9 ( 75 . 4 ) 5 . 84 ( 81 . 9 166 . 245 = + + 14. A cylindrical tank 2 m diameter, 3 m high is full of oil. If the specific gravity of oil is 0.9, what is the mass of oil in the tank?
• 30. K - kg KJ 114 . 0 R ) Rx ( 609 . 0 ) 297 . 0 ( 391 . 0 1854 . 0 K kg KJ 297 . 0 28 4143 . 8 R R X R x R K - kg KJ 1854 . 0 ) 340 ( 6 . 4 ) 1 ( 290 R m 1 1(1)(1) V mixture mRT PV % 9 . 60 609 . 0 x % 1 . 39 391 . 0 4.6 1.8 x kg 6 . 4 8 . 2 8 . 1 m x 2 N x x 2 N 2 N 3 x N2 = + = − = = + = = = = = → = = = = = = = + = % 22 . 37 3722 . 0 y % 78 . 62 6278 . 0 y 0223 . 0 Mi xi 0084 . 0 014 . 0 93 . 72 609 . 0 28 391 . 0 Mi xi kg/kg 93 . 72 0.114 8.3143 M x 2 N m x = = = = = + = + = = =   KPa 2 . 100 P 7 . 0 ) 291 )( 278 . 0 ( 867 . 0 V mRT P K - kg KJ 278 . 0 R ) 29 . 0 ( 4037 . 0 ) 27 . 0 ( 5963 . 0 R R x R x R 4037 . 0 867 . 0 35 . 0 x 5963 . 0 0.867 0.517 x kg 867 . 0 m 35 . 0 517 . 0 m m m kg 517 . 0 ) 291 ( 27 . 0 ) 7 . 0 ( 58 m T R m V P 2 2 1 1 2 1 2 1 1 1 1 1 1 = = = = + = + = = = = = = + = + = = = = kg 5 . 8482 ) 425 . 9 ( 900 m V m m 425 . 9 3 ) 2 ( 4 V m kg 900 ) 1000 ( 9 . 0 3 2 3 = = =  =  = = =  5. A cubical tank 1 m on a side, contains a mixture of 1.8 kg of nitrogen (M = 28; k = 1.399) and 2.8 kg of an unknown gas. The mixture pressure and temperature are 290 KPa and 340 K. Determine a) Molecular weight and gas constant of the unknown gas b) the volumetric analysis Given: mN2 = 1.8 kg ; mx = 2.8 kg P = 290 KPa ; T = 340K 6. A closed vessel of 0.7 m3 internal volume contains a gas at 58 KPa and 18C and with R = 0.27 KJ/kg- K.If now 0.35 kg of another gas at 18C and R = 0.29 KJ/kg-K is also admitted into the vessel. Calculate the final pressure of the mixture. Given: V = 0.7 m3 P1 = 58 KPa; T1 = 18 + 273 = 291 K R1 = 0.27 KJ/kg-K m2 = 0.35 kg T2 = 291 K R2 = 0.29 KJ/kg-K 5. In the figure shown the diameters of the two cylinders are 75 mm and 600 mm, the face of the piston is 6 m above the face of the weight “W” and the intervening passages is filled with oil (S = 0.80). What force F is required if W = 35 KN. F W = 35 KN
• 31. 1.5 m x 1.2 m 1.2 - x 6. If 1 100 N force F1 is applied to the piston with 5 cm diameter, what is the magnitude of the force F2 that can be resisted by the piston with the 10 cm diameter? Neglect the weights of the piston. 7. A glass tube 1.5 m long and 25 mm diameter with one end closed is inserted vertically with the open end down into a tank of water until the open end is submerged to a depth of 1.2 m. If the barometric pressure is 98 KPa, and neglecting vapor pressure, how high will water rise in the tube. (Assume isothermal conditions for air) MANOMETER 1. A mercury manometer shown is connected to a pipeline carrying water ( = 9.81 KN/m3 ). If the elevation at point B is 3 m above A and the mercury reading y = 1200 mm, what is the pressure in the pipe in KPa.( of mercury = 133.4 KN/m3 ) 2. A is water, Fluid B is oil (S = 0.80), z = 350 mm. Compute the pressure difference between m and n in KPa. A y = 1200 mm B 3 m Open m n Fluid A Fluid B z 5 cm 10 cm 2.2 m F1 F2 2 m oil
• 32. Example 2 What force must be exerted through the bolts to hold the dome in place? The metal dome and pipe weighs 6 KN and has no bottom. ( ) ( ) ( ) ( ) KN 351.7 6 - ) 9.81(36.46 bolts in Tension m 46 . 36 6 . 1 3 2 6 . 5 6 . 1 V 6 . 1 3 4 2 1 6 . 5 ) 6 . 1 ( V V - V V liquid of volume imaginary of Volume - V m 6 . 5 4 6 . 1 h 3 2 3 2 dome cylinder = = =       −  =        −  = = = + = Example 3 A hemispherical dome surrounds a closed tank as shown. If the tank and dome are filled with gasoline (S = 0.72) and the gage indicates a pressure of 60 KPa. What is the total tension in the bolts holding the dome in place. P =60 KPa 3.0 m Gasoline S = 0.72 R = 2 m 5.5 m ( ) ( ) ( ) KN 370 ) 4 . 52 )( 81 . 9 ( 72 . 0 W m 4 . 52 2 3 4 2 1 5 . 5 2 V V - V V prism fluid imaginary of weight the m 5.5 3 - 8.5 H prism l cylindrica imaginary of height - H m 8.5 h 60 h 0.72(9.81) 0 gage 0 with liquid the of surface free imaginary the to surface curved the extending By 3 3 2 dome cylinder = = =        −  = = = = = = + Example 4 A cubical block of wood 10 cm on a side floats at the interface between oil and water with its lower surface 1.50 cm below the interface. The density of the oil is 790 kg/m3 . What is the gage pressure at the lower face of the block? What is the mass and density of the block? ID = 20 cm 80 cm 400 cm 160 cm R = 160 cm water
• 33. FLUID MECHANICS (PRE – FINAL S4) March 15, 2017 Name ____________________________________ 1. A piece of wood of S = 0.651 is 8 cm square and 150 cm long. How many kilograms of lead weighing 11,200 kg/m3 must be fastened at one end of the stick so that it will float upright with 30.5 cm out of water?
• 34. kg 54 . 1 (11200) 10 x 1.4 W m 10 x 1.4 10,200 6.25 - 7.65 V 1000V 7.65 11,200V 6.25 BF BF W W 1000V BF2 kg 7.65 1000 ) 08 . 0 )( 305 . 0 5 . 1 ( BF 11,200V W kg 6.25 ) 651 ( 96 . 0 W m 96 . 0 ) 5 . 1 )( 08 . 0 ( 08 . 0 V m kg 651 ) 1000 ( 651 . 0 651 . 0 S 4 - 2 3 4 - 2 2 2 2 1 2 1 2 2 1 2 2 1 3 1 3 wood = = = = + = + + = + = = − = = = = = = = = = ρ 2. A barge is loaded with 150 Metric tons of coal. The weight of the empty barge in air is 35 Metric ton. If the barge is 5.5 m wide, 16 m long and 3 m high, what is its draft. (Depth below the water surface) m 1 . 2 h h ) 16 )( 5 . 5 ( 1000 ) 1000 )( 35 150 ( Vs W 1 W BF W 2 = = +  = + = 3. A prismatic object 20 cm thick by 20 cm wide by 40 cm long is weighed in water at a depth of 50 cm and found to weigh 50 N. What is its weight in air and its specific gravity? 312 . 1 50 - 206.96 206.96 S Newton 206.96 W 50 156.96 W ) 016 . 0 ( 810 , 9 W W W W 1 V 9810 1 W W W V 9810 W W W W 1000 gV W Newton g m W W W W 1000 V m S V m m 0.016 ) 4 . 0 )( 20 . 0 )( 20 . 0 ( V Newton 50 W W 1 1 2 1 2 1 1 2 1 1 1 1 2 1 1 1 1 1 1 2 1 1 1 1 1 1 1 3 1 2 1 = = = + = = − − = − = − = = − = = =  = = = = 4. A ship, with vertical sides near the water line, weighs 3630 Metric ton and draws 6.7 m in saltwater (S = 1.025). Discharge of 181 Metric ton of water ballast decreases the draft to 6.4 m. What would be the draft d of the ship in fresh water (S = 1) 2 . Eq ) 025 . 1 )( 4 . 6 ( 3449 L A ) 1000 )( 025 . 1 )( 4 . 6 ( L A ) 1000 )( 181 630 , 3 ( BF W 1 . Eq ) 025 . 1 ( 7 . 6 630 , 3 L A ) 1000 )( 025 . 1 )( 7 . 6 ( L A ) 1000 ( 3630 BF W 2 2 2 2 1 1 1 1 → = = − = → = = = FINAL EXAM (March 25, 2017) SET 2 NAME _____________________________________ 1. An object weighs 25.95 N when submerged in kerosene (S = 0.81) and weighs 26.6 N when submerged in gasoline (S = 0.68). Determine the specific weight of the object.
• 35. m 15 . 14 y A y A Ig yp m 2485 . 0 64 D Ig m 767 . 1 A m 142 . 14 45 sin 10 y y h sin45 KN 82 . 176 ) 5 . 1 ( 4 ) 10 )( 81 . 9 ( 02 . 1 A h F 2 4 4 2 2 = + = =  = = =  = = =  =  = 2. Determine the water power and mechanical efficiency of a centrifugal pump which has an input of 3 KW. If the pump has a 203 mm diameter suction line and a 152 mm diameter discharge line and handles 10 L/sec of water at 66C ( =980 kg/m3 ;  = 9.6 KN/m3 ). The suction line gauge shows 102 mm Hg vacuum and the discharge gauge shows 180 KPa. The Discharge gauge is located 61 cm above the center of the discharge pipeline and the pump inlet and discharge lines are at the same elevation. 3. A piece of wood of S = 0.651 is 8 cm square and 150 cm long. How many kilograms of lead weighing 11,200 kg/m3 must be fastened at one end of the stick so that it will float upright with 30.5 cm out of water? 4. At one point in a pipeline the water speed is 3 m/sec and the gage pressure is 50 KPa. Find the gage pressure at a second point in the line, 11 m lower than the first, if the pipe diameter at the second point is twice at the first. FLUID MECHANICS (ACTIVITY NO. 2/January 14, 2016) Group No. 1 1 ABRERA, JEHOYAH CHRISTI E. 2 ANDAM, ROBERTO JR. M. 3 Arañas, Kieth d. 4 BONAYOG, STEVEN ANGEL M. 5 Cabral, Christian Mark o. 6 COLIPANO, HUBERT C. 7 DOFELIZ, MEAGAN ROSE G. 8 LLERA, KRISTIAN RAY D. 9 Loking, James Ralph v. 10 LOPEZ, NORMELYN G. 11 MENORO, ALCRIS JOYCE B. 12 Munalem, Paul John B 1. A circular gate 1.5 m in diameter is inclined at an angle of 45. Sea water (S = 1.02) stands on one side of the gate to a height of 10 m above the center of the gate. Determine the total hydrostatic force F and the location of C.P. 64 D Ig 45 sin h y y A Ig e D 4 A A h F 4 2  =  = =  =  = 2. An airplane barometer reads 762 mm Hg at sea level and 736 mm Hg at some unknown elevation. If the temperature at sea level is 15 C, what is the elevation assuming: a. air at constant density (h = 287.8 m) b. air under isothermal condition(h = 292.8 m) c. air under isentropic conditions (h = 291.35 m) d. air condition follows standard atmosphere condition (h = 291.83 m) SOLUTION: ( ) ( ) KPa 13 . 98 325 . 101 760 736 P K 288 273 15 Ts KPa 6 . 101 325 . 101 760 762 Ps = = = + = = = FLUID MECHANICS (ACTIVITY NO. 2/January 14, 2016) Group no. 2 1 Abacahin, Ruvie Grace s.
• 36. 2 Acosta, Faradiban O. 3 BUARON, EMMAN REY T. 4 CALIXTRO, JUNIEL F. 5 Ellevera, Glenn Eric P. 6 Mabanta, Chris Dilon F. 7 Pactores, Garret Oliver r. 8 Tapitan, Romel y. 9 Virtudazo, Erick Jan b. 10 TALADO, JOHN KENNETH E. 11 REYES, REYVENCER T. 12 Onda, Ruffy q. 1. What force must be exerted through the bolts to hold the dome in place? The metal dome and pipe weighs 6 KN and has no bottom. ( ) ( ) ( ) ( ) KN 351.7 6 - ) 9.81(36.46 bolts in Tension m 46 . 36 6 . 1 3 2 6 . 5 6 . 1 V 6 . 1 3 4 2 1 6 . 5 ) 6 . 1 ( V V - V V liquid of volume imaginary of Volume - V m 6 . 5 4 6 . 1 h 3 2 3 2 dome cylinder = = =       −  =        −  = = = + = 2. A U tube mercury manometer is connected to a closed pressurized tank as shown. If the air pressure is 14 KPa, determine the differential reading h. The specific weight of the air is negligible. cm 9.5 m 095 . 0 ) 81 . 9 ( 6 . 12 ) 2 . 1 )( 81 . 9 ( h 0 h ) 81 . 9 ( 6 . 13 h 81 . 9 ) 6 . 0 ( 81 . 9 ) 6 . 0 ( 81 . 19 14 h ) 81 . 9 ( 6 . 13 h 81 . 9 ) 6 . 0 ( 81 . 9 ) 6 . 0 ( 81 . 9 14 = = = = − + + = − + + + FLUID MECHANICS (ACTIVITY NO. 2/January 14, 2016) Group no. 3 1 VENCER, KYLE IAN B. 2 Salac, James Rhode E. 3 Nuñez, Insan John L. 4 ESTOSO, LADY LEE I. 5 Golosino Jr., Virgilio G. 6 Levi, Daniel L. 7 Jimenez, James V. 8 Enanod, Lourence A. 9 CARIGA, SAMS M. 10 BACUÑATA, ERROLJOHN B. 11 Agbu, Jovan o. 12 ABERO, SWEETSEL MAE M. 1. A BOEING 747 flies at an altitude of 2500 m above sea level where Ps = 101.33 KPa and Ts = 288 K. Determine the pressure and temperature at this altitude considering a. air under isothermal condition (P2 = 75.31 KPa ; T = 288 K) b. air under isentropic conditions (P2 = 74.31 KPa ; T = 263.6 K)
• 37. c. air condition follows standard atmosphere condition (P2 = 74.66 KPa ; T = 271.75 K) K 288 Ts KPa 101.33 Ps m 2500 h = = = FLUID MECHANICS (ACTIVITY NO. 2/January 14, 2016) Group no. 4 1 ABUG, BON JASON A. 2 Amarille, Freniel Ian A. 3 BUNTAG, GILBERT G. 4 Dedumo, Francis Dominique C. 5 Estoya, Mark Louie s. 6 Joloyohoy, Ful Yuri a. 7 Mamucay, Jirah o. 8 Paran, Marc Joseph L. 9 Sobrepeña, Ferden s. 10 Tuble, Ian Stephen d. 11 Igot, Neeco Adrienn r. 12 1. The closed tank in the figure is filled with water. The pressure gage on the tank reads 48 KPa. Determine a. The height h in mm in the open water column b. The gage pressure acting on the bottom of the tank surface AB c. The absolute pressure of the air in the top of the tank if the local atmospheric pressure is 101 KPa absolute. KPa 149 101 48 P KPa 772 . 59 P P ) (0.6)(9.81 0.6(9.81) 48 m 5 . 5 h 0 ) 81 . 9 ( h ) 81 . 9 ( 60 . 0 48 abs AB AB = + = = = + + = = − + 2. The air above the liquid is under a pressure of 40 KPa gage, and the specific gravity of the liquid in the tank is 0.80. If the rectangular gate is 1 m wide and if y1 = 1 m and y2 = 3 m , What force P is required to hold the gate in place? SOLUTION ( ) KN 32 . 107 3 ) 8 . 1 ( 86 . 178 p 3P 0.3) F(1.5 0 M m 3 . 0 e m 5 . 7 ) 5 . 2 ( 3 y A m 25 . 2 12 ) 3 ( 1 Ig y A Ig e KN 86 . 178 ) 3 )( 1 ( ) 5 . 2 )( 81 . 9 ( 8 . 0 40 F Hinge @ 3 4 3 = = = + =  = = = = = = = + =
• 38. FLUID MECHANICS QUIZ NO. 3 1. A block of wood has a vertical projection of 15.24 cm when placed in water and 10.2 cm when placed in alcohol. If the specific gravity of alcohol is 0.82, find the specific gravity of wood gravity of wood. 601 . 0 382 . 0 2296 . 0 1000 S 0.382 6) 1000(0.229 4 eq. eq.3 4 A(0.382) W 3 A ) 2296 . 0 ( 1000 W m 382 . 0 h ) 0.82(0.102 1524 . 0 0.82h - h 0.102) - 0.82(h 0.1524) - (h 2 eq. and 1 eq. Equating 2 0.102) - A(h 0.82(1000) BF W Alcohol In 1 0.1524) - 1000A(h BF W water In B B B B = =  =  = = →  = → = = − = = → = = → = = 2. A cubical block of wood 10 cm on a side floats at the interface between oil and water with its lower surface 1.50 cm below the interface. The density of the oil is 790 kg/m3 . What is the gage pressure at the lower face of the block? What is the mass and density of the block?
• 39. P = -6.9 KPag • CG Free Surface h m 5 . 1 A • CP P F 45 3 m 3. Compartments A and B of the tank shown in the figure below are closed and filled with air and a liquid with S = 0.6. If the atmospheric pressure is 101 KPa (abs) and the pressure gage reads 3.5 KPa (gage), determine the manometer reading h in cm. m 9 . 1 ) 6 . 0 1 ( 81 . 9 ) 03 . 0 )( 81 . 9 ( 6 . 13 02 . 0 ) 81 . 9 ( 6 . 0 5 . 3 h 0 ) 03 . 0 )( 81 . 9 ( 6 . 13 02 . 0 ) 81 . 9 ( 6 . 0 h ) 81 . 9 ( 6 . 0 h 81 . 9 5 . 3 = − + + = = + + + − 4. The gate shown is hinged at A and rests on a smooth floor at B. The gate is 3 m square. Oil (S = 0.80) stands on the left side of the gate to a height of 1.5 m above A. Above the oil surface is a gas under a gage pressure of -6.9 KPa. Determine the amount of the vertical force P applied at B that would be required to open the gate.   KN 8 . 118 ) 3 )( 3 ( 56 . 2 ) 81 . 9 ( 80 . 0 9 . 6 F m 56 . 2 45 sin 5 . 1 5 . 1 h = + − = =  + = 5. A spherical buoy 2 m in diameter floats half submerge in a liquid with S = 1.5. What is the weight of the lead anchor weighing 7000 kg/m3 will completely submerged the buoy in the liquid.
• 40. kg 3990 W m 57 . 0 V V 1500 6270 V 7000 3135 V 1500 BF V 7000 W kg 3135 W kg 6270 1500(4.18) ) 1500(V BF BF BF W W : 2 figure from kg 3135 ) 09 . 2 ( 1500 W ) V ( BF W submerge volume m 09 . 2 ) 1 ( 3 4 2 1 V m 18 . 4 R 3 4 V m kg 1500 ) 1000 ( 5 . 1 2 3 2 2 2 2 2 2 2 1 sphere 1 2 1 2 1 1 s L 1 1 3 3 s 3 3 sphere 3 L = = + = + = = = = = = + = + = =  = = → =        = =  = = =  6. A glass tube 1.5 m long and 25 mm diameter with one end closed is inserted vertically with the open end down into a tank of water until the open end is submerged to a depth of 1.2 m. If the barometric pressure is 98 KPa, and neglecting vapor pressure, how high will water rise in the tube. (Assume isothermal conditions for air)        cm 14 m 14 . 0 x 2 98 . 12 7 . 12 x 0 8 . 1 x 7 . 12 x 0 658 . 17 x 487 . 124 x 81 . 9 x 81 . 9 x 715 . 14 x 772 . 109 658 . 164 147 ) x 5 . 1 )( x 81 . 9 772 . 109 ( 147 ) x 5 . 1 )( x 81 . 9 772 . 11 98 ( 147 ) x 5 . 1 ( ) x 2 . 1 ( 81 . 9 98 ) 5 . 1 ( 98 ) x 5 . 1 ( ) 025 . 0 ( 4 ) x 2 . 1 ( 81 . 9 98 ) 5 . 1 ( ) 025 . 0 ( 4 98 2 ) x 5 . 1 ( ) 025 . 0 ( 4 V 1 ) x 2 . 1 ( 81 . 9 98 P m 00074 . 0 ) 5 . 1 ( ) 025 . 0 ( 4 V V P V P : isothermal fOR 2 2 2 2 2 2 2 2 3 2 1 2 2 1 1 = =  − = = − + = − + − − − = − − = − − + = − − + =       −  − + =       → −  = → − + = =  = = 1. A brass cube 152.4 mm on a side weighs 298.2 N. We want to hold this cube in equilibrium under water by attaching a light foam buoy to it. If the foam weighs 707.3 N/m3 , what is the minimum required volume of the buoy?
• 41. 3 2 2 2 2 2 2 2 3 1 1 m 029 . 0 3 . 707 9810 72 . 34 2 . 298 V ) 9810 ( V 72 . 34 V 3 . 707 2 . 298 2 ) 9810 ( V BF 1 V 3 . 707 W N 72 . 34 ) 9810 ( ) 1524 . 0 ( BF N 2 . 298 W = − − = + = + → = → = = = = 2. A man dives into a lake and tries to lift a large rock weighing 170 kg. If the density of the granite rock is 2700 kg/m3, find the force that the man needs to apply to lift it from the bottom of the lake. Assume density of lake water to be 1000 kg/m3 . kg 107 63 - 170 T T BF W kg 63 ) 063 . 0 ( 1000 BF m 063 . 0 2700 170 V kg 170 V 2700 W 3 = = + = = = = = = = Example 1 A large pipe called a penstock in hydraulic work is 1.5 m in diameter. Here it is composed of wooden staves bound together by steel hoops each 3.23 cm2 in area, and is used to conduct water from a reservoir to a powerhouse. If the maximum tensile stress permitted in the hoops is 130 MPa, what is the maximum spacing between hoops under a head of 30.5 m. cm 18.7 m 187 . 0 L 0.000323 224.4L 130000 KPa 130000 S m 000323 . 0 A cm 3.23 A ; A T S hoop the on Stress Tensile KN L 4 . 224 2 F T KN L 81 . 448 5)L 299.205(1. F LD A ; A F P projection vertical the On m 1.5 D KPa 205 . 299 ) 5 . 30 ( 81 . 9 P 2 2 = = = = = = = = = = = = = = = = Example 2 Compute the wall stress in a 1200 mm diameter steel pipe 6 mm thick under a pressure of 970 KPa. Given: D = 1.2 m t = 0.006 m P = 970 KPa KPa 000 , 97 ) 006 . 0 ( 2 ) 2 . 1 ( 970 t 2 PD S = = = Example 3 What is the minimum allowable thickness of 600 mm diameter steel pipe under an internal pressure of 860 1.5 m L L
• 42. KPa with a working stress in the steel of 70,000KPa. mm 4 m 004 . 0 t t 2 ) 600 . 0 ( 860 000 , 70 t 2 PD S = = = = Example 4 A wood stave pipe is bound by steel rods which take the entire bursting stress. Find the proper spacing for 25 mm steel rods for a 1800 mm diameter wood stave pipe under a pressure of 590 KPa if the working stress in the steel is 105,000 KPa. Given Dr = 0.025 m D = 1.8 m P = 590 KPa S = 105,000 KPa ( ) ( ) cm 10 L m 10 . 0 590(1.8) 2(51.54) L KN 51.54 T 0.025 4T 0 00 , 105 025 . 0 4 T S rods the For DL T 2 A F P 2 2 = = = =  =  = = = Example 5 A vertical cylindrical tank, 2 m in diameter and 4 m high, is held together by means of two steel hoops, one at the top and one at the bottom. When molasses (S = 1.50) stands to a depth of 3 m in the tank, what is the tensile force in each hoop?
• 43. 2g v Q P POWER JET 2 2 Jet  = A. NOZZLE L 2 2 2 2 1 2 1 1 H Z g 2 v P Z g 2 v P + + +  = + +  From continuity equation: Q = Av ; for 1 = 2 =  2 2 1 1 v A v A Q = = For a nozzle the head loss HL is equal to: g 2 v 1 C 1 2 2 2 v         − where: Cv - velocity coefficient B. VENTURI METER
• 44.                 −       − + − =                 − = − = − + − − = = +         =         = = + + = + + 4 1 2 2 1 2 1 2 4 1 2 2 2 2 1 2 2 2 1 2 1 w Hg 2 1 2 Hg w 1 2 2 4 1 2 2 1 2 2 1 2 1 2 2 1 1 2 2 2 2 1 2 1 1 d d 1 ) z z ( γ P P g 2 v d d 1 g 2 v g 2 v v ) z z ( γ P P ) γ γ h ( P - P P (h ) γ - (h ) γ P meter th e In v d d v v d d v v A v A z g 2 v γ P z g 2 v γ P C. ORIFICE An orifice is an opening with a closed perimeter in which fluid flows. By applying Bernoulli's Energy theorem: 2 2 2 2 1 2 1 1 Z g 2 v P Z g 2 v P + +  = + +  But P1 = P2 = Pa and v1 is negligible, then 2 1 2 2 Z Z g 2 v − = and from figure: Z1 - Z2 = h, therefore h g 2 v 2 2 = gh 2 v2 = let v2 = vt gh 2 vt = where: vt - theoretical velocity, m/sec h - head producing the flow, meters g - gravitational acceleration, m/sec2
• 45. COEFFICIENT OF VELOCITY (Cv) velocity l theoretica velocity actual Cv = t v ' v = Cv COEFFICIENT OF CONTRACTION orifice the of area contracta vena @ jet of area Cc = A a = Cc where: a - area of jet at vena contracta, m2 A - area of the orifice, m2 COEFFICIENT OF DISCHARGE flow l theoretica flow actual Cd = Q ' Q = Cd c v d C C C = where: v' - actual velocity vt - theoretical velocity a - area of jet at vena contracta A - area of orifice Q' - actual flow Q - theoretical flow Cv - coefficient of velocity Cc - coefficient of contraction Cd - coefficient of discharge JET TRAJECTORY:
• 46. z 1 TANK z 2 A B • 1 • 2 5 g ) sin2 ( v R ) 2 sin( ) cos sin 2 ( g ) cos sin 2 ( v g ) cos v )( 2(vsin R ) t 2 ( cos v R 4 g ) (vsin t g ) (vsin t gt 2g ) (vsin 3 gt d gt 0 d gt t v d 2 to 0 at 2 g sin v t gt v v 1 2g ) (vsin d 2gd - v 0 2gd - v v 0 to 1 at nozzle of tip at water of velocity v 2 2 2 2 2 2 2 1 2 2 2 1 2 2 1 2 2 1 1 1 0 2 2 1 2 1 2 o →  =  =     =   =  = →  =  = =  → = + = + = →  = − = →  = = = − If the jet is flowing from a vertical orifice: If the jet is initially horizontal vx = v. D. PUMP Pump is a steady - state, steady - flow machine in which mechanical energy is added to the liquid from one point to another point of higher pressure.
• 47. Points 1 & 2 are the reference points at suction and discharge, respectively. a) Total Dynamic Head (Ht) ( ) L 1 2 2 1 2 2 1 2 H Z Z 2g v v γ P P t h + − +         − +         − = meters b) Fluid or Water Power (FP) t h Q FP  = KW c) Capacity or Discharge (Q) 2 2 1 1 v A v A Q = = m3 /sec d) Brake or Shaft Power (BP) 000 , 60 TN 2 BP  = KW e) Motor Power (MP) (For motor driven pump) 1. For single - phase motors 1000 ) (cos EI MP  = KW 2. For 3 - phase motors 1000 ) (cos EI 3 MP  = KW f) Pump Efficiency BP FP p =  x 100% g) Motor Efficiency MP BP m =  x 100% h) Combined Pump-Motor Efficiency (Overall Efficiency) IP MP c =  x 100 % where: ht - total dynamic head, m FP - Fluid or Water Power, Kw BP - Brake or Shaft Power, KW MP - Motor Power. KW IP - Power input to motor, KW Q - Capacity or Discharge, m3 /sec  - Specific weight of the liquid pumped, KN/m3 v - velocity, m/sec P - pressure, KPa g - gravitational acceleration, m/sec2 Z - elevation, m (Positive if measured above datum, and negative if measured below datum) HL - head loss (due to fluid friction, fittings and turbulence in pipes) T - Brake torque, N-m N - no. of RPM (revolutions per minute) E - Volts I - current drawn by the motor, Amperes cos - Power Factor E. HYDRAULIC TURBINE Impulse Type (Pelton Type)
• 48. v 041 . 0 0 v 0 0 v P g 2 v 041 . 0 g 2 v 1 ) 98 . 0 ( 1 g 2 v 1 C 1 H H z g 2 v P z g 2 v P equation s Bernoulli' Applying KPa 55 P 98 . 0 C sec m 3 v m 10 . 0 d 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 v L L 2 2 2 2 1 2 1 1 1 v 1 1 + + + = + + =       − =       − = + + +  = + +  = = = = Head Gates (Fully Open Y – Gross Head A B Dam Head Water Tail Water Penstock Draft Tube 1• 2• zA PA Reaction Type( Francis Turbine) Fundamental Equations 1. TOTAL DYNAMIC HEAD (h): a. For an Impulse Type h = (Z1 - Z2) - HL meters h = Y - HL meters Y = Z1 - Z2 meters b. For a Reaction Type h = (Z1 - Z2) - HL meters h = Y - HL meters Y = Z1 - Z2 meters Y - gross head at plant, m at A to 2 2 A A meters z g 2 v P h A + +  = 2. WATER POWER (WP): FP = Qh KW 3. DISCHARGE OR RATE OF FLOW (Q): Q = Av m3 /sec 4. BRAKE POWER (BP): BP = 2TN KW 60 000 T - brake torque, N-m N - rotative speed, RPM 5. TURBINE EFFICIENCY (e) e = BP x 100% FP e = evehem where: ev - volumetric efficiency eh - hydraulic efficiency em - mechanical efficiency 6. ROTATIVE SPEED (N): n f 120 N = where: f - frequency, cps or Hertz n - no. of generator poles (usually divisible by 4) QUIZ NO. 4 Problem No. 1 The velocity of water in a 10 cm diameter pipe is 3 m/sec. At the end of the pipe is a nozzle whose velocity coefficient is 0.98. If the pressure in the pipe is 55 KPa, what is the velocity in the jet? What is the diameter of the jet? What is the rate of discharge? What is the head loss?
• 49. KPa 86 . 33 760 101.325 -254 P m 6 z ; 0 z sec m 8 . 7 v ; sec m 4 . 4 v A Q v d 4 A m 018 . 0 A m; 0.152 d m 032 . 0 A ; m 203 . 0 d sec m 142 . 0 Q 1 2 1 2 1 2 2 2 2 2 1 1 3 − =       = = = = = =  = = = = = = ( ) ( ) KPa 45 . 414 P P H z z g 2 v v h P H z z g 2 v v P P h m 84 . 53 ) 81 . 9 )( 142 . 0 ( 75 h h Q WP 2 1 L 1 2 2 1 2 2 t 2 L 1 2 2 1 2 2 1 2 t t t =  +       − − − − −  = + − + − +  − = = =  = Problem No. 2 A centrifugal pump draws water from a well at the rate of 142 L/sec of water through a 203 mm ID suction line and a 152 mm ID discharge line. The suction gauge located on the pump centerline reads 254 mm Hg vacuum, while the discharge gauge is 6 m above the pump centerline. If the power input to the water is 75 KW, find the reading of the discharge gauge in KPa. Problem No. 3 A 15 KW suction pump draws water from a suction line whose diameter is 200 mm and discharges through a line whose diameter is 150 mm. The velocity in 150 mm line is 3.6 m/sec. If the pressure at point A in the suction line is 34.5 KPa below the atmosphere where A is 1.8 m below that of B on the 150 mm line, Determine the maximum elevation above B to which water can be raised assuming a head loss of 3 m due to friction. FP 15 KW d1 0.2 m d2 0.15 m v2 3.6 m/sec P1 -34.5 KPa P2 0 KPa HL 3 m A1 0.03 m^2 A2 0.02 m^2
• 50. Problem No. 4 A power nozzle throws a jet of water that is 50 mm in diameter. The diameter of the base of the nozzle and of the approach pipe is150 mm. If the power of the nozzle jet is 42 HP and the pressure head at the base of the nozzle is 54 m, compute the head lost in the nozzle. Problem No. 5 A fire pump delivers water through a 150 mm main to a hydrant to which is connected a 75 mm hose, terminating in a 25 mm nozzle. The nozzle is 1.5 m above the hydrant and 10 m above the pump. Assuming frictional losses of 3 m from the pump to the hydrant, 2 m in the hydrant and, and 12 m from the hydrant to the base of the nozzle, and a loss in the nozzle of 6% of the velocity head in the jet, to what vertical height can the jet be thrown if the gage pressure at the pump is 550KPa. Q 0.064 m^3/sec v1 2.025 m/sec SW 9.81 KN/m^3 P1/SW -3.52 m P2/SW 0.000 m v1^2/2g 0.209 m v2^2/2g 0.66 m z1 0.0 m z2 1.8 + h ht 24.04 m D(Phead) 3.52 D(Vhead) 0.45 D(Ehead) (1.8 + h) h 15.27 m
• 51. 93 . 0 67 . 0 625 . 0 Cv ) Cv ( Cc Cd 67 . 0 12 8 D ' d A a Cc 625 . 0 Q ' Q Cd sec m 12 . 0 ) 34 . 15 ( ) 10 . 0 ( 4 Q sec m 075 . 0 ' Q sec m 34 . 15 ) 2(9.81)(12 v velocity l theoretica gh 2 v 3 2 3 = = = = = = = = = =  = = = = → = Problem No. 6 Water issues from a circular orifice under a head of 12 m. The diameter of the orifice is 10 cm. If the discharge is found to be 75 L/sec, what is the coefficient of discharge? If the diameter at the vena cotracta is measured to be 8 cm, what is the coefficient of contraction and what is the coefficient of velocity. Problem No. 7 A jet discharges from an orifice in a vertical plane under a head of 3.65 m. The diameter of the orifice is 3.75 cm and the measured discharge is 6 L/sec. The coordinates of the centerline of the jet are 3.46 m horizontally from the vena contracta and 0.9 m below the center of the orifice. Find the coefficient of discharge, velocity and contraction.
• 52.                 −       − +  − = − +  − =                 − − +  − =                 − +         −         − + + +  = +         +          − + + +  = + +  + + +  = + +          =         = = = 4 1 2 2 v 2 1 2 1 2 2 2 1 2 1 4 1 2 2 v 2 2 2 1 2 1 4 1 2 2 v 2 2 2 2 2 v 2 2 2 2 1 2 2 4 1 2 1 2 2 2 v 2 2 2 2 1 2 1 1 L 2 2 2 2 1 2 1 1 2 2 4 1 2 2 1 2 2 4 1 2 2 1 2 2 1 1 d d C 1 ) z z ( P P g 2 v ) z z ( P P d d C 1 g 2 v ) z z ( P P d d 1 1 C 1 g 2 v g 2 v 1 C 1 z g 2 v P z g 2 v d d P g 2 v 1 C 1 z g 2 v P z g 2 v P H z g 2 v P z g 2 v P g 2 v d d g 2 v v d d v v A v A Q Problem No. 8 The inside diameters of the suction and discharge pipes of a pump are 20 cm and 15 cm, respectively. The discharge pressure is read by a gage at a point 2 m above the centerline of the pump, and the suction pressure is read by a gage 1 m below the pump centerline. If the pressure gage reads 145 KPa and the suction gage reads a vacuum of 250 mm Hg when diesel fuel (S = 0.82) is pumped at the rate of 30 L/sec, Find the KW power of the driving motor if overall pump efficiency is 75%. Problem No. 9 A jet of water 7.6 cm in diameter discharges through a nozzle whose velocity coefficient is 0.96. If the pressure in the pipe is 82.7 KPa and the pipe diameter is 20 cm and if it is assumed that there is no contraction of the jet, what is the velocity at the tip of the nozzle? What is the rate of discharge? d1 0.20 m d2 0.076 m Cv 0.96 P1 82.7 Kpa P2 0 KPa v1 1.80 m/sec v2 12.467 m/sec SW 9.81 KN/m^3 g 9.81 m/sec^2 v2 12.47 m/sec A1 0.031 m^2 A2 0.005 m^2 Q 0.057 m^3/sec d1 0.20 m d2 0.15 m z1 -1 m z2 2 m P1 -33.33 Kpa P2 145 KPa S 0.82 SW(water) 9.81 KN/m^3 SW 8.0442 KN/m^3 Q 0.03 m^3/sec A1 0.031 m^2 A2 0.018 m^2 v1 0.955 m/sec v2 1.70 m/sec P1/SW -4.14 m P2/SW 18.03 m v1^2/2g 0.05 m v2^2/2g 0.15 m D(Phead) 22.17 m D(Vhead) 0.10 m D(ElHead) 3.00 m HL 0.00 m ht 25.27 m WP 6.10 KW e 0.75 BP 8.13 KW
• 53. SAMPLE PROBLEMS APPLICATION OF BERNOULLI’S EQUATION ExampleNo. 1 The water in a 10 m diameter, 2 m high aboveground swimming pool is to be emptied by unplugging a 3 cm diameter, 25 m long horizontal pipe attached to the bottom of the pool. Determine the maximum discharge rate of water through the pipe. Example No. 2 A large tank open to the atmosphere is filled with water to a height of 5 m from the outlet tap. A tap near the bottom of the tank is now opened and water flows out from the smooth and rounded outlet. Determine the water velocity at the outlet. Example No. 3 The water level of a tank on a building roof is 20 m above the ground. A hose leads from the tank bottom to the ground. The end of the hose has a nozzle, which is pointed straight up. What is the maximum height to which the water could rise. h = 2 m 1 2 m/sec 2 3 . 6 ) 2 )( 8 1 . 9 ( 2 v g 2 v 2 ) 2 ( g 2 v 0 0 0 0 m -2 Z 0 ; Z 0 ; P 0 ; v ; 0 P Z g 2 v γ P Z g 2 v γ P 2 2 2 2 2 2 1 2 1 1 2 2 2 2 1 2 1 1 = = = − + + = + + = = = = = + + = + + /sec m 0044 . 0 ) 23 . 6 ( ) 03 . 0 ( 4 π Q Av Q 3 2 = = = 1 m/sec 6 2 3 . 9 5 ) 8 1 . 9 ( 2 v g 2 v 5 ) 5 ( g 2 v 0 0 0 0 m 5 - Z 0 ; Z 0 ; P 0 ; v ; 0 P Z g 2 v γ P Z g 2 v γ P 2 2 2 2 2 2 1 2 1 1 2 2 2 2 1 2 1 1 = = = − + + = + + = = = = = + + = + + nozzle the of tip at the velocity v heigh t imum max h g 2 v h h 0 0 0 2g v 0 0 v h; Z 0; Z 0; P ; 0 P Z g 2 v γ P Z g 2 v γ P 1 2 1 2 1 2 2 1 2 1 2 2 2 2 1 2 1 1 − − = + + = + + = = = = = + + = + +
• 54. Example no. 4 Water flows through a horizontal pipe at the rate of 1 Gal./sec. The pipe consist of two sections of diameter 4 in. and 2 in with a smooth reducing section. The pressure difference between the two pipe sections is measured by mercury manometer . Neglecting frictional effects, determine the differential height of mercury between the two pipe sections. Q = 1 gal/sec = 0.0038 m3 /sec D1 = 4 in. = 0.1016 m D2 = 2 in = 0.0508 m 1 2 h x Mercury S = 13.6 inches 53 . 0 h meters 0134 . 0 h 123.606 - 1.655 - h -1.655 123.606h - 2 eq. and 1 eq. Equating 2 KPa 655 . 1 ) 1687 . 0 ( 81 . 9 P P 1687 . 0 18 . 0 0113 . 0 P P 0 18 . 0 P 0 0113 . 0 P m 18 . 0 ) 81 . 9 ( 2 ) 875 . 1 ( 2g v m 0113 . 0 ) 81 . 9 ( 2 ) 47 . 0 ( 2g v m/sec 875 . 1 ) 0508 . 0 ( 4 0038 . 0 A Q v m/sec 47 . 0 ) 1016 . 0 ( 4 0038 . 0 A Q v Z g 2 v P Z g 2 v P 2 to 1 from 1 606 . 123 P P h ) 81 . 9 ( 6 . 13 h 81 . 9 P P P x 81 . 9 h ) 81 . 9 ( 6 . 13 h 81 . 9 x 81 . 9 P 1 2 1 2 2 1 2 2 2 2 2 1 2 2 2 2 1 1 2 2 2 2 1 2 1 1 1 2 1 2 2 1 = = = = → − = − = − − = − = − + + = + + = = = = = = = = = = + + = + + → − = − − = − = − − + + γ γ γ π π γ γ
• 55. 5 5 . 2 m 9 − Example No. 5 A 300 mm pipe is connected by a reducer to a 100 mm pipe. Points 1 and 2 are at the same elevation. The pressure at point 1 is 200 KPa. Q = 30 L/sec flowing from 1 to 2, and the energy lost between 1 and 2 is equivalent to 20 KPa. Compute the pressure at 2 if the liquid is oil with S = 0.80. (174.2 KPa) Example No. 6 A venturi meter having a diameter of 150 mm at the throat is installed in a 300 mm water main. In a differential gage partly filled with mercury (the remainder of the tube being filled with water) and connected with the meter at the inlet and at the throat, what would be the difference in level of the mercury columns if the discharge is 150 L/sec? Neglect loss of head. (h=273 mm)
• 56. K Pa 7 6 6 . 3 3 ) 8 1 . 9 ( 4 4 2 . 3 P P m 4 4 2 . 3 γ P P 2 3 0 . 0 6 7 2 . 3 g 2 v g 2 v γ P P 0 Z Z Z g 2 v γ P Z g 2 v γ P m 3 .6 7 2 2 g v sec; / m 4 8 8 . 8 ) 1 5 . 0 ( 4 π 1 5 0 . 0 v m 2 3 0 . 0 2 g v sec; / m 1 2 2 . 2 ) 3 0 . 0 ( 4 π 1 5 0 . 0 v A Q v 0 H sec / m 1 5 0 . 0 Q m 1 5 . 0 d m 3 0 . 0 d m K N 1 3 3 .4 1 6 1 3 .6 (9 .8 1 ) γ 1 3 .6 mercu ry o f S sec m 1 5 0 . 0 L 1 0 0 0 m 1 x sec L 1 5 0 Q 2 1 2 1 2 1 2 2 2 1 2 1 2 2 2 2 1 2 1 1 2 2 2 2 2 1 2 1 L 3 2 1 3 3 3 = = − = − − = − = − = = + + = + + = = = = = = = = = = = = = = = = mm 273 h m 0.273 h ) 81 . 9 416 . 133 ( 766 . 33 ) 81 . 9 416 . 133 ( P P h ) 81 . 9 416 . 133 ( h P P h 81 . 9 h 416 . 133 P P P h 416 . 133 h 81 . 9 P P x 81 . 9 h 416 . 133 h 81 . 9 ) x ( 81 . 9 P 2 1 2 1 2 1 2 1 2 1 = = − = − − = − = − − = − = − + = − − + +
• 57. Example no. 7 A mechanical engineer of an industrial plant is required to install a centrifugal pump to lift 15 L/sec of water from a sump to a storage tank on a tower. The water is to be delivered into a 105 KPag tank and the water level in the tank is 20 m above the water level in the sump. Pump centerline is 4 m above the water level in the sump. The suction pipe is 100 mm diameter and discharge pipe is 65 mm diameter. Head loss at suction is 3 times the velocity head in the suction line and head loss at discharge is 20 times the velocity head in the discharge pipeline. Other data are as follows: p = 75 % m = 80 % E = 220 Volts Motor – 3-Phase Power Factor = 0.92 Requirements: a. Sketch of the problem b. Total dynamic head in m c. Water Power in KW d. Motor power in KW e. Line current drawn by the motor in amperes f. Total power cost per day for 10 hours a day continuous operation and a power costs of P 5.00/KW-hr m 8 3 . 2 0 g 2 v 2 0 h m 5 6 . 0 g 2 v 3 h m 0 4 1 . 1 ) 8 1 . 9 ( 2 2 ) 5 2 . 4 ( g 2 v m 1 8 6 . 0 ) 8 1 . 9 ( 2 2 ) 9 1 . 1 ( g 2 v sec m 5 2 . 4 ) 0 6 5 . 0 ( 4 π 0 1 5 . 0 A Q v sec m 9 1 . 1 ) 1 0 . 0 ( 4 π 0 1 5 . 0 A Q v A Q v A v Q 2 d Ld 2 s Ls 2 d 2 s 2 s d 2 s s =         = =         = = = = = = = = = = = = = 20 m 4 m 100 mm 65 mm 1 • 2 • 105 KPa
• 58. 3 m 0 4.5 m m/sec 5 . 9 2 . 1 7 1 5 . 1 4 1 5 3 5 ) 8 1 . 9 ( 2 v v m K N 7 1 5 . 1 4 1 .5 (9 .8 1 ) γ v elo city cal th eo reti h γ P P g 2 v Z Z γ P P g 2 v 0 v h ; - Z ; 0 Z : 1 p o in t at D atu m Z g 2 v γ P Z g 2 v γ P lo st h ead g co n sid erin With o u t t 2 3 2 1 2 2 1 2 1 2 2 1 2 1 2 2 2 2 1 2 1 1 =       + + = = = = →       + − = − + − = = = = + + = + + 0 0 . 6 3 9 P ) 5 )( 1 0 ( 8 . 1 2 Co st amp eres 4 4 . 3 6 I 1 0 0 0 2 ) (2 2 0 )I(0 .9 3 1 2 .8 1 0 0 0 EI(P.F.) 3 MP K W 8 . 1 2 0 .7 5 1 0 .2 2 MP K W 2 2 . 1 0 0 .7 5 7 .6 7 BP K W 6 7 . 7 )(5 2 .0 9 1 ) 0 .0 1 5 (9 .8 1 WP m 0 9 1 . 5 2 h 8 3 . 2 0 ) 0 2 0 ( 0 9 .8 1 0 - 1 0 5 h 1 p o in t th ro u g h lin e d atu m 2 , to 1 at H Z Z 2 g v v γ P P h m 3 9 . 2 1 2 0 .8 3 0 .5 6 H t t L 1 2 2 1 2 2 1 2 t L = = = = = = = = = = = = + − + + = + − + − + − = = + = Example No. 8 Figure below shows a siphon discharging oil (sp gr 0.90). The siphon is composed of 8 cm. pipe from A to B followed by 10 cm. pipe from B to the open discharge at C. The head losses are from 1 to 2, 0.34 m; from 2 to 3, 0.2 m; from 3 to 4;0.8 m. Compute the discharge in L/sec Example No. 9 The liquid in the figure has a specific gravity of 1.5. The gas pressure PA is 35 KPa and PBis -15 KPa. The orifice is 100 mm in diameter with Cd = Cv = 0.95. Determine the velocity in the jet and the discharge when h = 1.2. (9.025 m/sec; 0.071 m3 /sec) •1 •2 PA PB h
• 59. pump V-1 I-1 6 m I-2 V-2 72 m 1• 2• 150 mm 100 mm A B m 63 . 78 83 . 6 72 h H Z Z h ic) (Atmospher gage 0 P P negligible are v and v KPa 827 P meters; 3 . 84 γ P H Z g 2 v γ P ht Z g 2 v γ P 63 . 6 78 0 0 0 ) 81 . 9 ( 2 ) 55 . 2 γ P 2 to 1 point At h Z g 2 v γ P Z g 2 v γ P m 83 . 6 h h H datum) on B (Point 2 to B At KPa 2 . 56 P ; meters 73 . 5 γ P m 63 . 6 g 2 v 20 h 20 . 0 6 ) 81 . 9 ( 2 ) 13 . 1 ( γ P 0 0 0 m 20 . 0 g 2 v 3 h h Z g 2 v γ P Z g 2 v γ P sec / m 55 . 2 ) 100 . 0 ( 4 π 020 . 0 v datum) on 1 (Point A to 1 At m/sec 13 . 1 ) 150 . 0 ( 4 π 020 . 0 v HP 7 . 20 KW 43 . 15 Power Water A Q v )(78.63) 0.020(9.81 Power Water sec m 020 . 0 1000 20 Q t L 1 2 t 2 1 2 1 B B L 2 2 2 2 1 2 1 1 2 B Ld 2 2 2 2 B 2 d B Ld Ls L A A 2 d Ld 2 A 2 s Ls Ls B 2 S A 1 2 1 1 2 d 2 s 3 = + = + − = = = = = + + + = + + + + + + = + + + + + = + + = + = = = =         = + − + = + + =         = + + + = + + = = = = = = = = = = Example no. 10 A pump draws water from reservoir A and lifts it to reservoir B as shown in the figure. The loss of head from A to 1 is 3 times the velocity head in the 150 mm pipe and the loss of head from 2 to B is 20 times the velocity head in the 100 mm pipe. Compute the horsepower output of the pump and the pressure heads at 1 and 2 when the discharge is 20 L/sec. (FP= 20.73 HP; 5.74 m ; 84.3 m)
• 60. Example No. 11 The diameters of the suction and discharge pipe of a fuel pump for a day tank of a gasoline engine are 150 mm and 100 mm, respectively. The discharge pressure gauge located 10 m above the pump centerline reads 320 KPa and the suction pressure gauge which is 4 m below the pump centerline reads a vacuum of 60 KPa. Head losses due to pipe friction, turbulence and fittings amounts to 15 m. If gasoline with a relative density S = 0.75 is pumped at the rate of 35 L/sec, find a. The total dynamic head developed by the pump b. The fluid power in KW c. The brake or shaft power delivered to the fluid for a pump efficiency of 75% d. The brake torque if the pump speed is 1200 RPM e. The electrical power input to the pump motor for a motor efficiency of 92% f. The line current drawn by the motor if the motor is 3 – phase, 240 volts, and 0.9 Power Factor g. The cost of power for 5 hours operation, if electricity costs P 1.50/KW-hr Q 0.035 m^3/sec S 0.75 SW(water) 9.81 KN/m^3 SW 7.3575 KN/m^3 g 9.81 m/sec^2 d1 0.150 m d2 0.100 m P1 -60.00 KPa P2 320.00 KPa z1 -4 m z2 10 m HL 15 m FP 75 KW A1 0.018 m^2 A2 0.008 m^2 v1 2.0 m/sec v2 4.5 m/sec v1^2/2g 0.20 m v2^2/2g 1.01 m P1/sw -8.15 m P2/sw 43.49 m D(Phead) 51.65 m D(vhead) 0.81 m D(Zhead) 14 m ht 81.46 m Water Power 20.98 KW Pump Efficiency 75.00 % Brake Power 27.97 KW Motor Efficiency 92.00 % Motor Power 30.40 KW E (Volts) 220.00 Volts Power Factor 0.90 Phase 3.00 Phase N(RPM) 1200.00 RPM Torque(N-m) 222.57 N-m Line Current 88.65 Amperes PowerCost 1.50 Pesos/KW-hr Cost 228.01 Pesos
• 61. Example No. 12 A jet of liquid is directly vertically upward. At A (Nozzle tip) its diameter is 75 mm and its velocity is 10 m/sec. Neglecting air friction, determine its diameter at a point 4 m above A. d1 0.075 m v1 10 m/sec P1 0 Kpa P2 0 KPa g 9.81 m/sec^2 A1 0.0044 m^2 Q 0.0442 m^3/sec SW 9.81 KN/m^3 P1/SW 0 m P2/SW 0 m v1^2/2g 5.097 m z1 0 m z2 4 m HL 0 m v2^2/2g 9.097 m v2 13.36 m/sec A2 0.0033 m^2 d2 0.065 m d2 6.5 cm Example No. 13 A closed vessel contains water up to a height of 2 m and over the water surface there is air having a pressure of 8.829 N/cm2 above atmospheric pressure. At the bottom of the vessel there is an orifice of diameter 15 cm. Find the rate of flow of water from orifice if Cd = 0.6. Example No. 14 The 600 mm pipe shown in the figure conducts water from a reservoir A to a pressure turbine, which discharges through another 600 mm pipe into tailrace B. The loss of head from A to 1 is 5 times the velocity head in the pipe and the loss of head from 2 to B is 0.2 times the velocity head in the pipe. If the discharge is 700 L/sec , what horsepower is being given up by the water to the turbine and what are the pressure heads at 1 and 2.(FP = 537.4 HP; 53.628 m; -4.75 m) HP 537 KW 400.6 (58.336) 0.70(9.81) h Q WP meters 336 . 58 664 . 1 60 h H ) Z Z ( h 0 V v gage 0 P P m 60 Z 0; Z B poin through line datum With h H Z g 2 v P Z g 2 v P B to A At meters 1.664 0.064 1.6 H m 064 . 0 ) 32 . 0 ( 20 . 0 h m 6 . 1 5(0.320 h m 2 3 . 0 ) 81 . 9 ( 2 ) 5 . 2 ( 2g v m/sec 5 . 2 ) 6 . 0 ( ) 70 . 0 ( 4 A Q v /sec m 0.70 L/sec 700 Q L B A B A B A A B L B 2 B B A 2 A A L B - L2 1 - LA 2 2 2 3 = = =  = = − = − − = = = = = = = + + + +  = + +  = + = = = = = = = =  = = = = h 2 m P1 88.29 Kpa g 9.81 m/sec^2 SW 9.81 KN/m^3 P1/SW 9.0 m d(orifice) 0.15 m A(Orifice) 0.018 m^2 Theoretical Velocity 14.69 m/sec Cd 0.60 Q(m^3/sec) 0.1558 m^3/sec Q(L/sec) 155.8 L/sec
• 62. v acu u m K Pa 4 6 .7 K Pa -4 6 .7 1 ) -4 .7 5 6 (9 .8 P meters 7 5 6 . 4 0 6 4 . 0 3 2 . 0 5 . 4 γ P 0 6 4 . 0 5 . 4 0 0 0 3 2 . 0 γ P h Z g 2 v γ P Z g 2 v γ P m -4 .5 Z ; 0 Z 2 p o in t th ro u g h d atu m w ith B, to 2 A t K Pa 5 4 1 .3 ) 5 5 .1 8 (9 .8 1 P meters 1 8 . 5 5 3 2 . 0 5 . 5 5 γ P 5 . 5 5 3 2 . 0 γ P 0 0 0 m -5 5 .5 Z1 ; 0 ZA A p o in t th ro u g h d atu m w ith h Z g 2 v γ P Z g 2 v γ P 1 A to A t 2 2 2 B 2 L B 2 B B 2 2 2 2 B 2 1 1 1 1 LA 1 2 1 1 A 2 A A = = = − = + − − = + − + = + + + + + = + + = = = = = − = − + + = + + = = + + + = + + − − FLUID MECHANICS (PRE – FINAL S4) March 15, 2017 Name ____________________________________ 5. A piece of wood of S = 0.651 is 8 cm square and 150 cm long. How many kilograms of lead weighing 11,200 kg/m3 must be fastened at one end of the stick so that it will float upright with 30.5 cm out of water? k g 6 2 5 .9 3 W m 0 .0 5 6 1 0 ,2 0 0 6 2 4 .9 6 - 1 1 9 5 V 1 0 0 0 V 1 1 9 5 1 1 ,2 0 0 V 6 2 4 .9 6 BF BF W W 1 0 0 0 V BF2 k g 1 1 9 5 1 0 0 0 ) 3 0 5 . 0 5 . 1 ( BF 1 1 ,2 0 0 V W k g 6 2 4 .9 6 ) 6 5 1 ( 9 6 . 0 W m 9 6 . 0 ) 5 . 1 )( 0 8 . 0 ( 0 8 . 0 V m k g 6 5 1 ) 1 0 0 0 ( 6 5 1 . 0 ρ 6 5 1 . 0 S 2 3 2 2 2 2 1 2 1 2 1 2 2 1 3 1 3 wood = = = + = + + = + = = − = = = = = = = = = 6. A barge is loaded with 150 Metric tons of coal. The weight of the empty barge in air is 35 Metric ton. If the barge is 5.5 m wide, 16 m long and 3 m high, what is its draft. (Depth below the water surface) m 1 . 2 h h ) 16 )( 5 . 5 ( 1000 ) 1000 )( 35 150 ( Vs ρ W 1 W BF W 2 = = + = + = 7. A prismatic object 20 cm thick by 20 cm wide by 40 cm long is weighed in water at a depth of 50 cm and found to weigh 50 N. What is its weight in air and its specific gravity?
• 63. KW 751 22, GP m N 147,198 T 60,000 TN 2 BP KW 27,746.2 ) 90 . 0 )( 9 . 126 )( 81 . 9 ( 76 . 24 h Q' e BP m 126.9 4.9 122 h head dynamic total h sec m 76 . 24 2(9.81)4.9 91x0.91) 0.61(5)(0. Q' flow Actual Q C Q' gates 5 for flow l theoretica 5(Av) Q velocity l theoretica 2gh v flow the producing head m 9 . 4 4 . 70 3 . 75 h t Turbine t t 3 d = − =  = = =  = = + = − = = → = → = → = → = − = 3 1 2 . 1 5 0 - 2 0 6 .9 6 2 0 6 .9 6 S N ew to n 2 0 6 .9 6 W 5 0 1 5 6 .9 6 W ) 0 1 6 . 0 ( 8 1 0 , 9 W W W W 1 V 9 8 1 0 1 W W W V 9 8 1 0 W W W W 1 0 0 0 g V W N ew to n g m W W W W 1 0 0 0 V m S V m ρ m 0 .0 1 6 ) 4 . 0 )( 2 0 . 0 )( 2 0 . 0 ( V N ew to n 5 0 W W 1 1 2 1 2 1 1 2 1 1 1 1 2 1 1 1 1 1 1 2 1 1 1 1 1 1 1 3 1 2 1 = = = + = = − − = − = − = = − = = = = = = = 8. A ship, with vertical sides near the water line, weighs 3630 Metric ton and draws 6.7 m in saltwater (S = 1.025). Discharge of 181 Metric ton of water ballast decreases the draft to 6.4 m. What would be the draft d of the ship in fresh water (S = 1) 2 . Eq ) 025 . 1 )( 4 . 6 ( 3449 L A ) 1000 )( 025 . 1 )( 4 . 6 ( L A ) 1000 )( 181 630 , 3 ( BF W 1 . Eq ) 025 . 1 ( 7 . 6 630 , 3 L A ) 1000 )( 025 . 1 )( 7 . 6 ( L A ) 1000 ( 3630 BF W 2 2 2 2 1 1 1 1 → = = − = → = = = In a hydroelectric power plant, the water surface on the crest of the dam is at elevation 75.3 m while the water surface just at the outlet of the head gate is at elevation 70.4 m. The head gate has 5 gates of 0.91 m x 0.91 m leading to the penstock and are fully opened. Assume 61% as coefficient of discharge, determine a. The quantity of water that enters the hydraulic turbine in m3 /sec b. The KW power that the turbine will developed, assuming eturbine = 90% efficiency and the turbine is 122 m below the entrance of the penstock c. The brake torque for a speed N = 1800 rpm d. The number of generator poles if f = 60 Hertz e. The electrical power developed by the generator if electrical and windage loses amounts to 18%
• 64. FINAL EXAM (March 25, 2017) SET 2 NAME _____________________________________ 1. An object weighs 25.95 N when submerged in kerosene (S = 0.81) and weighs 26.6 N when submerged in gasoline (S = 0.68). Determine the specific weight of the object. 3 m KN 116 . 61 ) 81 . 9 ( 23 . 6 = =  2. Determine the water power and mechanical efficiency of a centrifugal pump which has an input of 3 KW. If the pump has a 203 mm diameter suction line and a 152 mm diameter discharge line and handles 10 L/sec of water at 66C ( =980 kg/m3 ;  = 9.6 KN/m3 ). The suction line gauge shows 102 mm Hg vacuum and the discharge gauge shows 180 KPa. The Discharge gauge is located 61 cm above the center of the discharge pipeline and the pump inlet and discharge lines are at the same elevation. BP 3 KW D1 0.203 m D2 0.152 m Q 0.01 m^3/sec A1 0.032 m^2 A2 0.018 m^2 v1 0.309 m/sec v2 0.551 m/sec g 9.810 m/sec^2 v1^2/2g 0.005 m v2^2/2g 0.015 m SW 9.6 KN/m^3 P1 -13.60 Kpa P2 180 KPa z2 0.61 m z1 0 m (P2-P1)/sw 20.17 m (v2^2- v1^2)/2g 0.01 m (z2-z1) 0.61 m HL 0.00 m Ht 20.79 m WP 1.996 KW em 66.52 % 3. A piece of wood of S = 0.651 is 8 cm square and 150 cm long. How many kilograms of lead weighing 11,200 kg/m3 must be fastened at one end of the stick so that it will float upright with 30.5 cm out of water? k g 6 2 5 .9 3 W m 0 .0 5 6 1 0 ,2 0 0 6 2 4 .9 6 - 1 1 9 5 V 1 0 0 0 V 1 1 9 5 1 1 ,2 0 0 V 6 2 4 .9 6 BF BF W W 1 0 0 0 V BF2 k g 1 1 9 5 1 0 0 0 ) 3 0 5 . 0 5 . 1 ( BF 1 1 ,2 0 0 V W k g 6 2 4 .9 6 ) 6 5 1 ( 9 6 . 0 W m 9 6 . 0 ) 5 . 1 )( 0 8 . 0 ( 0 8 . 0 V m k g 6 5 1 ) 1 0 0 0 ( 6 5 1 . 0 ρ 6 5 1 . 0 S 2 3 2 2 2 2 1 2 1 2 1 2 2 1 3 1 3 wood = = = + = + + = + = = − = = = = = = = = =
• 65. 4. At one point in a pipeline the water speed is 3 m/sec and the gage pressure is 50 KPa. Find the gage pressure at a second point in the line, 11 m lower than the first, if the pipe diameter at the second point is twice at the first. FLUID MECHANICS (PRE – FINAL S2) March 04, 2017 Name ____________________________________ 1. The diameters of the suction and discharge pipe of a fuel pump for a day tank of a gasoline engine are 150 mm and 100 mm, respectively. The discharge pressure gauge located 10 m above the pump centerline reads 320 KPa and the suction pressure gauge which is 4 m below the pump centerline reads a vacuum of 60 KPa. Head losses due to pipe friction, turbulence and fittings amounts to 15 m. If gasoline with a relative density S = 0.75 is pumped at the rate of 35 L/sec, find a. The total dynamic head developed by the pump b. The fluid power in KW c. The brake or shaft power delivered to the fluid for a pump efficiency of 75% d. The brake torque if the pump speed is 1200 RPM e. The electrical power input to the pump motor for a motor efficiency of 92% f. The line current drawn by the motor if the motor is 3 – phase, 240 volts, and 0.9 Power Factor g. The cost of power for 5 hours operation, if electricity costs P 1.50/KW-hr Q 0.035 m^3/sec S 0.75 SW(water) 9.81 KN/m^3 SW 7.3575 KN/m^3 g 9.81 m/sec^2 d1 0.150 m d2 0.100 m P1 -60.00 KPa P2 320.00 KPa z1 -4 m z2 10 m HL 15 m FP 75 KW A1 0.018 m^2 A2 0.008 m^2 v1 2.0 m/sec v2 4.5 m/sec v1^2/2g 0.20 m v2^2/2g 1.01 m P1/sw -8.15 m P2/sw 43.49 m D(Phead) 51.65 m D(vhead) 0.81 m D(Zhead) 14 m ht 81.46 m Water Power 20.98 KW Pump Efficiency 75.00 % Brake Power 27.97 KW Motor Efficiency 92.00 % Motor Power 30.40 KW E (Volts) 220.00 Volts Power Factor 0.90 Phase 3.00 Phase N(RPM) 1200.00 RPM Torque(N-m) 222.57 N-m Line Current 88.65 Amperes PowerCost 1.50 Pesos/KW-hr Cost 228.01 Pesos
• 66. 2. A jet of liquid is directly vertically upward. At A (Nozzle tip) its diameter is 75 mm and its velocity is 10 m/sec. Neglecting air friction, determine its diameter at a point 4 m above A. d1 0.075 m v1 10 m/sec P1 0 Kpa P2 0 KPa g 9.81 m/sec^2 A1 0.0044 m^2 Q 0.0442 m^3/sec SW 9.81 KN/m^3 P1/SW 0 m P2/SW 0 m v1^2/2g 5.097 m z1 0 m z2 4 m HL 0 m v2^2/2g 9.097 m v2 13.36 m/sec A2 0.0033 m^2 d2 0.065 m d2 6.5 cm 3. A closed vessel contains water up to a height of 2 m and over the water surface there is air having a pressure of 8.829 N/cm2 above atmospheric pressure. At the bottom of the vessel there is an orifice of diameter 15 cm. Find the rate of flow of water from orifice if Cd = 0.6. Example No. 15 h 2 m P1 88.29 Kpa g 9.81 m/sec^2 SW 9.81 KN/m^3 P1/SW 9.0 m d(orifice) 0.15 m A(Orifice) 0.018 m^2 Theoretical Velocity 14.69 m/sec Cd 0.60 Q(m^3/sec) 0.1558 m^3/sec Q(L/sec) 155.8 L/sec
• 67. A Francis turbine is installed with a vertical draft tube. The pressure gauge located at the penstock leading to the turbine casing reads 372.6 KPa and velocity of water at inlet is 6 m/sec. The discharge is 2.5 m3 /sec. The hydraulic efficiency is 85%, and the overall efficiency is 82%. The top of the draft tube is 1.5 m below the centerline of the spiral casing, while the tailrace level is 2.5 m from the top of the draft tube. There is no velocity of whirl at the top or bottom of the draft tube and leakage losses are negligible. Calculate, a) the net effective head in meters (43.817 m) b) the brake power in kw. (881.2 kw) c) the plant output for a generator efficiency of 92%. (810.7 kw) d) the mechanical efficiency (96.550) GIVEN: P1 = 372.6 KPa v = 6 m/sec Q = 2.5 m3 /sec eh = 0.85 e = 0.82 ZB = (1.5 + 2.5) = 4 m % 5 . 9 6 9 6 5 . 0 8 5 . 0 8 2 . 0 e ) 1 )( 8 5 . 0 ( e 8 2 . 0 e e e e KW 7 . 8 1 0 ) 8 8 1 .2 (0 .9 2 OUTPUT PLANT KW 2 . 8 8 1 2 ) 1 0 7 4 .6 (0 .8 BP k w 6 . 1 0 7 4 ) 8 1 7 . 4 3 )( 8 1 . 9 ( 5 . 2 WP h γ Q WP m 8 1 7 . 4 3 h 4 ) 8 1 . 9 ( 2 ) 6 ( 8 1 . 9 6 . 3 7 2 h Z g 2 v γ P h e) (n eg lig ib l 0 h 0 Z 0 v 0 P h h Z g 2 v γ P Z g 2 v γ P 2 ) p o in t th ro u g h lin e (d atu m 2 to b p o in t at m m v h m 2 A 2 A A 2 A L 2 2 2 2 A L 2 2 2 2 A 2 A A = = = = = = = = = = = = = + + = + + = = = = + + + + = + + − = − Example no. 16 A 4 m3 /hr pump delivers water to a pressure tank. At the start, the gauge reads 138 KPa until it reads 276 KPa and then the pump was shut off. The volume of the tank is 160 Liters. At 276 KPa, the water occupied 2/3 of the tank volume. a) Determine the volume of water that can be taken out until the gauge reads 138 KPa. b) If 1 m3 /hr of water is constantly used, in how many minutes from 138 KPa will the pump run until the gauge reads 276 KPa.
• 68. meters 2gD fLv h 2 f = sec 8 4 . 3 6 min 6 1 4 . 0 h rs 0 .0 1 0 2 3 t 0 3 0 7 . 0 t 1 t 4 Liters 7 . 3 0 0 3 0 7 . 0 0 7 6 . 0 1 0 6 7 . 0 o u t V tak en Liters 7 . 1 0 6 1 0 6 7 . 0 ) 1 6 . 0 ( 3 2 2 7 6 @ V w Liters 7 6 0 7 6 . 0 0 .0 8 4 ) - (0 .1 6 0 V w @ 1 3 8 m3 0 8 4 . 0 V V P V P m3 0 5 3 3 . 0 ) 1 6 0 . 0 ( 3 1 V K Pa 3 7 7 .3 2 5 1 0 1 .3 2 5 2 7 6 P K Pa 3 2 5 . 2 3 9 3 2 5 . 1 0 1 1 3 8 P 1 2 2 1 1 2 2 1 = = = = − = = − = = = = = = = = = = = = + = = + = HEAD LOSSES HL = Major loss + Minor losses Major Loss: Head loss due to friction and turbulence in pipes Minor Losses: Minor losses include, losses due to valves and fittings, enlargement, contraction, pipe entrance and pipe exit. Minor losses are most easily obtained in terms of equivalent length of pipe "Le".the advantage of this approach is that both pipe and fittings are expressed in terms of "Equivalent Length" of pipe of the same relative roughness. Considering Major loss only Darcy-Weisbach Equation Considering Major and Minor losses where; f - friction factor from Moody's Chart L - length of pipe, m Le - equivalent length in straight pipe of valves and fittings, m V - velocity, m/sec D - pipe inside diameter, m g - gravitational acceleration, m/sec2 0 REYNOLD'S NUMBER: Reynold's Number is a non dimensional one which combines the physical quantities which describes the flow either Laminar or Turbulent flow. The friction loss in a pipeline is also dependent upon this dimensionless factor. where;  - absolute or dynamic viscosity, Pa-sec  - kinematic viscosity, m2 /sec For a Reynold's Number of less 2100 flow is said to Laminar For a Reynold's Number of greater than 3000 the flow is Turbulent FRICTION FACTOR: Moody's Chart f /D NR  - absolute roughness D - inside diameter /D - relative roughness VALUES OF ABSOLUTE ROUGHNESS  FOR NEW PIPES Type of Material Feet Millimeter Drawn tubing, brass, lead, glass centrifugally spun cement, bituminous lining, transite 0.000005 0.0015 Commercial Steel, Wrought iron 0.00015 0.046 Welded steel pipe 0.00015 0.046 Asphalt-dipped cast iron 0.0004 0.12 Galvanized iron 0.0005 0.15 Cast iron, average 0.00085 0.25 Wood stave 0.0006 to 0.18 to 0.003 0.9 meters 2gD )v e L f(L h 2 f  + = vD vD NR ν μ ρ = =