Question 8.1: Steam generated in a power plant at a pressure of 8600 kPa a...

(a) The turbine operates under the same conditions as that of Ex. 7.6 where, on the basis of 1 kg of steam:
(\Delta H)_S=-1274.2 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

Thus                        W_s(\text { isentropic })=(\Delta H)_S=-1274.2 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

Moreover, the enthalpy at the end of isentropic expansion, called H_{2}^{′} in Ex. 7.6, is here:

H_3^{\prime}=2117.4 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

Subscripts refer to Fig. 8.4. The enthalpy of saturated liquid condensate at 10 kPa (and  t^{sat} = 45.83° C ) is:

H_4=191.8 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

By Eq. (8.2) applied to the condenser,

Q = ΔH          (8.2)

Q(\text { condenser })=H_4-H_3^{\prime}=191.8-2117.4=-1925.6 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

where the minus sign indicates heat flow out of the system.

The pump operates under essentially the same conditions as the pump of Ex. 7.10, where:

W_s(\text { isentropic })=(\Delta H)_S=8.7 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

and H_1=H_4+(\Delta H)_S=191.8+8.7=200.5 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

The enthalpy of superheated steam at 8600 kPa and 500°C is:

H_2 = 3391.6 kJ⋅ kg^{−1}

By Eq. (8.2) applied to the boiler,

Q \text { (boiler) }=H_2-H_1=3391.6-200.5=3191.1 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

The net work of the Rankine cycle is the sum of the turbine work and the pump work:

W_s(\text { Rankine })=-1274.2+8.7=-1265.5 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

This result is of course also:

W_s(\text { Rankine })=-Q(\text { boiler })-Q(\text { condenser })

 = −3191.1 + 1925.6 = −1265.5  kJ⋅kg^{−1}

The thermal efficiency of the cycle is:

\eta=\frac{-W_s(\text { Rankine })}{Q}=\frac{1265.5}{3191.1}=0.3966

(b) With a turbine efficiency of 0.75, then also from Ex. 7.6:

W_s(\text { turbine })=\Delta H=-955.6 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

and  H_3=H_2+\Delta H=3391.6-955.6=2436.0 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

For the condenser,

Q(\text { condenser })=H_4-H_3=191.8-2436.0=-2244.2 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

As shown in Ex. 7.10 for the pump,

W_s(\text { pump })=\Delta H=11.6 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

The net work of the cycle is therefore:

W_s(\text { net })=-955.6+11.6=-944.0 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

and              H_1=H_4+\Delta H=191.8+11.6=203.4 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

Then           Q \text { (boiler) }=H_2-H_1=3391.6-203.4=3188.2 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

The thermal efficiency of the cycle is therefore:

\eta=\frac{-W_s(\text { net })}{Q(\text { boiler })}=\frac{944.0}{3188.2}=0.2961

which can be compared with the result of part (a). The main difference is that due to the turbine efficiency because the energy required by the pump is small in either case.

(c) For a power rating of 80,000 kW:

\dot{W}_s(\text { net })=\dot{m} W_s(\text { net })

or              \dot{m}=\frac{\dot{W}_s(\text { net })}{W_s(\text { net })}=\frac{-80,000 \mathrm{~kJ} \cdot \mathrm{s}^{-1}}{-944.0 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}}=84.75 \mathrm{~kg} \cdot \mathrm{s}^{-1}

Then by Eq. (8.1),

\dot{Q}=\dot{m} \Delta H         (8.1)

\dot{Q}(\text { boiler })=(84.75)(3188.2)=270.2 \times 10^3 \mathrm{~kJ} \cdot \mathrm{s}^{-1}

\dot{Q} \text { (condenser) }=(84.75)(-2244.2)=-190.2 \times 10^3 \mathrm{~kJ} \cdot \mathrm{s}^{-1}

Note that

\dot{Q} \text { (boiler) }+\dot{Q} \text { (condenser) }=-\dot{W}_s \text { (net) }