THERMODYNAMICS, FLUID AND PLANT PROCESSES. The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial.
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1 THERMODYNAMICS, FLUID AND PLANT PROCESSES The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial. THERMODYNAMICS TUTORIAL 2 THERMODYNAMIC PRINCIPLES SAE SOLUTIONS SELF ASSESSMENT EXERCISE No kg/s of steam flows in a pipe 40 mm bore at 200 bar pressure and 400oC. i. Look up the specific volume of the steam and determine the mean velocity in the pipe. ii. Determine the kinetic energy being transported per second. iii. Determine the enthalpy being transported per second. A πd 2 /4 π x (0.04) 2 / m 2 p 200bar θ 400 o C V m v m /s u V/A / m/s KE mu 2 /2 1 x /2 1. W h 2819 kj/kg (from the tables) SELF ASSESSMENT EXERCISE No.2 1. The shaft of a steam turbine produces 600 Nm torque at 0 rev/s. Calculate the work transfer rate from the steam. SP 2πNT 2π x 0 x W 2. A car engine produces 0 kw of power at 000 rev/min. Calculate the torque produced. T P/2πN 0000/(2π x 0) 9. Nm SELF ASSESSMENT EXERCISE No. 1. A non-flow system receives 80 kj of heat transfer and loses 20 kj as work transfer. What is the change in the internal energy of the fluid? U Q + W 80 + (-20) 60 kj 2. A non-flow system receives 100 kj of heat transfer and also 40 kj of work is transferred to it. What is the change in the internal energy of the fluid? U Q + W kj. A steady flow system receives 00 kw of heat and loses 200 kw of work. What is the net change in the energy of the fluid flowing through it? Φ + P E/s 00 + (-200) 00 kw E/s 4. A steady flow system loses 2 kw of heat also loses 4 kw of work. What is the net change in the energy of the fluid flowing through it? Φ + P E/s -2 + (-4) -6kW E/s
2 . A steady flow system loses kw of heat also loses 20 kw of work. The fluid flows through the system at a steady rate of 70 kg/s. The velocity at inlet is 20 m/s and at outlet it is 10 m/s. The inlet is 20 m above the outlet. Calculate the following. i. The change in K.E./s (-10. kw) ii. The change in P.E/s (-1.7 kw) iii. The change in enthalpy/s (1.2 kw) Φ - kw KE/s (m/2)( v v 2 1 ) P -20 kw KE/s (70/2)( ) -10. kw m 70 kg/s v 1 20 m/s PE/s mg(z 2 - z 1 ) v 2 10 m/s PE/s 70 x 9.81(0-20) kw z 1 20 m z 2 0 m Φ + P H/s + KE/s + PE/s (-) + (-20) H/s + (-10.) + (-1.74) H/s 1.24 kw SELF ASSESSMENT EXERCISE No.4 1. Gas is contained inside a cylinder fitted with a piston. The gas is at 20oC and has a mass of 20 g. The gas is compressed with a mean force of 80 N which moves the piston 0 mm. At the same time Joules of heat transfer occurs out of the gas. Calculate the following. i. The work done.(4 J) ii. The change in internal energy. (-1 J) iii. The final temperature. (19.9 o C) Take cv as 718 J/kg K Q + W U W F x 80 x J U -1 J U m c v (θ 2 20) 0.02 x 718 x (θ 2 20) θ o C 2. A steady flow air compressor draws in air at 20oC and compresses it to 120oC at outlet. The mass flow rate is 0.7 kg/s. At the same time, kw of heat is transferred into the system. Calculate the following. i. The change in enthalpy per second. (70. kw) ii. The work transfer rate. (6. kw) Take cp as 100 J/kg K. Φ + P H/s (ignore PE and KE) H/s m c p T 0.7 x 1.00 (120 20) 70. kw + P 70. P 6. kw. A steady flow boiler is supplied with water at 1 kg/s, 100 bar pressure and 200oC. The water is heated and turned into steam. This leaves at 1 kg/s, 100 bar and 00oC. Using your steam tables, find the following. i. The specific enthalpy of the water entering. (86 kj/kg) ii. The specific enthalpy of the steam leaving. (7 kj/kg) iii. The heat transfer rate. (7.7 kw) H/s 1 (h 2 h 1 ) 1(7 86) 77 kw Φ + P H/s P 0 Φ 7.7 MW
3 4. A pump delivers 0 dm/min of water from an inlet pressure of 100 kpa to an outlet pressure of MPa. There is no measurable rise in temperature. Ignoring K.E. and P.E, calculate the work transfer rate. (2.42 kw) Φ + P H/s + KE/s + PE/s KE/s PE/s 0 Φ 0 P H/s FE + U U 0 P FE p 2 V 2 p 1 V 1 V 1 V 2 0 x 10 - m /s P 0 x 10 - ( x x 10 ) 144 kw. A water pump delivers 10 dm/minute (0.1 m/min) drawing it in at 100 kpa and delivering it at 00 kpa. Assuming that only flow energy changes occur, calculate the power supplied to the pump. (860 W) Φ + P E/s FE/s Φ 0 P FE/s (0.1/60)(00 100) x kw 6. A steam condenser is supplied with 2 kg/s of steam at 0.07 bar and dryness fraction 0.9. The steam is condensed into saturated water at outlet. Determine the following. i. The specific enthalpies at inlet and outlet. (21 kj/kg and 16 kj/kg) ii. The heat transfer rate. (46 kw) h 1 h f + x h fg at 0.07 bar h (2409) 21.1 kj/kg h 2 h f 16 kj/kg Φ m(h 2 - h 1 ) 2( ) kj kg/s of gas is heated at constant pressure in a steady flow system from 10oC to 180oC. Calculate the heat transfer rate Φ. (7.4 kw) C p 1.1 kj/kg K Q 0.2 x 1.1 (180 10) 7.4 kj kg of gas is cooled from 120oC to 0oC at constant volume in a closed system. Calculate the heat transfer. (-16.8 kj) Cv 0.8 kj/kg. Q 0. x 0.8 (0 120) kj
4 SELF ASSESSMENT EXERCISE No. 1. A vapour is expanded from 12 bar and 0 cm to 10 cm and the resulting pressure is 6 bar. Calculate the index of compression n. (0.6) p 1 12 bar p 2 6 bar V 1 0 cm V 2 10 cm 12 x 0 n 6 x 10 n 12/6 2 (10/0) n n 2 n ln2 n ln n ln2/ ln a.A gas is compressed from 200 kpa and 00 cm to 800 kpa by the law pv1.4c. Calculate the new volume. (111.4 cm ) 200 x V V 2 00(200/800) 1/ cm 2.b.The gas was at 0oC before compression. Calculate the new temperature using the gas law pv/t C. (207 o C) T2 p 2 V 2 T 1 /(p 1 V 1 ) 800x x 2/(200x 00) K or 207 o C.a. A gas is expanded from 2 MPa and 0 cm to 10 cm by the law pv1.2 C. Calculate the new pressure. (06 kpa) 2 x p 2 x p 2 06 kpa.b. The temperature was 00oC before expansion. Calculate the final temperature. (14 o C) T2 p 2 V 2 T 1 /(p 1 V 1 ) 0.06 x 10 x 77/(2 x 0) 86.7 K or 14 o C SELF ASSESSMENT EXERCISE No A gas is expanded from 1 MPa and 1000oC to 100 kpa. Calculate the final temperature when the process is i. Isothermal (n1) (1000 o C) ii Polytropic (n1.2) (94 o C) iii. Adiabatic (γ 1.4) (86 o C) iv. Polytropic (n 1.6) (264 o C) p 1 1 MPa p MPa T K ISOTHERMAL POLYTROPIC ADIABATIC POLYTROPIC T 2 T o C or 127 k T 2 T 1 (p 2 / p 1 ) 1-1/n 127(0.1) K or 94 o C T 2 T 1 (p 2 / p 1 ) 1-1/γ 127(0.1) K or 86 o C T 2 T 1 (p 2 / p 1 ) 1-1/n 127(0.1) K or 264 o C
5 2. A gas is compressed from 120 kpa and 1oC to 800 kpa. Calculate the final temperature when the process is i. Isothermal (n1) (1 o C) ii. Polytropic (n1.) (17 o C) iii Adiabatic (γ1.4) (222 o C) iv. Polytropic (n 1.) (269 o C) p kpa p kpa T K ISOTHERMAL POLYTROPIC ADIABATIC POLYTROPIC T 2 T K o C or 1 o C T 2 T 1 (p 2 / p 1 ) 1-1/n 288(800/120) K or 17 o C T 2 T 1 (p 2 / p 1 ) 1-1/γ 288(800/120) K or 222 o C T 2 T 1 (p 2 / p 1 ) 1-1/n 288(800/120) K or 269 o C. A gas is compressed from 200 kpa and 20oC to 1.1 MPa by the law pv1.c. The mass is 0.02 kg. cp100 J/kg K. c v 718 J/kg K. Calculate the following. i. The final temperature. (44 K) ii. The change in internal energy (2.0 kj) iii. The change in enthalpy (2.84 kj) p kpa p MPa T 1 29 K T 2 T 1 (p 2 / p 1 ) 1-1/n 29(1.1/0.2) K U 0.02 x ( ) kj H 0.02 x 1.00( ) 2.88 kj 4. A gas is expanded from 900 kpa and 1200oC to 120 kpa by the law pv1.4 C. The mass is 0.01 kg. cp1100 J/kg K cv 70 J/kg K Calculate the following. i. The final temperature. ii. The change in internal energy iii. The change in enthalpy p kpa p kpa T K T 2 T 1 (p 2 / p 1 ) 1-1/n 147(120/900) K U 0.01 x 0.7 ( ) -7.2 kj H 0.01 x 1.1( ) kj
6 SELF ASSESSMENT EXERCISE No.7 1. kg/s of steam is expanded in a turbine from 10 bar and 200oC to 1. bar by the law pv1.2c. Determine the following. i. The initial and final volumes. (0.618 m and m ) ii. The dryness fraction after expansion. (0.86) iii. The initial and final enthalpies. (2829 kj/kg and 288 kj/kg) iv. The change in enthalpy kw) p 1 10 bar p 2 1. bar T 1 47 K θ oc V 1 m v 1 x m 10 x (V 2 ) 1.2 V 2 m v 2 / 1 m /kg V 2 x m v g x /( x 1.19) 0.86 h kj/kg h (2226) 288 kj/kg H ( ) MW kg/s of steam is expanded from 70 bar and 40oC to 0.0 bar by the law pv1. C. Determine the following. i. The initial and final volumes. (0.066 m /kg and 17.4 m /kg) ii. The dryness fraction after expansion. (0.411) iii. The initial and final enthalpies. (287 kj/kg and 11 kj/kg) iv. The change in enthalpy. (-228 kw) p 1 70 bar p bar θ 1 40 oc v m /kg 70 x (V 2 ) 1. v 2 m /kg x 28.2 x h kj/kg h (242) 11.9 kj/kg H 1. ( ) kw. A horizontal cylindrical vessel is divided into two sections each 1m volume, by a nonconducting piston. One section contains steam of dryness fraction 0. at a pressure of 1 bar, while the other contains air at the same pressure and temperature as the steam. Heat is transferred to the steam very slowly until its pressure reaches 2 bar. Assume that the compression of the air is adiabatic (γ1.4) and neglect the effect of friction between the piston and cylinder. Calculate the following. i. The final volume of the steam. ii. The mass of the steam. iii. The initial internal energy of the steam. iv The final dryness fraction of the steam. v. The final internal energy of the steam. vi. The heat added to the steam. t s at 1 bar 99.6 o C so the initial air temperature is 99.6 o C AIR Adiabatic compression T 2 T 1 (p 2 / p 1 ) 1-1/γ 72.6(2/1) K m pv/rt 1 x 10 x 1/(287 x 72.6) 0.9 kg U 0.96 x 718 x( ) 4.86 J Q + W U Q 0 hence W 4.86 J p 1 V 1 /T 1 p 2 V 2 /T 2 V m V 0.9 m p V p V 2x10 x x10 x 1 W J γ 1 0.4
7 STEAM V m W kj Q U W u 1 u f + x u fg at 1 bar ( ) kj/kg V 1 1 m x m v g at 1 bar m 1/(0. x 1.694) 1.97 kg V x m v g at 2 bar x 1.9/(0.886 x 1.97) 0.8 u 2 u f + x u fg at 2 bar ( ) kj/kg Q kj SELF ASSESSMENT EXERCISE No g of steam at 10 bar and 0oC expands reversibly in a closed system to 2 bar by the law pv1.c. Calculate the following. i. The initial volume. ( m ) ii. The final volume. ( m ) iii. The work done. (-2.92 kj) p 1 10 bar p 2 2 bar θ 1 0 oc m 10 g From the tables v m /kg V x m p 1 V 1 p2 V 2 10 x V 2 V m p2v2 p1v1 2x10 x x 10 x W 2921J g of gas at 20oC and 1 bar pressure is compressed to 9 bar by the law pv1.4 C. Taking the gas constant R 287 J/kg K calculate the work done. (Note that for a compression process the work will turn out to be positive if you correctly identify the initial and final conditions). (.67 kj) m 20 g T 1 29 K p 1 1 bar p 2 9 bar p 1 V p2 V /1.4 p2 T2 T p 1 mr( T1 T ) 0.02 x n W () 49 K ( ).68 kj. Gas at 600 kpa and 0.0 dm is expanded reversibly to 100 kpa by the law pv1. C. Calculate the work done. V m p kpa p kpa 1/ x 0.0 p 1 V 1 p2 V 2 V m 100 p V2 p1v1 100 x10 x x 10 x W J
8 4. 1 g of gas is compressed isothermally from 100 kpa and 20oC to 1 MPa pressure. The gas constant is 287 J/kg K. Calculate the work done. (2.9 kj) m 0.01 kg T 1 29 K p kpa p 2 1 MPa p 1 V 1 p 2 V 2 W p V ln(p 2 / p 1 ) m R T ln(p 2 / p 1 ) 0.01 x 287 x 29 x ln(10) 2.9 kj. Steam at 10 bar with a volume of 80 cm is expanded reversibly to 1 bar by the law pvc. Calculate the work done. ( kj) V 1 80 cm p 1 10 bar p 2 1 bar p 1 V 1 p 2 V 2 V 2 p 1 V 1 / p cm W p V ln(v 1 / V 2 ) 10 x 10 x 80 x 10-6 ln(0.1) J 6. Gas fills a cylinder fitted with a frictionless piston. The initial pressure and volume are 40 MPa and 0.0 dm respectively. The gas expands reversibly and polytropically to 0. MPa and 1 dm respectively. Calculate the index of expansion and the work done. (1.46 and -.24 kj) p 1 40 MPa V dm p 2 0. MPa V 2 1 dm p 1 V 1 n p 2 V 2 n (40/0.) (1/0.0) n n n x10 p V2 p1v1 W x 1 x x 10 x 0.0 x KJ 7. An air compressor commences compression when the cylinder contains 12 g at a pressure is 1.01 bar and the temperature is 20 o C. The compression is completed when the pressure is 7 bar and the temperature 90 o C. (1.124 and 1944 J) The characteristic gas constant R is 287 J/kg K. Assuming the process is reversible and polytropic, calculate the index of compression and the work done. p bar T 1 29 K p 2 7 bar T 2 6 K T T 1 1/n 1 1/n 2 p /n p /n ln(1.29)/ln n mr(t T 1) x 287(6 29) W J
9 SELF ASSESSMENT EXERCISE No.9 Take Cv 718 J/kg K and R 287 J/kg K throughout dm of gas at 100 kpa and 20oC is compressed to 1.2 MPa reversibly by the law pv1.2 C. Calculate the following. i. The final volume. (0.126 dm ) ii. The work transfer. (27 J) iii. The final temperature. (170oC) iv. The mass. (1.189 g) v. The change in internal energy. (128 J) vi. The heat transfer. (-128 J) V 1 1 dm T 1 29 K p kpa p MPa r 1/ x 1 p 1 V 1 p2 V 2 V dm x p V p V 1.2 x10 x x10 10 x 10 W J p2v2 1.2 x 10 x x 29 T2 T1 44 K p1v1 10 x 1 m pv/rt 10 x 10 - /(287 x 29) kg U m c v T x 718 x (44-29) 128 J Q + W U Q kj kg of gas at 20 bar and 1100oC is expanded reversibly to 2 bar by the law pv1. C in a closed system. Calculate the following. i. The initial volume. (9.8 dm ) ii. The final volume. (8 dm ) iii. The work transfer. (-27 kj) iv. The change in internal energy. (-20. kj) v. The heat transfer. (6.7 kj) p 1 20 bar T 1 17 K p 2 2 bar m 0.0 kg pv m R T V x 287 x 17/20 x m 1/ x 10 x p 1 V 1 p2 V 2 V m 2 x 10 p V p V 2 x10 x x 10 x W kj 0. p2v2 T2 T1 807 K p1v1 U m c v T 0.0 x 718 x (807-17) -20. kj Q + W U Q kj
10 kg of air at 700 kpa and 800oC is expanded adiabatically to 100 kpa in a closed system. Taking γ 1.4 calculate the following. i. The final temperature. (61.4 K) ii. The work transfer. (26. kj) iii. The change in internal energy. (-26. J) m 0.08 kg T K p kpa p kpa 1 1/ γ p2 100 T2 T K p m R T 0.08 x 287(61 107) W kj γ U m c v T 0.08 x 718(61 107) kj 4. A horizontal cylinder is fitted with a frictionless piston and its movement is restrained by a spring as shown. a. The spring force is directly proportional to movement such that F/ x k Show that the change in pressure is directly proportional to the change in volume such that p/ V k/a 2 b. The air is initially at a pressure and temperature of 100 kpa and 00 K respectively. Calculate the initial volume such that when the air is heated, the pressure volume graph is a straight line that extends to the origin. (0. dm ) c. The air is heated making the volume three times the original value. Calculate the following. i. The mass. (0.8 g) ii. The final pressure. (00 kpa) iii. The final temperature. (2700 K) iv. The work done. (-200 kj) v. The change in internal energy. (917 J) vi. The heat transfer. (1.12 kj) p 1 1 bar T 1 00 K V 1 1 m p 2 bar p 2 bar V 2 x m V m W F x/2 A p x/2 V p/2 2 x 10 x 0.4/2 40 kj out of system W - 40 kj p2v2 x 1.4 x 00 T2 T1 p1v1 1x K m pv/rt 1 x 10 x 1/(287 x 00) kg U m c v T x 718( ) 800kJ Q kj
11 SELF ASSESSMENT EXERCISE No kg of dry saturated steam at 10 bar pressure is expanded reversibly in a closed system to 1 bar by the law pv1.2 C. Calculate the following. i. The initial volume. (8.9 dm ) ii. The final volume. (264 dm ) iii. The work transfer. (-62 kj) iv. The dryness fraction. (0.779) v. The change in internal energy. (-108 kj) vi. The heat transfer. (-46 kj) m 0.2 kg dry saturated steam p 1 10 bar p 2 1 bar V 1 m v g 0.2 x m x 10 x p 1 V 1 p2 V 2 V m 1x 10 p V p V 1 x10 x x 10 x W kj 0.2 V 2 m x v g at 1 bar x(1.694) x u 1 u g at 10 bar 284 kj/kg K U 1 m u x kj U 2 m u x kj U kj Q kj 2. Steam at 1 bar and 20 o C is expanded reversibly in a closed system to bar. At this pressure the steam is just dry saturated. For a mass of 1 kg calculate the following. i. The final volume. ii. The change in internal energy. iii. The work done. iv. The heat transfer. p 1 1 bar θ 1 20 o C p 2 bar 1/1.2 v m /kg v 2 v g at bar m /kg u kj/kg u 2 u g at bar 262 kj/kg U ( ) -1 kj/kg n n p1 V p 2 V ( 2.466) n n ln()/ln(2.466) p2v2 p1v1 x10 x x 10 x 0.12 W 187 kj Q U - W kj/kg
THERMODYNAMICS, FLUID AND PLANT PROCESSES. The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial.
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