Question 5.6: An inventor claims to have devised a process that takes in o...
An inventor claims to have devised a process that takes in only saturated steam at 100°C and which by a complicated series of steps makes heat continuously available at a temperature level of 200°C, with 2000 kJ of energy available at 200°C for every kilogram of steam taken into the process. Show whether or not this process is possible. To give this process the most favorable conditions, assume cooling water available in unlimited quantity at a temperature of 0°C.
For any process to be even theoretically possible, it must meet the requirements of the first and second laws of thermodynamics. The detailed mechanism need not be known to determine whether this is true; only the overall result is required. If the claims of the inventor satisfy the laws of thermodynamics, fulfilling the claims is theoretically
possible. The determination of a mechanism is then a matter of ingenuity. Otherwise, the process is impossible, and no mechanism for carrying it out can be devised.
In the present instance, a continuous process takes in saturated steam at 100°C, and makes heat Q continuously available at a temperature level of 200°C. Because cooling water is available at 0°C, the maximum possible use is made of the steam if it is condensed and cooled (without freezing) to this exit temperature and discharged at atmospheric pressure. This is a limiting case (most favorable to the inventor), and requires a heat-exchange surface of infinite area.
It is not possible in this process for heat to be liberated only at the 200°C temperature level, because as we have shown no process is possible that does nothing except transfer heat from one temperature level to a higher one. We must therefore suppose that some heat Q_σ is transferred to the cooling water at 0°C. Moreover, the process must satisfy the first law; thus by Eq. (2.32):
\Delta H=Q+W_s (2.32)
\Delta H=Q+Q_\sigma+W_s
where ΔH is the enthalpy change of the steam as it flows through the system. Because no shaft work accompanies the process, W_s = 0. The surroundings consist of cooling water, which acts as a heat reservoir at the constant temperature of T_\sigma = 273.15 K, and a heat reservoir at T = 473.15 K to which heat in the amount of 2000 kJ is transferred for each kilogram of steam entering the system. The diagram of Fig. 5.5 indicates overall results of the process based on the inventor’s claim.
The values of H and S shown on Fig. 5.5 for liquid water at 0°C and for saturated steam at 100°C are taken from the steam tables (App. E). Note that the values for liquid water at 0°C are for saturated liquid (P^{sat} = 0.61 kPa), but the effect of an increase in pressure to atmospheric pressure is insignificant. On the basis of 1 kg of entering steam, the first law becomes:
\Delta H=H_2-H_1=0.0-2676.0=-2000+Q_\sigma \quad \text { and } \quad Q_\sigma=-676.0 kJ
The negative value for Q_\sigma indicates heat transfer is from the system to the cooling water.
We now examine this result in the light of the second law to determine whether the entropy generation is greater than or less than zero for the process. Equation (5.18) is here written:
\Delta S-\frac{Q_\sigma}{T_\sigma}-\frac{Q}{473.15}=S_G
For 1 kg of steam,
\Delta S=S_2 – S_1=0.0000 – 7.3554=-7.3554 kJ \cdot K ^{-1}
The entropy generation is then:
S_G=-7.3554 – \frac{-676.0}{273.15} – \frac{-2000}{473.15}
=-7.3554+4.2270+2.4748=-0.6536 kJ \cdot K ^{-1}
This negative result means that the process as described is impossible; the second law in the form of Eq. (5.18) requires S_G ≥ 0.
