Question 13.6: A two-stage steam turbine receives dry saturated steam at 20...
A two-stage steam turbine receives dry saturated steam at 200. psia. It has an interstage pressure of 80.0 psia and a condenser pressure of 1.00 psia. Determine
a. The isentropic Rankine cycle thermal efficiency of the system without regeneration present.
b. The isentropic Rankine cycle thermal efficiency of the system and the mass fraction of regeneration steam required with an open loop boiler feedwater regenerator at a pressure of 80.0 psia.
a. First, draw a sketch of the configuration of the isentropic Rankine cycle system before boiler feedwater regeneration is added (Figure 13.21).
Its thermal efficiency is
(η_T)_{\substack{\text{isentropic}\\\text{Rankine}\\}} = \frac{(h_1 − h_{2s}) – (h_{4s} − h_3)}{(h_1 −h_{4s})}
where, assuming an incompressible liquid condensate, Eq. (13.6) gives h_{4s}= h_3 + v_3(p_4 – p_3) and v_3 = v_4 (1.00 \text{psia}) = 0.01614 \text{ft³/lbm}. The monitoring station data are as follows:
| \underline{ \text{Station 1}} | \underline{ \text{Station 2}s} |
| p_1 = 200. \text{psia} | |
| \underline{x_1 = 1.00} | p_{2s} = p_{2} = 1.00 \text{psia} |
| h_1 =1199.3 \text{Btu/lbm} | \underline{s_{2s} – s_1 = 1.5466 \text{Btu/(lbm.R)}} |
| s_1 = 1.5466 \text{Btu/(lbm.R)} | h_{2s} = 863.5 \text{Btu/lbm} |
| \underline{ \text{Station 3}} | \underline{ \text{Station 4}s} |
| p_3 = 1.00 \text{psia} | p_{4s} = p_{4} = 200. \text{psia} |
| \underline{x_3 = 0.00} | \underline{s_{4s} – s_3 = 0.1326 \text{Btu/(lbm.R)}} |
| h_3 = 69.7 \text{Btu/lbm} | h_{4s} = h_3 + v_3 (p_4 – p_3) |
| s_3 = 0.1326 \text{Btu/(lbm.R)} | = 69.70 + (0.01614) (200. – 1.00) (144/778.16) |
| = 69.70 + 0.594 = 70.3 \text{Btu/lbm} |
where, at station 2s, we use
x_{2s} = \frac{s_{2s} −s_{f2}}{s_{fg2}} = \frac{s_1 − s_{f2}}{s_{fg2}} = \frac{1.5466 − 0.1326}{18455} = 0.7662
then,
h_{2s} = 69.70 + (0.7662) (1036.0) = 863.5 \text{Btu/lbm}
The thermal efficiency is now
(η_T)_{\substack{\text{isentropic}\\\text{Rankine}\\}} = \frac{1199.3 − 863.5 − 0.594}{1199.3 − 70.3} = 0.297 = 29.7\%
b. Now, draw a sketch of the configuration of the isentropic Rankine cycle system with one open loop boiler feedwater regenerator (Figure 13.22).
The properties at monitoring stations 1, 2, and 3 are the same as they were in part a, but pump 1 brings the condensate pressure up to only 80.0 psia (to match the vapor inlet pressure), and pump 2 brings the pressure the rest of the way up from 80.0 to 200. psia. Then, v_3 = v_f(1.00 \text{psia}) = 0.01614 \text{ft³ /lbm} and v_6 = v_f(80.0 \text{psia}) = 0.01757 \text{ft³ /lbm}. The additional monitoring station data needed are
| \underline{ \text{Station 4}s} | \underline{ \text{Station 5}s} |
| p_{4s} = p_4 = 80.0 \text{psia} | p_{5s} = p_4 = 80.0 \text{psia} |
| \underline{s_{4s} = s_3 = 0.1326 \text{Btu/(lbm.R)}} | \underline{s_{5s} = s_1 = 1.5466 \text{Btu/(lbm.R)}} |
| h_{4s} = h_3 + v_3 (p_4 – p_3) | x_{5s} = 0.9358 |
| = 69.7+ (0.01614) (80.0 – 1.00)(144/778.16) | h_{5s} = 1125.7 \text{Btu/lbm} |
| = 69.7 + 0.236 = 69.9 \text{Btu/lbm} | |
| \underline{ \text{Station 6}} | \underline{ \text{Station 7}s} |
| p_6 = 80.0 \text{psia} | p_{7s} = p_7 = 200. \text{psia} |
| \underline{x_6 = 0.00} | \underline{s_{7s} = s_6 = 0.4535 \text{Btu/(lbm.R)}} |
| h_6 = 282.2 \text{Btu/lbm} | h_{7s} = h_6 + v_6 (p_7 – p_6) |
| s_6 = 0.4535 \text{Btu/(lbm.R)} | = 282.2 + (0.01757)(200. – 80.0)(144/778) |
| = 282.2 + 0.390 = 282.6 \text{Btu/lbm} |
where, at station 5s, we determine that x_{5s} = (1.5466 – 0.4535)/1.1681 = 0.9358 and h_{5s} = 282.2 + (0.9358)(901.4) = 1125.7 \text{Btu/lbm}. Equation (13.11) now gives the value of y as
\text{y} = \frac{h_6 − h_{4s}}{h_{5s} − h_{4s}} = \frac{282.6 − 69.9}{1125.7 − 69.9} = 0.201
then, the isentropic thermal efficiency of the cycle is given by Eq. (13.12b) with all the η_s = 1.0 as
(η_T)_{\substack{\text{Rankine cycle}\\\text{with 1 regenerator }\\}} = 1− (\frac{h_{2s} − h_3}{h_1 − h_{7s}}) (1 − \text{y}) = 1 – (\frac{863.5 − 69.7}{1199.3 − 0282.6} ) (1 − 0.201) = 0.308 = 30.8\%
By altering the value of \text{y} (the amount of steam bled from the turbine) and recomputing the value of h_{7s}, it becomes clear that the cycle thermal efficiency is maximized at 31.2% when \text{y} = 0.138. This is shown in Figure 13.23.
