Question 7.14: Consider a cogeneration plant of a large hospital, consistin......
Consider a cogeneration plant of a large hospital, consisting of a steam boiler and a back-pressure turbine. The boiler produces a mass flow rate of 26.1 kg/s of steam at 40 bar and 400°C, the exergy of the feed water being negligible, as well as the heat lost through its surface. The exergy of the fuel supplied to the boiler is 102 MW, with a unit price of € 8/GJ (exergy) and the exergy of the gases at the boiler outlet is 5.5 MW. The turbine has an isentropic efficiency of 78%, with the back pressure at the output being 7 bar. The investment in the boiler is 0.8 M€ and in the turbine 0.6 M€. The number of equivalent hours of operation of the plant per year is 6000, its useful life is 20 years and the annual interest rate is 4%. If the ambient temperature is T_{0}=288\,\mathrm{K}, determine:
(a) The rate of exergy destruction in the boiler.
(b) The rate of exergy destruction in the turbine.
(c) The cost (€c/kg) of the steam produced in the boiler.
(d) The cost (€c/MJ) of the work produced in the turbine.
(a) We use \underline1 for the state of the steam at the outlet of the boiler and \underline2 for the state after the expansion in the turbine. The exergy of the steam at the boiler outlet is calculated by means of Eq. (3.9) of Chapter 3. From the superheated steam tables we see h_{1}=3.213.6\,\rm k J/\mathrm{kg} and s_{1}=6.769\;{\mathrm{kJ/kg}}\cdot\mathrm{K}. From the liquid water tables, we have h_{0}=63\ \rm kJ/k{g} and s_{0}=0.2245\mathrm{~kJ/kg}\cdot\mathrm{K}. This then gives that
b_{1}=h_{1}-h_{0}-T_{0}(s_{1}-s_{0})=1265.8\ {\frac{\mathrm{kJ}}{\mathrm{kg}}} \\ {\dot{B}}_{1}=\dot m_{1}b_{1}=33.0\,\mathrm{MW}From the exergy balance in the boiler, we have
\dot{D}_{b}=\dot{F}_{b}-\dot{B}_{1}-\dot{L}_{b}=63.5\,\mathrm{MW}(b) According to the definition of isentropic efficiency, we have
\eta_{s}={\frac{h_{1}-h_{2}}{h_{1}-h_{2s}}}=0.78To know {\mathrm{h}}_{2s}, we look in the superheated steam tables for the state of pressure {\mathrm{p}}_{2}=7 bar and for the entropy s_{2s}=s_{1}=6.769\;\mathrm{kJ/kg}\cdot\mathrm{K}. The enthalpy of this state is h_{2s}=2799.1\ \mathrm{kJ/kg}. Returning to the expression of isentropic efficiency, we now calculate the enthalpy of state \underline2
h_{2}=h_{1}-\eta_{s}(h_{1}-h_{2s})=2890.3\ \frac{\mathrm{kJ}}{\mathrm{kg}}Therefore, the steam turbine power is
\dot{W}_{S T}=\dot{m}_{1}(h_{1}-h_{2})=8.44\:\mathrm{MW}From the exergy balance in the steam turbine, we have
F_{S T}=\dot{B}_{1}-\dot{B}_{2}\quad P_{T V}=\dot{W}_{T V} \\ \dot{D}_{S T}=\dot{B}_{1}-\dot{B}_{2}-\dot{W}_{S T}and as
{\dot{B}}_{1}-{\dot{B}}_{2}=9.61\ \mathrm{MW}we get
\dot{D}_{S T}=1.2\,\mathrm{MW}(c) To calculate the exergoeconomic cost of the steam produced in the boiler, we carry out an exergoeconomic costs balance in the boiler
c_{1}\dot{B}_{1}=c_{F,b}\dot{B}_{F,b}+\dot{Z}_{b}We calculate the capital recovery factor for the boiler
a_{b}=\frac{i(1+i)^{n}}{(1+i)^{n}-1}=\frac{0.04\times1.04^{20}}{1.04^{20}-1}=0.0735so that its capital cost rate is
Z_{b}={\frac{a_{b}I_{b}}{H}}=9.8{\frac{€}{h}}Returning to the equation of the monetary cost balance in the boiler, we have
c_{1}{\dot{B}}_{1}=81.87{\frac{€ c}{s}}Therefore, the unit cost of the steam produced in the boiler is
c_{1}=2.48{\frac{€ c}{\mathrm{MJ}}}=3.14{\frac{€ c}{\mathrm{kg}}}(d) To calculate the monetary cost of the work produced in the steam turbine, we undertake an exergoeconomic costs balance in the turbine.
c_{S T}\dot{W}_{S T}=c_{1}\dot{B}_{1}-c_{2}\dot{B}_{2}+\dot{Z}_{S T}Taking into account that the turbine fuel is {\dot{B}}_{1}-{\dot{B}}_{2} the following equality is true {{c}}_{1}-{{c}}_{2}. We calculate the capital cost rate of the turbine
Z_{S T}=\frac{a_{S T}I_{S T}}{H}=7.35\frac{€}{h}So that
c_{S T}=\frac{2.48\ .\ 9.61+\frac{7.35}{36}}{8.44}=2.85\;\frac{€ c}{\mathrm{MJ}}=10.26\frac{€ c}{\mathrm{kWh}}