Determine the heat transfer (in J mol^-1) when water is heated in steady flow pr
ID: 475808 • Letter: D
Question
Determine the heat transfer (in J mol^-1) when water is heated in steady flow process from 25 500 degree C at atmospheric pressure. Show that the entropy change of an ideal gas, with a gas constant R and constant heat capacity undergoing a mechanical reversible work in a closed system from (T_a, P_a) to (T, P) is given expression:^3 [Delta S = integral_r^r C_r^s dT/T - R ln [P/P_0]] One kg of air (M_a = 29 g mol^-1, C_p = 21 J mol^-1 K^-4) is compressed from (25 degree C, 10 bar) to in a frictionless piston cylinder. Assuming that air behaves as an ideal gas, calculate the energy, enthalpy and entropy changes. An inventor claims to have devised a cyclic engine which exchanges heat with reservoirs at 25 250 degree C, and which produces 0.45 kJ of work for each kJ of heat extracted from the hot reservoir claim believable? Why or why not?^3 A heat engine receives 800 kJ of heat from a high temperature source at 900 degree C during a converts 250 kJ of this heat to net work (W) and rejects the remaining quantity of energy temperatures sink at 30 degree C. How much heat is rejected to the low temperatures reservoir?^2 Determine if this heat engine violates the second law of thermodynamics on the basis The Carnot principle, ^2 The Clausius inequality, ^2Explanation / Answer
1.
3.
Assuming the compression to be adiabatic
For adiabatic compression
T2/T1= (P2/P1) R/CP, R= CP-CV= 29-21 = 8 J/mole.K
T1= 25 deg.c =25+273= 298K, P1=10 bar, P2= 225 bar
T2/T1= (225/10) 8/29 = 2.36
T2= 2.36*298 =703 K= 430 deg.c
Work done = n*R*(T2-T1)/ (Y-1), where Y= CP/CV = 29/21= 1.38
Where n= no of moles of air = 1000g/29 g.mole ( Molar mass of air =29 g/mole)=34.48
Work done =34.48* 8.314*( 703-298)/(1.38-1) joules= 305526.4 joules
Since the process ia adiabatic, entropy change = 0
Enthalpy = n*CP*(T2-T1)= 34.48*29*( 703-298) =404968 joules
Internal energy = n*CV*(T2-T1)= 34.48*21*( 703-298) =293252 joules
.
4.
Efficiency of heat engine = (1-T2/T1)= 1-(25+273)/(250+273)= 1-298/523=0.43
Efficiency of heat engine is also equal to = W/Q = 0.45/1 . This is less than carnot efficiency, hence the claim is false, since any engine cannot have efficiency > Carnot efficiency.
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