Heat Engine Problem

Gloy to Gool Q. A special with heat 3 Engine thoonal is to veservon be designed at 1000K, while iteracting 500K produces 2500 KJ from of work by receiving reseguri at 1000 K reversible If mode this Engine 700K and The 4360 F5 of heat energy the performal to be operated is between the other much happens heat interaction occurs two theomal reserved , 700K 1000k Q2 YQ, = 4300 KJ W -2500k = H.E Q3 500 K W By energy balance W Q1 + Q2 = Q3 W Q, tQ2 - Q3 Qit Q2 - Q3 W = NHE = Heat input Q,+Q2 scale In T. t temp T2-T3 = 1000 1000 +700 +700-500 NHE = T,+T2 1200 = 1700 NHE=0.705 = NHE = W PitQ2 = N Q,++2 N.HE

QitQ2 = 2500 0.705 Q,+Q2 3542 KJ Q2 = 3542 Q1 Q2 = 3542-4300 Q2= 7 758 KJ - Now, Q,+Q2 = WtQ3 4300 + -758 = 2500+ Q3 Q3 4300-758-2509 - Q3= 1042 W. Now, Checking W- Q,+Q2 - Q3 4300 + -758 -1042 = W= 2500KJ Q. cold stovage is maistened +35°C at -5°C the while the A is kept at to the heat surrounding leakage estimated surrousding to be Cop from the 29 KW cold storage Actual COP if the Actual Cop bhw of the is Find the is frd of the Ideal Calculat the the plant working temp. Also plant power same required to drive the

Cover data NO to 36°C 5 = COR = 35-(5) IT x 12-35°C If 5 T2 35 = Cop- -TI RE 268 72 Cop= = Q2 = 308-20 268 T1- T2 Q1-Q2 (cop) 6.7 2.2 Actual ideal Cop = 16COP) as ideal 3x67 i Now / power required to KW drive imp plant; Actual cop, = 2.2 29 Q = IN Cop = W W= 29 = 13 KW 2.2 I (Cop), Grar Qr True W2 (doring o Address