A gas is taken through the cycle ABCA showin in the PV-diagram, A gas is taken t
ID: 3899578 • Letter: A
Question
A gas is taken through the cycle ABCA showin in the PV-diagram,
A gas is taken through the cycle ABCA shown in the PV - diagram. In the diagram, Pi = 100000 Pa, = 200000 Pa, P2 = 200000 Pa, V1 = 0.03 m3, and V2 = 0.06 m3. Determine the internal energy of the gas at each state in the cycle. Determine the work done by the gas for each process. Determine the heat added to the gas for each process. (Heat removed from the gas is negative.) Determine the following for one full cycle based on your answers to (b) and (c): The net work done by the gas. The heat added to the gas. The efficiency of the engine (as a fraction).Explanation / Answer
a)
Ua = m*Cv*Ta
= mR/(gama - 1)*Ta
= Pa*Va / (gamma - 1)
= 100000*0.03 / (1.4 - 1)
= 7500 J
Ub = m*Cv*Tb
= mR/(gama - 1)*Tb
= Pb*Vb / (gamma - 1)
= 200000*0.03 / (1.4 - 1)
= 15000 J
Uc = m*Cv*Tc
= mR/(gama - 1)*Tc
= Pc*Vc / (gamma - 1)
= 100000*0.06 / (1.4 - 1)
= 15000 J
b)
Wab = 0 (snce volume change = 0)
Wbc = Area below BC
= 1/2*(100000 + 200000)*(0.06 - 0.03)
= 4500 J
Wca = P*(Va - Vc)
= 100000*(0.06 - 0.03)
= -3000 J
c)
Qab = Wab + (Ub - Ua)
= 0 + (15000 - 7500)
= 7500 J
Qbc = Wbc + (Uc - Ub)
= 4500 + (15000 - 15000)
= 4500 J
Qca = Wca + (Ua - Uc)
= -3000 + (7500 - 15000)
= -10500 J
d)
Net work = Wab + Wbc + Wca
= 0 + 4500 - 3000
= 1500 J
Net heat added = Qab + Qbc
= 7500 + 4500
= 10500 J
Efficiency = Net work / Heat added
= 1500 / 10500
= 0.1428
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