Question 15.4: Refrigerant 134a vapor enters the suction header of a single......
Refrigerant 134a vapor enters the suction header of a single-stage reciprocating compressor at 45 psia and 40 F. The discharge pressure is 200 psia. Pressure drop in the suction valve is 2 psi, and the pressure loss in the discharge valve is 4 psi. The vapor is superheated 12 F during the intake stroke. The clearance volume is 5 percent of piston displacement. Determine the (a) volumetric efficiency, (b) compressor pumping capacity if the piston displacement is 10 in.³ and the crankshaft rotates at 1725 rpm, and (c) shaft horsepower required for a mechanical efficiency of 70 percent.
The P–v diagram of Fig. 15-6 will be used to aid in the problem solution. From the given data,
p_{4} = 200 psia; P_{c} = P_{d} = P_{4} + 4 = 204 psia
p_{3} = 45 psia; P_{b} = P_{3} − 2 = 43 psia
From Chart 3 in Appendix E, v_{3} = 1.09 ft^{3}/lbm, and at point b, t_{b} = 52 F. We have P_{b} = 43 psia and v_{b} = 1.16 ft^{3}/lbm. The clearance factor is given as 0.05, and it will be assumed that n = k = 1.26. Then using Eq. 15-8
\eta _{v} = \left[1 + C – C \left\lgroup\frac{P_{c}}{P_{b}} \right\rgroup^{1/n} \right] \frac{v_{3}}{v_{b}} (15-8)
\eta _{v} = \left[1 + 0.05 – (0.05) \left\lgroup\frac{204}{43} \right\rgroup^{1/1.26} \right] \frac{1.09}{1.16} = 0.83
Now from Eq. 15-10
\eta _{v} = \frac{\dot{m} v_{3}}{PD} (15-10)
\dot{m} = \frac{\eta _{v} (PD)}{v_{3}} = \frac{0.83}{7.6} \frac{(10)(1725)}{1728} = 1.09 lbm/min
The shaft power required will be computed using Eqs. 15-12 and 15-13:
w = \frac{n}{n – 1} P_{b} v_{b} \left[\left\lgroup\frac{P_{c}}{P_{b}} \right\rgroup^{(n – 1)1/n} – 1\right] (15-12)
\dot{W} = \frac{\dot{m} w}{\eta_{m}} (15-13)
\dot{W} = \frac{\dot{m} w}{\eta _{m}} = \left\lgroup\frac{\dot{m}}{\eta _{m}} \right\rgroup \frac{n}{(n – 1)} P_{b} v_{b} \left[\left\lgroup\frac{P_{c}}{P_{b}} \right\rgroup^{(n – 1)1/n} – 1\right]
\dot{W} = \left\lgroup\frac{7.6}{0.7} \right\rgroup \frac{1.26}{(1.16 – 1)} (43) (144) 1.26 \left[\left\lgroup\frac{204}{43} \right\rgroup^{(1.26 – 1)/1.26} – 1\right]
\dot{W} = 143,000 (ft-lbf)/min = 4.3 hp
