A cogeneration plant is considered. The maximum rate of process heat supply, the power produced and the utilization factor when no process heat is supplied, and the rate of process heat supply when steam is extracted from the steam line and turbine at specified ratios are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Pressure drops and heat losses in piping are negligible. 3 Kinetic and potential energy changes are negligible.
Analysis The schematic of the cogeneration plant and the T-s diagram of the cycle are shown in Fig. 10–23. The power plant operates on an ideal
cycle and thus the pumps and the turbines are isentropic; there are no pressure drops in the boiler, process heater, and condenser; and steam leaves the condenser and the process heater as saturated liquid.
The work inputs to the pumps and the enthalpies at various states are as follows:
\begin{aligned}w_{\text {pump I,in }} &=v_{8}\left(P_{9}-P_{8}\right)=\left(0.001005 m ^{3} / kg \right)[(7000-5) kPa ]\left(\frac{1 kJ }{1 kPa \cdot m ^{3}}\right) \\&=7.03 kJ / kg \\w_{\text {pump II,in }} &=v_{7}\left(P_{10}-P_{7}\right)=\left(0.001093 m ^{3} / kg \right)[(7000-500) kPa ]\left(\frac{1 kJ }{1 kPa \cdot m ^{3}}\right) \\&=7.10 kJ / kg \\h_{1} &=h_{2}=h_{3}=h_{4}=3411.4 kJ / kg \\h_{5} &=2739.3 kJ / kg \\h_{6} &=2073.0 kJ / kg \\h_{7} &=h_{f @ 500 kPa }=640.09 kJ / kg \\h_{8} &=h_{f @ 5 kPa }=137.75 kJ / kg \\h_{9} &=h_{8}+w_{\text {pump } I , \text { in }}=(137.75+7.03) kJ / kg =144.78 kJ / kg \\h_{10} &=h_{7}+w_{\text {pump II,in }}=(640.09+7.10) kJ / kg =647.19 kJ / kg\end{aligned}
(a) The maximum rate of process heat is achieved when all the steam leaving the boiler is throttled and sent to the process heater and none is sent to the turbine \text { (that is, } \dot{m}_{4}=\dot{m}_{7}=\dot{m}_{1}=15 kg / s \text { and } \dot{m}_{3}=\dot{m}_{5}=\dot{m}_{6}=0 \text { ) } . Thus,
\dot{Q}_{p, \max }=\dot{m}_{1}\left(h_{4}-h_{7}\right)=(15 kg / s )[(3411.4-640.09) kJ / kg ]=41,570 kW
The utilization factor is 100 percent in this case since no heat is rejected in the condenser, heat losses from the piping and other components are assumed to be negligible, and combustion losses are not considered.
(b) When no process heat is supplied, all the steam leaving the boiler passes through the turbine and expands to the condenser pressure of 5 kPa (that is, \dot{m}_{3}=\dot{m}_{6}=\dot{m}_{1}=15 kg / s \text { and } \dot{m}_{2}=\dot{m}_{5}=0 ). Maximum power is produced in this mode, which is determined to be
\begin{aligned} \dot{W}_{\text {turb,out }} &=\dot{m}\left(h_{3}-h_{6}\right)=(15 kg / s )[(3411.4-2073.0) kJ / kg ]=20,076 kW \\ \dot{W}_{\text {pump,in }} &=(15 kg / s )(7.03 kJ / kg )=105 kW \\ \dot{W}_{\text {net,out }} &=\dot{W}_{\text {turb,out }}-\dot{W}_{\text {pump,in }}=(20,076-105) kW =19,971 kW \cong 20.0 MW \\ \dot{Q}_{\text {in }} &=\dot{m}_{1}\left(h_{1}-h_{11}\right)=(15 kg / s )[(3411.4-144.78) kJ / kg ]=48,999 kW \end{aligned}
Thus,
\epsilon_{u}=\frac{\dot{W}_{\text {net }}+\dot{Q}_{p}}{\dot{Q}_{\text {in }}}=\frac{(19,971+0) kW }{48,999 kW }=0.408 \text { or } 40.8 \%
That is, 40.8 percent of the energy is utilized for a useful purpose. Notice that the utilization factor is equivalent to the thermal efficiency in this case.
(c) Neglecting any kinetic and potential energy changes, an energy balance on the process heater yields
\begin{aligned} \dot{E}_{\text {in }} &=\dot{E}_{\text {out }} \\ \dot{m}_{4} h_{4}+\dot{m}_{5} h_{5} &=\dot{Q}_{p, \text { out }}+\dot{m}_{7} h_{7} \end{aligned}
or
\dot{Q}_{p,out}= \dot{m}_{4}h_4+\dot{m}_{5}h_5-\dot{m}_{7}h_7
where
\begin{aligned}&\dot{m}_{4}=(0.1)(15 kg / s )=1.5 kg / s \\&\dot{m}_{5}=(0.7)(15 kg / s )=10.5 kg / s \\&\dot{m}_{7}=\dot{m}_{4}+\dot{m}_{5}=1.5+10.5=12 kg / s\end{aligned}
Thus
\begin{aligned}\dot{Q}_{p, \text { out }}=&(1.5 kg / s )(3411.4 kJ / kg )+(10.5 kg / s )(2739.3 kJ / kg ) \\&-(12 kg / s )(640.09 kJ / kg ) \\=& 26.2 MW\end{aligned}
Discussion Note that 26.2 MW of the heat transferred will be utilized in the process heater. We could also show that 11.0 MW of power is produced in this case, and the rate of heat input in the boiler is 43.0 MW. Thus the utilization factor is 86.5 percent.
