Question 5.8: Rework Ex. 5.6, making use of the equation for ideal work.

The procedure here is to calculate the maximum possible work W_{ideal}, which can be obtained from 1 kg of steam in a flow process as it undergoes a change in state from saturated steam at 100°C to liquid water at 0°C. The problem then reduces to the question of whether this amount of work is sufficient to operate a Carnot refrigerator rejecting 2000 kJ as heat at 200°C and taking heat from the unlimited supply of cooling water at 0°C. From Ex. 5.6, we have

\Delta H=-2676.0  kJ \cdot kg ^{-1} \quad        \text { and }       \quad \Delta S=-7.3554  kJ \cdot K ^{-1} \cdot kg ^{-1}

With negligible kinetic- and potential-energy terms, Eq. (5.22) yields:

W_{\text {ideal }}=\Delta H  –  T_\sigma \Delta S=-2676.0  –  (273.15)(-7.3554)=-666.9  kJ \cdot kg ^{-1}

If this amount of work, numerically the maximum obtainable from the steam, is used to drive the Carnot refrigerator operating between the temperatures of 0°C and 200°C, the heat rejected is found from Eq. (5.5), solved for Q_H:

\frac{W}{Q_H}=\frac{T_C}{T_H}-1     (5.5)

Q_H=W_{\text {ideal }}\left(\frac{T}{T_\sigma-T}\right)=666.9\left(\frac{200+273.15}{200-0}\right)=1577.7  kJ

As calculated in Ex. 5.6, this is the maximum possible heat release at 200°C; it is less than the claimed value of 2000 kJ. As in Ex. 5.6, we conclude that the process as described is not possible.