YF_ISM_20

YF_ISM_20 - THE SECOND LAW OF THERMODYNAMICS 20 20.1....

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20-1 T HE S ECOND L AW OF T HERMODYNAMICS 20.1. IDENTIFY: For a heat engine, HC . WQ Q =− H . W e Q = H 0, Q > C 0. Q < SET UP: 2200 J. W = C 4300 J. Q = EXECUTE: (a) 6500 J. QWQ =+ = (b) 2200 J 0.34 34%. 6500 J e == = EVALUATE: Since the engine operates on a cycle, the net Q equal the net W . But to calculate the efficiency we use the heat energy input, H . Q 20.2. IDENTIFY: For a heat engine, . H . W e Q = H 0, Q > C 0. Q < SET UP: H 9000 J. Q = C 6400 J. Q = EXECUTE: (a) 9000 J 6400 J 2600 J. W =−= (b) H 2600 J 0.29 29%. 9000 J W e Q = = EVALUATE: Since the engine operates on a cycle, the net Q equal the net W . But to calculate the efficiency we use the heat energy input, H . Q 20.3. IDENTIFY and SET UP: The problem deals with a heat engine. 3700 W W =+ and H 16,100 J. Q Use Eq.(20.4) to calculate the efficiency e and Eq.(20.2) to calculate C . Q Power / . Wt = EXECUTE: (a) H work output 3700 J 0.23 23%. heat energy input 16,100 J W e Q = = = (b) WQQ Q == − Heat discarded is CH 16,100 J 3700 J 12,400 J. QQ W = = (c) H Q is supplied by burning fuel; Hc Qm L = where c L is the heat of combustion. H 4 c 16,100 J 0.350 g. 4.60 10 J/g Q m L = × (d) 3700 J W = per cycle In 1.00 s t = the engine goes through 60.0 cycles. / 60.0(3700 J)/1.00 s 222 kW PWt = 5 (2.22 10 W)(1 hp/746 W) 298 hp P = EVALUATE: C 12,400 J. Q In one cycle tot C H 3700 J. Q =+= This equals tot W for one cycle. 20.4. IDENTIFY: . H . W e Q = H 0, Q > C 0. Q < SET UP: For 1.00 s, 3 180 10 J. W EXECUTE: (a) 3 5 H 180 10 J 6.43 10 J. 0.280 W Q e × = × (b) 55 5 6.43 10 J 1.80 10 J 4.63 10 J. W −× EVALUATE: Of the 5 6.43 10 J × of heat energy supplied to the engine each second, 5 1.80 10 J × is converted to mechanical work and the remaining 5 4.63 10 J × is discarded into the low temperature reservoir. 20
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20-2 Chapter 20 20.5. IDENTIFY: HC . WQ Q =− H . W e Q = H 0, Q > C 0. Q < Dividing by t gives equivalent equations for the rate of heat flows and power output. SET UP: / 330 MW. Wt = H / 1300 MW. Qt = EXECUTE: (a) HH /3 3 0 M W 0.25 25%. /1 3 0 0 M W WW t e QQ t == = = = (b) CH W so / / / 1300 MW 330 MW 970 MW. QtQtW t = = EVALUATE: The equations for e and W have the same form when written in terms of power output and rate of heat flow. 20.6. IDENTIFY: Apply 1 1 1. e r γ C H Q e Q SET UP: In part (b), H 10,000 J. Q = The heat discarded is C . Q EXECUTE: (a) 0.40 1 1 0.594 59.4%. 9.50 e = = (b) (1 ) (10,000 J)(1 0.594) 4060 J. e = = EVALUATE: The work output of the engine is 10,000 J 4060 J 5940 J =−= = 20.7. IDENTIFY: 1 1 e r SET UP: 1.40 = and 0.650. e = EXECUTE: 1 1 10 . 3 5 0 . e r 0.40 1 0.350 r = and 13.8. r = EVALUATE: e increases when r increases. 20.8. IDENTIFY: 1 1 γ er SET UP: r is the compression ratio. EXECUTE: (a) 0.40 1 (8.8) 0.581, e = which rounds to 58%. (b) 0.40 1 (9.6) 0.595 e = an increase of 1.4%. EVALUATE: An increase in r gives an increase in e . 20.9. IDENTIFY and SET UP: For the refrigerator 2.10 K = and 4 C 3.4 10 J. Q =+ × Use Eq.(20.9) to calculate W and then Eq.(20.2) to calculate H . Q (a) EXECUTE: Performance coefficient C / KQW = (Eq.20.9) 44 C / 3.40 10 J/2.10 1.62 10 J WQK × = × (b) SET UP: The operation of the device is illustrated in Figure 20.9 EXECUTE: QWQ 4 H 1.62 10 J 3.40 10 J 5.02 10 J Q × × × (negative because heat goes out of the system) Figure 20.9 EVALUATE . QW Q The heat H Q delivered to the high temperature reservoir is greater than the heat taken in from the low temperature reservoir.
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This note was uploaded on 03/14/2012 for the course MAE 162D taught by Professor Shaefer during the Spring '11 term at UCLA.

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YF_ISM_20 - THE SECOND LAW OF THERMODYNAMICS 20 20.1....

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