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Unformatted text preview: Solution Two reversible heat engines operate in series between a source at 527 °C and a sink at
17°C. If the engines have equal efﬁciencies and the ﬁrst rejects 400 U to the second, calculate: (i) the temperature at which heat is supplied to the second engine; (ii) the heat taken from the source;
(iii) the work done by each engine. Assume that each engine operates on the Carnot cycle. (1) Let the temperature at which heat is supplied to the second engine be T, then 1  (T/800) : 1  (29o/T) ie T = [(290x800) = 481.7'K (ii) 208.7 °C Let the heat from the source be 0. then 1 * (290/481.7) = 1 ‘ (400/0) ie 0 = 400x482/29O = 664.4 RJ (iii) For the first engine: W i 400  664.4 3 * 264.4 kJ ie work output 264.4 kJ For the second engine:  W/400 = 1  (290/481.7) w = ‘ 159.2 kJ ie work output = 159.2 kJ , SE”! 3% Asst/Viki 94 W W! 5.6 A fourcylinder petrol engine has a swept volume of 2000 cm 3, and the clearance volume
in each cylinder is 60 cm". Calculate the air standard cycle efﬁciency. If the introduction
conditions are 1 bar and 24 °C, and the maximum cycle temperature is 1400 °C, calculate
the mean effective pressure based on the air standard cycle. Sohﬁbn Swept volume per cylinder : 2000/4
= 500 cm3
Total volume = 500 + 60 = 560 cm3 560/60 = 9.333 ie compression ratio
Then,
cycle efficiency = 1 — (1/9.333)°4 = 0.591 or 59.1% \/ 53 Now,
T2 = 297X(9.333)°4 = 725.7 K
\ heat supplied = CV(T3 ~ T2)
2 0.718(1673 — 725.7)
2 680.2 kJ/kg
Then, net work output = 6802x0.591 = 402.0 kJ/kg
The mean effective pressure is given by the area on the p~v diagram divided by the length of the diagram.
ie mean effective pressure, pm = Q§§_work,bgtgut
v1 — V2
Then,
V1 = RTi/px = 287x297/1x105 = 0.852 m3/kg
and, V2 1 v1/9.333 = 0.852/9.333 0.091 m3/kg Therefore, pm = 402x1000 = 5.28 bar W 2 “Cr” (We—'6 > + Cram)
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 Spring '09
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