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Rotating machines part 1
1.
027 Electrical &
Electronics Principles Pg 1 8.0A ROTATING MACHINES 8.A.0 Torque A torque is defined as the turning moment of a force and its SI Unit is Newton metre (Nm). The torque about a point is equal to the force times the distance measured perpendicular to the force. Examples Torque in a couple Ex 4: t = F x L sinθ © Seychelles Institutes of Technology Prepared by Mr. Dinesh
2.
027 Electrical &
Electronics Principles Pg 2 8.0A ROTATING MACHINES 8.A.1 Torque on a coil in a magnetic field Fig 1 Fig 1 (a) is the field between two magnets, (b) the field due to a current in a straight wire and (c) the resulting field if they are put together. This last field is known as the "catapult" field because it tends to catapult the wire out of the field in the direction shown by the arrow. Fig 2 Fig 2(a) show If a coil carrying a current is placed in a magnetic field it will experience a force on two of its sides in such a way as to make the coil rotate. Fig 2(b) shows coil will rotate from the 'double catapult' field diagram. Since the current moves along the two opposite sides of the coil in opposite directions the two sides receive a force in opposite directions also, thus turning the coil. © Seychelles Institutes of Technology Prepared by Mr. Dinesh
3.
027 Electrical &
Electronics Principles Pg 3 8.0A ROTATING MACHINES 8.A.2 Torque on a current carrying conductor Consider a rectangular coil in Fig 3 when the plane of the coil is lying along the field lines (θ = 90o) Where, τ = torque (Nm) L = length (m) W = width (m) B = magnetic field (Wb) I = current (A) Fig 3 τ = F x perpendicular distance τ = F x W = BIL x W = BI(LW) τ = BIA © Seychelles Institutes of Technology Prepared by Mr. Dinesh
4.
027 Electrical &
Electronics Principles Pg 4 8.0A ROTATING MACHINES Consider a number of rectangular coil in fig 4 when the plane of the coil is rotating along the field lines Where, τ = torque (Nm) L = length (m) W = width (m) d = perpendicular distance (m) θ = angle (°) B = magnetic field (Wb) I = current (A) N = number of turns Fig 4 τ = F x perpendicular distance τ = F x d = BILN x Wsinθ = BI(LW)Nsinθ τ = BIANsinθ © Seychelles Institutes of Technology Prepared by Mr. Dinesh
5.
027 Electrical &
Electronics Principles Pg 5 8.0A ROTATING MACHINES 8.A.3 Solved work examples on “torque on a current carrying conductor” Work example 1 Calculate the torque needed to hold a coil of 20 turns and area 8cm2 at an angle of 60o to a magnetic field of flux density 0.1T if the coil carries a current of 0.5 A. Solution Torque, τ = BANIsinθ = 0.1x20x8x10-4sin 60° τ = 8x10-4 Nm Work example 2 a) What is the maximum torque on a 150-turn square loop of wire 18.0 cm on a side that carries a 50.0A current in a 1.60-T field? b) What is the torque when θ is 10.9°? Solution a) Maximum torque, τ = BANIsinθ = 1.60x150x0.0182 x50.0sin 90° τ = b) When θ is 10.9°, τ = BANIsinθ = 1.60x150x0.182 x50.0sin 10.9° τ = © Seychelles Institutes of Technology Prepared by Mr. Dinesh
6.
027 Electrical &
Electronics Principles Pg 6 8.0A ROTATING MACHINES 8.A.4 E.m.f. induced in a straight conductor in a uniform magnetic field When a straight conductor is moved through a magnetic field an e.m.f. is induced between its ends. This movement must be in such a direction that the conductor cuts through the lines of magnetic flux, and will be a maximum when it moves at right angles to the field (Fig 5). Where, E = e.m.f. induced (V) L = length (m) B = magnetic field (Wb) v = velocity (m/s) Fig 5 If the conductor moves with velocity v at right angles to the field then the flux cut per second will be BvL (since the conductor will sweep out an area vL every second). But the rate of cutting flux is equal to the e.m.f. induced in the conductor. Therefore E = BLv © Seychelles Institutes of Technology Prepared by Mr. Dinesh
7.
027 Electrical &
Electronics Principles Pg 7 8.0A ROTATING MACHINES If the conductor cuts through the flux at an angle θ ( in Fig 5), where θ is the angle between the magnetic field and the direction of motion, the equation becomes E = BLv sinθ 8.A.5 Solved work examples on Emf induced on a straight conductor Work example 1 Calculate a) The e.m.f. generated between the wing tips of an aircraft that is flying horizontally at 200 ms-1 in a region where the vertical component of the Earth's magnetic field is 4.0x10-5 T, if the aircraft has a wingspan of 25 m. b) What will be the e.m.f. if the θ = 30°. Solution a) Maximum e.m.f., E = BLvsinθ = 4.0x10-5x25x200 sin 90° E = 0.2V b) When θ is 30°, E = BANIsinθ = 4.0x10-5x150x200 sin 30° E = 0.1V © Seychelles Institutes of Technology Prepared by Mr. Dinesh
8.
027 Electrical &
Electronics Principles Pg 8 8.0A ROTATING MACHINES 8.A.4 Emf induced in a straight conductor rotating in a uniform magnetic field Consider the Fig 6, a diagram of an electrical generator. A generator with a single rectangular coil rotated at constant angular velocity in a uniform magnetic field produces an e.m.f. that varies sinusoidally in time. Note the generator is similar to a motor, except the shaft is rotated to produce a current rather than the other way around. Where, E = e.m.f. induced (V) L = length (m) B = magnetic field (Wb) v = velocity (m/s) θ = angle (°) ω = angular velocity (rad/s) Fig 6 The induced e.m.f. by considering only the side wires. Motional e.m.f. is given to be E = BLv, where the velocity v is perpendicular to the magnetic field B Here, the velocity is at an angle θ with B, so that its component perpendicular to B is vsinθ. Thus in this case the e.m.f. induced on each side is E = BLvsinθ, and they are in the same direction. The total EMF ε around the loop is then: E = 2BLvsinθ This expression is valid, but it does not give EMF as a function of time. © Seychelles Institutes of Technology Prepared by Mr. Dinesh
9.
027 Electrical &
Electronics Principles Pg 9 8.0A ROTATING MACHINES To find the time dependence of EMF, we assume the coil rotates at a constant angular velocity ω. The angle θ is related to angular velocity by θ = ωt, so that: E = 2Blvsinωt Now, linear velocity v is related to angular velocity by v = rω. Here r = w/2, so that v = (w/2)ω, and: E = 2BLw2ωsinωt E = (Lw)Bωsinωt Noting that the area of the loop is A = Lw, and allowing for N loops, we find that: E = NABω sinωt The maximum value of the e.m.f (Emax) is when θ (= ωt) = 90o, is given by E = NABω © Seychelles Institutes of Technology Prepared by Mr. Dinesh
10.
027 Electrical &
Electronics Principles Pg 10 8.0A ROTATING MACHINES 8.A.5 Solved work examples on Emf induced in a rotating in a uniform magnetic field Work example 1 Calculate a) The maximum value of the e.m.f generated in a coil with 200 turns and of area 10 cm2 rotating at 60 radians per second in a field of flux density 0.1 T. b) The e.m.f. when θ = 60o Solution a) Maximum e.m.f., E = BANωsinθ = 0.1x10-5x200x60 sin 90° E = 1.2V b) When θ is 30°, E = BANωsinθ = 0.1x10-5x200x60 sin 60° E = 1.04V Notice the use of radians per second. © Seychelles Institutes of Technology Prepared by Mr. Dinesh
11.
027 Electrical &
Electronics Principles Pg 11 8.0A ROTATING MACHINES 8.A.6 Power generated in rotating machines The linear motion of a conductor through a magnetic field, the mechanical power supplied is equal to the electrical power generated. 8.1.7 Derivation of power in rotating machines For a rotating object, the linear distance covered at the circumference of rotation is the product of the radius with the angle covered. That is: Linear distance = radius × angular distance. By definition, Linear distance = linear speed × time = radius × angular speed × time. By the definition of torque: Torque = radius × force. These two values can be substituted into the definition of power: Power = Force × Linear distance Time = (Torque⁄Radius)× (Radius × Angular speed) Time = Torque × Angular speed or Power = Torque × 2π × Rotational speed Where, Torque, τ (N/m) Angular speed, ω (rad/s) Rotational speed, n (rev/s) Radius, r (m) © Seychelles Institutes of Technology Prepared by Mr. Dinesh
12.
027 Electrical &
Electronics Principles Pg 12 8.0A ROTATING MACHINES Rotational power is given by, P = ω. τ Where, ω, angular velocity is defined as the rate of change of angular displacement which specifies the angular speed (rotational speed) of an object and the axis about which the object is rotating. Measured in radian per second (rad/s) τ, torque is defined as the moment of force is the tendency of a force to rotate an object about an axis. Measured in newton per metre (N/m) P, In rotational systems, power is defined as the product of the torque and angular velocity. Measured in watts (W) 8.A.8 Power efficiency Power efficiency is defined as the ratio of output power divided by input power. Ƞ = 100% · Where, ƞ = efficiency (%) Pout = output power (W) Pin = input power (W) 퐏퐨퐮퐭 퐏퐢퐧 © Seychelles Institutes of Technology Prepared by Mr. Dinesh
13.
027 Electrical &
Electronics Principles Pg 13 8.0A ROTATING MACHINES 8.A.9 Solved work examples on power in rotating machines Work example 1 A machine is rotating with speed of 3000 rev/min (rpm) consumes 5kW. Calculate the torque of the shaft. Solution Given, ƞ = 3000 rpm P = 5kW τ = ? Using τ = 퐏/ퟐ훑ƞ = (5 × 1000) ÷ [(2π × 3000 60 )] τ = 15.9 N/m Work example 2 An electric motor has an input power consumption of 50 watts. The motor was activated for 60 seconds and produced work of 2970 joules. Find the efficiency of motor? Solution Given, Pin = 50W t = 60s E = 2970J ƞ = ? Pout = ? Using Pout = E / t = 2970/60 Pout = 49.45W Now ƞ = ퟏퟎퟎ% × 퐏퐨퐮퐭 / 퐏퐢퐧 = ퟏퟎퟎ × (ퟒퟗ. ퟗퟓ/ퟓퟎ) ƞ = ퟗퟗ% © Seychelles Institutes of Technology Prepared by Mr. Dinesh
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