MT1-cheatsheet

# 2046 a ideal gases do not interact so each gas expands

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 6 J/gal)(1 gal)/(7.2 104 J/s) = 1.81 103 s = 30 min. 20.6 (a) Work is positive for an expansion. Because the work done is represented by the area under the PV curve, Wbc > Wac. The net work is the sum of the works done in each leg. For a positive net work, Wbc > 0, so we have an expansion from b to c, and the path must be clockwise. P b Tb = Tc (b) We find the ratio of volumes from the ideal gas equation: a c (Pc/Pa)/(Vc/Va ) = Tc/Ta ; V (1)(Vc/Va ) = Tc/Ta , so Vc/Va = Vc/Vb = Tc/Ta = Tb /Ta . 0 The work done during the isothermal expansion is Wbc = nRTb ln(Vc/Vb) = nRTb ln(Tb/Ta) = (1.0 mol)(8.315 J/mol K)(423 K) ln(423 K/273 K) = 1.54 103 J. The work done during the constant pressure compression is Wca = Pa(Va Vc) = PaVa[1 (Vc/Va)] = nRTa [1 (Vc/Va)] = (1.0 mol)(8.315 J/mol K)(273 K)[1 (423 K/273 K)] = 1.25 103 J. For the isothermal expansion Ubc = 0, so Qbc = Wbc = 1.54 103 J. For the constant volume compression Wab = 0, so Qab = Uab = nCV(Tb Ta) = (1.0 mol)8(8.315 J/mol K)(423 K 273 K) = 1.87 103 J. The efficiency is e = Wnet/Qadded = (Wbc + Wca)/(Qbc + Qab) = (1.54 103 J 1.25 103 J)/(1.54 103 J + 1.87 103 J) = 0.085 = 8.5%. 20.18 For the efficiencies of the engines, we have TL2 QL2 Engine W2 QL1 = QH2 TH2 TL1 Engine W1 QH1 TH1 = 0.65eCarnot = 0.60[1 (TL1/TH1)] = 0.65{1 [(703 K)/(953 K)]} = 0.171; e2 = 0.65eCarnot = 0.60[1 (TL2/TH2)] = 0.65{1 [(553 K)/(688 K)]} = 0.128. Because coal is burned to produce the input heat to the first engine, we need to find Q H1. We relate W1 to QH1 from the efficiency: e1 = W1/QH1 , or W1 = 0.171QH1 . Because the exhaust heat from the first engine is the input heat to the second engine, we have e2 = W2/QH2 = W2/QL1 , or W2 = 0.128QL1 . For the first engine we know that QL1 = QH1 W1 , so we get W2 = 0.128QL1 = 0.128(QH1 0.171QH1) = 0.106QH1 . For the total work, we have W = W1 + W2 = 0.171QH1 + 0.106QH1 = 0.277QH1 . When we use the rate at which this work is done, we get 900 106 W = 0.277(QH1/t), which gives QH1/t = 3.25 109...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online