Answer:
a. Mass flow rate through the boiler = 5.462lbm/s
b. Power produced by the turbine = 2525.8kW
c. The rate of heat supply in the boiler = 6901.42Btu/s
d. Thermal efficiency of the cycle = 34.3%
Explanation:
In order to provide a solution, we must assume that ;
- The system is operating at a steady condition
- Kinetic and potential energy changes are negligible
Now from steam tables, we calculate specific volume [tex]v[/tex] and enthalpy [tex]h[/tex] as,
[tex]h_1 = 95.96Btu/lb[/tex] ( [tex]h_1 = h_f[/tex] at [tex]2psia[/tex] )
[tex]v_1 = 0.016238ft^3/lb[/tex] ( [tex]v_1 = v_f[/tex] at [tex]2psia[/tex] )
[tex]w_{p,in} = v_1(P_2-P_1) = 0.016238(1500-2) * \frac{1}{5.404} = 4.501 Btu/lb[/tex]
[tex]w_p = h_2 - h_1\\h_2 = w_p+h_1=4.501+95.96=100.461Btu/lb[/tex]
[tex]h_3 = 1364.0Btu/lb[/tex]
[tex]s_3 = 1.5073Btu/lb.R[/tex]
( at [tex]P_3 = 1500psia[/tex] & [tex]T_3 = 800^0F[/tex] )
[tex]P_4 = 2psia\\S_4 = S_3\\x_4S = \frac{S_4-S_f}{S_{fg}}=\frac{1.5073-0.1783}{1.7374}=0.765[/tex]
( [tex]S_f[/tex] & [tex]S_{fg}[/tex] when pressure is 2psia)
[tex]h_4S = h_f+x_4S*h_{fg}=95.96+(0.765)(1021.0)=877.025Btu/lb[/tex]
[tex]n_T= \frac{h_3-h_4}{h_3-h_4S}\\ h_4=h_3-n_T(h_3-h_4S)=1364.0-0.90(1364.0-877.025)=925.7Btu/lb[/tex]
Therefore,
[tex]q_{in}=h_3-h_2=1364.0-100.461=1263.54Btu/lb\\q_{out}=h_4-h_1=925.7-95.96=829.74Btu/lb\\w_{net}=q_{in}-q_{out}=1263.54-829.74=433.8Btu/lb[/tex]
To calculate the mass flow rate of steam in the cycle, we use the formula
[tex]W_{net}=mw_{net}\\m=\frac{W_{net}}{w_{net}} =\frac{2500}{433.8}=5.763*(\frac{0.94782Btu}{1Kj} )=5.462lb/s[/tex]
where [tex]1Kj = 0.947817 Btu[/tex]
The power output and the rate of heat addition are calculated thus,
[tex]W_{T,out}=m(h_3-h_4)=(5.462lb/s)*(1364-925.7)Btu/lb*(\frac{1Kj}{0.94782Btu} )\\=5.462*438.3*1.055=2525.8KW[/tex]
[tex]Q_{in}=mq_{in}=5.462(1263.54)=6901.46Btu/s[/tex]
The thermal efficiency of the cycle can be found thus;
[tex]n_{th}=\frac{W_{net}}{Q_{in}} =\frac{2500}{6901.46}*(\frac{0.94782Btu}{1Kj} ) =0.343[/tex]
= 34.3%
