fundamentals of thermal fluid sciences

1. UNIVERSITY OF ENGINEERING AND
TECHNOLOGY LAHORE
MECHATRONICS & CONTROL ENGINEERING
DEPARTMENT
MCT-114 Fundamentals of Thermal Sciences
Assignment # 2 (HW-2)
Submitted To: Ms. Aisha Shoaib
Name:
MUHAMMAD ANEES
Regd. No.:
2021-MC-03
Section:
A
Date of Submission:
14/04/2022

2. HOMEWORK # 2
2–14C What are the ordinary and absolute temperature scales in the SI and the English
system?
In the SI system, the ordinary temperature scale is the Celsius scale, and the absolute temperature
scale is the Kelvin scale.
In the English system, the ordinary temperature scale is the Fahrenheit scale and the absolute
temperature scale is the Rankine scale.
2–15C Consider an alcohol and a mercury thermometer that read exactly 0°C at the ice point
and 100°C at the steam point. The distance between the two points is divided into 100 equal
parts in both thermometers. Do you think these thermometers will give exactly the same
reading at a temperature of, say, 60°C? Explain.
The two thermometers will have different readings at the same temperature.
Reason:
The operation of these two thermometers is based on the thermal expansion of a fluid. Both
mercury and alcohol have different thermal expansion. If both fluids have a thermal expansion
coefficient that vary linearly with temperature, then they will always show the same reading.
Otherwise, the readings will deviate. The expansion of liquids is not linear and there is no way the
expansion curves of two substances will be equal. Therefore, the temperature reading on a mercury
thermometer will differ from the reading on the alcohol thermometer although the difference is
usually small.
2–16C Consider two closed systems A and B. System A contains 3000 kJ of thermal energy
at 20°C, whereas systemB contains 200 kJ of thermal energy at 50°C. Now the systems are
brought into contact with each other. Determine the direction of any heat transfer between
the two systems.
When two systems with different temperature are in contact the thermal energy will always transfer
from the hotter one to the colder one. This case is the same, thermal energy will travel from the
system B to the system A until they reach equilibrium.
The fact that the system A has more thermal energy does not change the thermal energy flow, for
example that could just mean that the system A is a lot bigger and therefore has more thermal
energy than system B.
2–17 The deep body temperature of a healthy person is 37°C. What is it in kelvins?
We can obtain the deep body temperature in kelvins by converting the temperature from Celsius
scale to Kelvin scale using the following relation:
Result
T(K) = T(°C) + 273.15 = 37 + 273.15 = 310.15 K
310.15 K
Note:

3. The normal body temperature in Fahrenheit is 98.6° F which is equal to 37° C. Fahrenheit can be
converted to Kelvin by first changing it to Celsius then adding 273.15.
2–19 The temperature of a systemrises by 70°C during a heating process. Express this rise
in temperature in kelvins.
Given:
T = 70 °C
SOLUTION:
The conversion of T(°C) to T(K) is shown below:
T(K) = T(°C) + 273.15
So,
T = (70 + 273.15) K = 343.15 K
T = 343.15 K
The conversion of T(°C) to T(F) is shown below:
T (°F) = (T (°C) × 9/5) + 32
So,
T (°F) = (70 × 9/5) + 32 = 158 °F
T = 158 °F
The conversion of T(°C) to T(R) is shown below:
T (R) = (T (°C) × 9/5) + 491.67
So,
T (R) = (70 × 9/5) + 491.67 = 617.67 R
T = 617.67 R
2–22E The temperature of a system drops by 45°F during a cooling process. Express this
drop in temperature in K, R, and °C.
given:
Temperature in degree Fahrenheit = 45°F
 To convert this temperature into degree Celsius, we use the conversion factor:
°C= (45°F -32) x 5/9
Putting values in above equation, we get:
°C= (45°F -32) x 5/9
T (°C) =7.22°C

4. Temperature in degree Celsius = 7.22°C
 To convert this temperature into Kelvins, we use the conversion factor:
(T(K)-273.15) = (T(°F) – 32) x 5/9
Putting values in above equation, we get:
T(K) = (45°F -32) x 5/9 + 273.15
T(K) = 280.37K
Temperature in Kelvins = 280.37 K
 To convert this temperature into Rankine, we use the conversion factor:
T(R) = T(°F) + 459.67
Putting values in above equation, we get:
T(R) = T(°F) + 459.67
T(R) = 504.67 R
Temperature in Rankine = 504.67 R
Hence, the temperature drop in K is 280.37 K, in degree Celsius is 7.22°C and in Rankine is 504.67
R
2-23C Explain why some people experience nose bleeding and some others experience
shortness of breath at high elevations.
As we move high up the altitude, the atmospheric pressure goes on decreasing. Our blood also
exerts pressure on the walls of the blood vessels through which it flows, which is commonly known
as the blood pressure. At sea level, both these pressures balance each other. But at high altitudes,
due to a reduction in atmospheric pressure, the pressure of the blood becomes more thereby causing
a rupturing of the blood vessels lining the nasal passage. This is what leads to bleeding. As far as
shortness of breath is concerned, it may be due to the low oxygen content in the upper layers of
the atmosphere.
2–24C A health magazine reported that physicians measured 100 adults’ blood pressure
using two different arm positions: parallel to the body (along the side) and perpendicular to
the body (straight out). Readings in the parallel position were up to 10 percent higher than
those in the perpendicular position, regardless of whether the patient was standing, sitting,
or lying down. Explain the possible cause for the difference.
When a person spreading his arms in the parallel to the body the blood vessels are more restricted
when the arm is perpendicular to the body.
In parallel position of arms, the pressure exerted by the heart to pump the blood will be more to
overcome the increased resistance to flow compared to the perpendicular position of arms. It is
invariant of the position of a person whether sitting, lying or standing.

5. 2–25C Someone claims that the absolute pressure in a liquid of constant density doubles
when the depth is doubled. Do you agree? Explain.
Not quite. The gauge pressure doubles, since that is directly proportional to depth, but the added
pressure of the atmosphere is constant, so it will not double, so the absolute pressure will always
be off by 14.7 psi when the depth is doubled (if the depth is h and absolute pressure is p then for a
depth of 2h the absolute pressure is 2p - 14.7 psi).
2–26C Express Pascal’s law, and give a real-world example of it.
Pascal's principle states that the pressure applied to a confined fluid increases the pressure
throughout by the same amount. This is a consequence of the pressure in a fluid remaining constant
in the horizontal direction. A real-world example of Pascal's principle is the operation of the
hydraulic car jack.
Hydraulic Brakes:
One of the most common examples of Pascal's Law is the hydraulic braking system present in the
automobiles. Every time you see a car come to a halt; the principle of Pascal's Law comes into
action. A number of components form the braking system in cars.
2–27C Consider two identical fans, one at sea level and the other on top of a high mountain,
running at identical speeds. How would you compare (a) the volume flow rates and (b) the
mass flow rates of these two fans?
Fans work by pushing the volume of air from one place to another. With the elevation change the
volume does not change and the volume flow rate will not change either.
The density does change with the height Density is defined as a ratio of mass and volume that mass
occupies:
P= V
We have already stated that the volume stays the same, so if the density p reduces with height the
mass m also must reduce. That means that the high-altitude fan will push less mass in a time
interval and therefor the mass flow rate will be lesser at the high mountain than at the see level.
2–28 A vacuum gage connected to a chamber reads 35 kPa at a location where the
atmospheric pressure is 92 kPa. Determine the absolute pressure in the chamber.
GIVEN DATA:
Pgage = 35 kPa
Patm = 92 kPa
FORMULA:
Pabs = Patm- Pgage
SOLUTION:
Pabs = 92 kPa-35 kPa

6. Pabs = 57kPa
2–33 The water in a tank is pressurized by air, and the pressure is measured by
a multifluid manometer as shown in Fig. P2–33. Determine the gage pressure of
air in the tank if h1 = 0.2 m, h2 = 0.3 m, and h3 = 0.4 m. Take the densities of
water, oil, and mercury to be 1000 kg/m3, 850 kg/m3, and 13,600 kg/m3,
respectively.
Given:
h1 = 0.1 m
h2 = 0.2 m
h3 =0.4 m
The density of water = 1000 kg/m³
The density of oil = 850 kg/m³
The density of mercury = 13,600 kg/m³
To Find:
Pressure of Air Pair =?
Solution:
⇒ Pair + ρgh1 + ρgh2 = Patm + ρgh3
Pair = Patm + ρgh3 - ρgh2 - ρgh1
Substituting the values, (taking g= 10m/s²)
Pair =101325+53312−1666−980
= 151991 Pa
Hence, the air pressure in the tank is 151.99kPa.
2–36 The gage pressure in a liquid at a depth of 3 m is read to be 42 kPa. Determine the gage
pressure in the same liquid at a depth of 9 m.
Given:
Depth 1=h1=3m
gage pressure at depth 1= 42kPa
Depth 2=h2=9m
Solution:
Total pressure in liquid is given by,
P=Pa+ρgh
Where,
Pa= atmospheric pressure
ρ= Density of liquid
g= Gravitational acceleration
h= depth
Gage pressure is given by,

7. Pgage=P−Pa=ρgh
Gage pressure at depth 1 is given by,
Pgage1=ρgh1….(eq.i)
Gage pressure at depth 2 is given by,
Pgage2=ρgh2….(eq.ii)
Taking ration of equaion (i) and (ii),
Pgage1/Pgage2=ρgh1/ρgh2
Pgage1/Pgage2=h1/h2
Pgage1/Pgage2=3/9
Pgage2=42×9/3
Pgage2=126kPa
Gage pressure at depth of 9m=126kPa
2–37 The absolute pressure in water at a depth of 9 m is read to be 185 kPa. Determine (a)
the local atmospheric pressure, and (b) the absolute pressure at a depth of 5 m in a liquid
whose specific gravity is 0.85 at the same location.
The absolute pressure is the sum of the atmospheric and gauge pressure this can be expressed as;
Pabs = Pgage + Patm ……………………….1
Part A
The local atmospheric pressure can be calculated by making the atmospheric pressure in equation
1 the subject formula;
Pabs - Pgage = Patm ………………………2
the gauge pressure can be obtained with the expression;
Pgage = ρgh ………………….3
where ρ is the density of water = 1000 kg/ m3;
g is the acceleration due to gravity = 9.81 m/ s2;
h is the depth = 9 m
Therefore,
Pgage = 1000 kg/ m3 x 9.81 m/s2 x 9 m
Pgage = 88290 Pa
Now we take our value for the gauge pressure and substitute in equation 2;
Given,
Pgage = 88290 Pa
= 185 kPa = 185000 Pa (1000 Pa = 1 kPa)

8. Patm =185000Pa-82590Pa
Patm =96710Pa
The local atmospheric pressure is 96.71 kPa
Part B
To obtain the gauge pressure at a depth of 5 m in the liquid, we have to calculate the density of the
liquid;
Specific gravity (s.p) = Density of liquid/ Density of water
making the density of the liquid the subject formula we have;
The density of the liquid = s.p x Density of water
= 0.85 x 1000 kg/m3 = 850 kg/ m3
the density of the liquid is substituted into the equation shown below;
Pgage = ρ1 x g x h
where ρ1 is the density of the liquid
h is the depth = 5 m
g is the acceleration due to gravity.
Pgage = 850 kg/m3 x 9.81 m/s2 x 5
Pgage = 41692.5 Pa
The absolute temperature is calculated using equation 1
since the liquid is at the same location as the water, the atmospheric temperature is the same;
Pabs = Pgage + Patm ……………………….1
Pabs = 96710Pa + 41692.5Pa
Pabs = 138402.5 = 138.4 kPa
Therefore, the absolute pressure is 138.4 kPa
2–41 The barometer of a mountain hiker reads 750 mbars at the beginning of a hiking trip
and 650 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration,
determine the vertical distance climbed. Assume an average air density of 1.20 kg/m3.
A mountain hiker records the barometric reading before and after a hiking trip. The vertical
distance climbed is to be determined.
Assumptions:
The variation of air density and the gravitational acceleration with altitude is negligible.
Properties:

9. The density of air is given to be  = 1.20 kg/m3`.
Analysis:
Taking an air column between the top and the bottom of the mountain and writing a force balance
per unit base area, we obtain,
Wair/A = Pbottom - Ptop
(gh) air = Pbottom - Ptop
(0.750 - 0.650) (1 bar/ 100,000Nm-2) (1 N /1 kg/m2) (1.20) (9.81)
It yields h = 850 m
which is also the distance climbed.
2–44 The hydraulic lift in a car repair shop has an output diameter of 30 cm and is to lift
cars up to 2000 kg. Determine the fluid gage pressure that must be maintained in the
reservoir.
Assumptions
The weight of the piston of the lift is negligible.
Analysis
Pressure is force per unit area, and thus the gage pressure required is simply the ratio of the weight
of the car to the area of the lift,
Pgage = W/A = mg/ D2 / 4
Pgage = (1 kN /1000 kg/m2) x 4 x 2000 x 9.81/(0.3)2
Pgage = 277.57kPa
Discussion
Note that the pressure level in the reservoir can be reduced by using a piston with a larger area.
2–47 Both a gage and a manometer are attached to a gas tank to measure its pressure. If the
reading on the pressure gage is 80 kPa, determine the distance between the two fluid levels
of the manometer if the fluid is (a) mercury (r 5 13,600 kg/m3) or (b) water (r 5 1000 kg/m3).
The gage pressure and the vertical distance between the two fluid levels can be related
through pgage=gh
h={pgage}/{g}
Mercury: To find the distance between the two fluid levels, we will substitute values into our
equation.
h={pgage}/{g}
={80kPa}/ {(13600kg/m3) (9.81m/s2)} (1kN/m2/1kPa) (1000kg/m2s2/1kN) =(13600kg/m3)
(9.81m/s2)80kPa(1kPa1kN/m2) (1kN1000kg/m2⋅s2)
=0.60m
b) Water:
h={pgage}/{g}
=(1000kg/m3) (9.81m/s2)80kPa(1kPa1kN/m2) (1kN1000kg/m2⋅s2)
=8.15m

10. 2–49 A manometer containing oil (ρ=850 kg/m3) is attached to a tank filled with air. If the oil
level difference between the two columns is 80 cm and the atmospheric pressure is 98 kPa,
determine the absolute pressure of the air in the tank.
To find the absolute pressure we need to find the pressure in the opposite arm, and that will be
summed with the given atmospheric pressure.
Illustration is attached.
So,
Pabs = Pₐ + ρgh
Substituting the values,
Pabs = 98 kPa + (850 × 10 × 0.80)
Pabs = 98000 + 6800
Pabs = 16600
Pabs= 1.66 × 10⁵ Pa
2–50 A mercury manometer (r 5 13,600 kg/m3) is connected to an air duct to measure the
pressure inside. The difference in the manometer levels is 15 mm, and the atmospheric
pressure is 100 kPa. (a) Judging from Fig. P2–50, determine if the pressure in the duct is
above or below the atmospheric pressure. (b) Determine the absolute pressure in the duct.
a) From the picture we can conclude that the pressure inside of the air duct is higher than the
atmospheric pressure. We know that because the mercury height at the air duct side of the
manometer is lower than on the side with the free atmosphere, which means that the pressure is
higher on the air duct side.
b) The pressure in a system is
P = Patm+ ρg (y2-y1)
Where Patm is the atmospheric pressure in the open part of the tube, ρ is the liquid density and (y2-
y1) is the height difference between the two sides of the tube.
Let's reduce to SI units
(y2-Y1) = 15mm (1m / 1000mm) = 15 x 10-3m
Patm = 100 KPa = 100 x 103Pa
Let's calculate the pressure
P = Patm + 13.6 x 9.8 x 15 x 10⁻ ³
P = 100k+ 2.00k
P = 102kPa
2–52E The pressure in a natural gas pipeline is measured by the manometer shown in Fig.
P2–52E with one of the arms open to the atmosphere where the local atmospheric pressure
is 14.2 psia. Determine the absolute pressure in the pipeline.

11. The given parameters:
atmospheric pressure, P₀ = 14.2 Psia (lb/in² atm)
the height of the pipeline, h = 26 in
density of water, ρ = 62.4 lbm/ft³
To find:
 the absolute pressure in the pipeline
The absolute pressure in the pipeline is calculated as follows:
Pabs=Po+ ρgh
Where;
g is gravitational acceleration
h is height=26in
note: 1ft = 12in
pressure (psi) = c0nstant x (lb/gallon)
(lb/gallon) = lb/231in3
pressure (psi) = lb/231in3 x 12in
pressure (psi) = 12lb/231in2
pressure (psi) = 0.05(lb/ in2)
constant = 0.052
To use this constant;
 density must be in lb/gallon and,
 height must be in ft
density, ρ=62.4 x 1/7.48 = 8.342 lb/gallon
height, h= 26/12=2.16 ft
The absolute pressure is calculated as:
Pabs=Po + (constant x lb/gallon x feet)
Pabs=14.2 + (0.052 x 8.342 x 2.167 )
Pabs=14.2 + 0.94
Pabs=15.14 psi
Thus, the absolute pressure in the pipeline is 15.14 psi
2–56 Determine the pressure exerted on a diver at 45 m below the free surface of the sea.
Assume a barometric pressure of 101 kPa and a specific gravity of 1.03 for seawater.

12. We need to know the absolute pressure exerting on a diver at 45 m below the sea surface. Knowing
that the atmospheric pressure equals 101 kPa we could use the following equation to get the
absolute pressure:
P = Patm + p g h
As seen in Eq. (1) to get the pressure we should firstly know the seawater density. Knowing the
specific gravity of the seawater equals 1.03 we could obtain the density using the following
relation:
P = SC * Pwater (2)
Calculations
Using Eq. (2) to get the density of seawater P = SG * Pwater = 1.03 * 1000(Kg/m3)
=1030 Kg/m'
Substituting into Eq. (1) to get the absolute pressure exerted on the diver:
P = Patm + p g h
= 101 * 103(Pa) 1030(Kg/m3) * 9.81(m/s2) * 45(m)
= 555693.5 Pa
=555.7 kPa
2-57 Consider a Utube whose arms are open to the atmosphere. Now water is poured into
the Utube from one arm, and light oil (r 5 790 kg/m3) from the other. One arm contains
70cmhigh water, while the other arm contains both fluids with an oil water height ratio of 4.
Determine the height of each fluid in that arm.
Assumption:
Both water and oil are incompressible substances.
Properties:
The density of oil is given to be ρ = 790 kg/m3. We take the density of water to be ρ =1000 kg/m3.
Analysis:
The height of water column in the left arm of the manometer is given to be hw1 = 0.70 m. We let
the height of water and oil in the right arm to be hw2 and ha, respectively. Then, ha = 4hw2. Noting
that both arms are open to the atmosphere, the pressure at the bottom of the U-tube can be
expressed as
Pbottom =Patm+ρwghw1 and Pbottom= Patm + ρwghw2 + ρagha
Setting them equal to each other and simplifying,
ρwghw1= ρwghw2+ ρagha
ρwhw1= ρwhw2+ ρaha

13. hw1= hw2+(ρa/ρw) ha
Noting that ha = 4hw2, the water and oil column heights in the second arm are determined to be
0.7 m =hw2+4 hw2 (790/1000)
hw2= 0.168 m
0.7 m = 0.168m +(790/1000) ha
ha = 0.673
Adding both ha and hw2 we get:
Height=0.168+0.673
Height=0.841m=84.1cm
2–65 A hydraulic lift is to be used to lift a 2500 kg weight by putting a weight of 25 kg on a
piston with a diameter of 10 cm. Determine the diameter of the piston on which the weight
is to be placed.
m1= 2500kg
m2=25kg
d1=?
d2=10cm=0.1m
Calculation:
F1/A1 = F2/A2
m1g/ r1
2 = m2g/r2
2
m1/r1
2 = m2/r2
2
r1
2 = m1/m2 x r2
2
r1
2 = 0.25
r1 = 0.5
d1 = 1m = 100cm
2–69 The average temperature of the atmosphere in the world is approximated as a function
of altitude by the relation Tatm 5 288.15 2 6.5z where Tatm is the temperature of the
atmosphere in K and z is the altitude in km with z 5 0 at sea level. Determine the average
temperature of the atmosphere outside an airplane that is cruising at an altitude of 12,000
m.
Analysis:
Using the relation given, the average temperature of the atmosphere at an altitude of 12,000 m is
determined to be

14. Tatm = 288.15 -6.5z
= 288.15-6.5 x 12
= 210.15 K= - 63°C
Discussion:
This is the “average” temperature. The actual temperature at different times can be different.
2–73 A vertical piston–cylinder device contains a gas at a pressure of 100 kPa. The piston
has a mass of 5 kg and a diameter of 12 cm. Pressure of the gas is to be increased by placing
some weights on the piston. Determine the local atmospheric pressure and the mass of the
weights that will double the pressure of the gas inside the cylinder.
Given data
Pi = 100 kPa
mpiston = 5 Kg
D = 12 cm
We need to Know the atmospheric pressure and the mass should be added to double the gas
pressure.
Analysis
Considering the piston cylinder which has a mass of 5 Kg contains a gas under pressure. Since we
know the absolute pressure of the gas, we could get the atmospheric pressure by subtracting the
pressure exerted by the piston mass.
Patm + Ppiston = P gas
Patm = P gas - Ppiston
= P gas – mpg/r2
= 100kPa - 4.3kPa
= 95.67kPa
If the pressure is double then
Pdouble = Patm + g(mp+mx)/ r2
200K = 95.67k+866(5kg+mx)
104400/866 – 5 = mx
mx = 115.55kg

15. 2–74 An air conditioning system requires a 35mlong section of 15cm diameter duct work to
be laid underwater. Determine the upward force the water will exert on the duct. Take the
densities of air and water to be 1.3 kg/m3 and 1000 kg/m3, respectively.
Given Data
H = 35m
D=15cm=0.15m
Density of air= 1.33kgm-3
Density of water= 1000 kgm-3
To find
Upward force=?
Calculations:
V=AL
=(D2/4) L
= [  (0.15 m)2 /4] (35 m) = 0.6185 m3
Then the buoyancy force becomes
Fb= ρgV= (1000 kg/m3) (9.81 m/s2) (0.6185 m3) (1 kN /1000 kg m/s2)
Fb = 6.067kN
2–79 The lower half of a 6mhigh cylindrical container is filled with water (r = 1000 kg/m3)
and the upper half with oil that has a specific gravity of 0.85. Determine the pressure
difference between the top and bottom of cylinder.
Properties
The density of water is given to be r = 1000 kg/m3. The specific gravity of oil is given to be 0.85.
Analysis
The density of the oil is obtained by multiplying its specific gravity by the density of water,
  SGxρH2O= (0.85) (1000 kg/m3) = 850 kg/m3
The pressure difference between the top and the bottom of the cylinder is the sum of the pressure
differences across the two fluids,
Ptotal Poil Pwater
= (850 kg/m3) (9.81 m/s2) (3 m) + (1000 kg/m3) (9.81 m/s2) (3 m) (1 kPa/1000 N/m2)
= 54.4 kPa

16. 2–82 A glass tube is attached to a water pipe, as shown in Fig. P2–82. If the water pressure
at the bottom of the tube is 110 kPa and the local atmospheric pressure is 99 kPa, determine
how high the water will rise in the tube, in m. Take the density of water to be 1000 kg/m3.
Pbottom=110kPa
Patm = 99kPa
h=?
=1000 kg/m3
Calculations:
Pbottom = Patm+ gh
10000 = 99000 + (1000x9.8xh)
h = 11000/98000= 1.12m
Liquid rises up to 1.12m in the column due to change in pressure (hydrostatic pressure).S