Figure 8-13, which is similar to Fig. 8-9(b), shows a Carnot cycle in the liquid

Question

Figure 8-13, which is similar to Fig. 8-9(b), shows a Carnot cycle in the liquid-vapor region. The working substance is 1 kg of water, and T2 = 453 K, T1 = 313 K. Steam tables list values of T, P, u, s, and h at points on the saturation lines and these are tabulated below, in MKS units, for points a, b, e, and f. We wish to make a complete analysis of the cycle. Show that in the process a-b, qab = hb - ha, wab = hb - ha - ub + ua. Show that in the process b-c, qbc = 0, wbc = ub - uc. Show that in the process c-d, qcd = hd - hc, wcd = hd - hc - ud + uc Show that in the process d-a, qda = 0, wda = ud - ua.

Explanation / Answer

a) a-b process is a constant temperature process(isothermal)

There fore Q=CpdT = hb - ha =19.88 *10^5 J/Kg

Q=U+w

U=CvdT= Ub-Ua

Therefore w=hb-ha-ub+ua=1.68 J/Kg

b)b-c is an adiabatic process , so Q=0

Q=U+W

W=-U => W=-( Uc-Ub) =Ub-Uc =24.13 J/Kg

c)c-d process is a constant temperature process (isothermal)

There fore Q=CpdT = hd - hc =23.93 *10^5 J/Kg

Q=U+w

U=CvdT= Ud-Uc

Therefore w=hd-hc-ud+uc=1.3J/Kg

d)d-a is an adiabatic process , so Q=0

Q=U+W

W=-U => W=-( Ua-Ud) =Ud-Ua =16.7 J/Kg

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