Question 13.4: In 1876, George H. Corliss (1817–1888) built what was then t...
In 1876, George H. Corliss (1817–1888) built what was then the largest steam engine ever made (Figure 13.12), for the United States Centennial Exposition in Philadelphia, Pennsylvania. It had two cylinders, each with a 40.0-inch bore and a 10.0 ft stroke. Steam entered the engine as a saturated vapor at 100. psia, producing 1400. hp at only 36.0 rpm. The engine’s flywheel was 30.0 ft in diameter and weighed 56.0 tons. If the steam was condensed at 14.7 psia, determine the Rankine cycle thermal efficiency of this engine assuming (a) isentropic prime mover and pump, (b) an engine isentropic efficiency of 55.0% and a pump isentropic efficiency of 65.0%, and (c) the steam mass flow rate required to produce 1400. hp.
a. The thermodynamic states at the four main points around the Rankine cycle for this system are (Figure 13.13)
| \underline{ \text{Station 1—Engine inlet}} | \underline{ \text{Station 2}s\text{—Engine exit}} |
| p_1 = 100. \text{psia} | p_{2s} = 14.7 \text{psia} |
| x_1 = 1.00 | \underline{s_{2s} = s_1} |
| h_1 = h_g(100. \text{psia}) = 1187.8 \text{Btu/lbm} | x_{2s} = [s_{2s} – s_f (14.7 \text{psia})] / s_{fg} (14.7 \text{psia}) |
| s_1 = s_g(100. \text{psia}) = 1.6036 \text{Btu/lbm.R} | = [1.6036 − 0.3122] / 1.4447 = 0.8939 |
| h_{2s} = h_f (14.7 \text{psia}) + x_{2s} h_{fg} (14.7 \text{psia}) | |
| = 180.1 + 0.8939 (970.4) | |
| = 1047.5 \text{Btu/lbm} | |
| \underline{ \text{Station 3—Condenser exit}} | \underline{ \text{Station 4}s\text{—Boiler inlet}} |
| p_3 = p_{2s} = 14.7 \text{psia} | p_{4s} = p_{1} = 100 \text{psia} |
| \underline{x_3 = 0} | \underline{s_{4s}= s_3} |
| h_3 = h_f(14.7 \text{psia}) = 180.1 \text{Btu/lbm} | |
| v_3 = v_f(14.7 \text{psia}) =0.01672 \text{ft³/lbm} |
Then, the isentropic efficiency of this system is given by Eq. (13.9b) as
(η_T)_{\substack{\text{maximum}\\\text{Rankine}\\}} = \frac{h_1 − h_{2s} − v_3(p_4 − p_3)}{h_1 − h_3 − v_3 (p_4 − p_3)}
= \frac{(1187.8 − 1047.5 \text{Btu/lbm}) – (0.01672 \text{ft³/lbm})(100. − 14.7 \text{ lbf /in²})(\frac{144 \text{in²/ft²}}{118.16 \text{ ft.lbf /Btu}})}{(1187.8 − 180.1 \text{Btu/lbm}) – (0.01672 \text{ft³/lbm})(100. − 14.7 \text{ lbf /in²})(\frac{144 \text{in²/ft²}}{118.16 \text{ ft.lbf /Btu}})}
= 0.139 = 13.9\%
b. Here, we use Eq. (13.9a) with (η_s)_{pm} = 0.550. and (η_s)_p = 0.650.
(η_T)_{\text{Rankine}} = \frac{(h_1 − h_{2s}) (η_s)_{pm} − v_3(p_4 − p_3)/(η_s)_p }{h_1 − h_3 − v_3 (p_4 − p_3)/(η_s)_p}
= \frac{(1187.8 − 1047.5 \text{Btu/lbm}) (0.550) – (0.01672 \text{ft³/lbm})(100. − 14.7 \text{ lbf /in²})(\frac{144 \text{in²/ft²}}{118.16 \text{ ft.lbf /Btu}})(\frac{1}{0.650} )}{(1187.8 − 180.1 \text{Btu/lbm}) – (0.01672 \text{ft³/lbm})(100. − 14.7 \text{ lbf /in²})(\frac{144 \text{in²/ft²}}{118.16 \text{ ft.lbf /Btu}})(\frac{1}{0.650} )}
= 0.0762 = 7.62\%
c. If the actual power output from the engine is 1400 hp and the isentropic efficiency of the engine is 55%, then the mass flow rate of steam required is
\dot{m} = \frac{\dot{W_\text{actual}}}{(h_1 − h_{2s})(η_s)_{pm}} = \frac{(1400. \text{hp})(2545 \text{Btu/hp.h})}{(1187.8 − 1047.5 \text{Btu/lbm})(0.550)} = 46,200 \text{lbm/h}
