Question 15.5: A 10-ton refrigeration unit operates on R-22, has a single-s......
A 10-ton refrigeration unit operates on R-22, has a single-stage reciprocating compressor with a clearance volume of 3 percent, and operates at 1725 rpm. The refrigerant leaves the evaporator at 70 psia and 40 F. Pressure loss in the evaporator is 5 psi. Pressure loss in the suction valve is 2 psi, and there is 20 F superheat at the beginning of the compression stroke. The compressor discharge pressure is 300 psia, a pressure loss of 5 psi exists in the discharge valve, and the vapor is desuperheated 10 F between the compressor discharge and the condenser inlet. The pressure loss in tubing between the compressor and condenser is negligible, but the condenser pressure loss is 5 psi.
The liquid has a temperature of 110 F at the expansion valve. Determine the following: (a) compressor volumetric efficiency, (b) compressor piston displacement, (c) shaft power input with mechanical efficiency of 75 percent, (d) heat rejected in the condenser, and (e) power input to compressor per ton of cooling effect.
Most of the state points as shown in Fig. 15-16 may be located from the given data:
P_{1} = 290 psia, T_{1} = 110 F, i_{2} = i_{1}
P_{3^{′}} = 70 psia, t_{3^{′}} = 40 F, P_{2} = P_{3^{′}} + 5 = 75 psia
P_{a} = P_{b} = P_{3^{′}} – 2 = 68 psia, t_{b} = 50 F
P_{c} = P_{d} = 300 psia, P_{4} = P_{4^{′}} = P_{c} – 5 = 295 psia
States c, d, and 4 may be completely determined following the compressor and evaporator analysis. From Chart 4 in Appendix E, v_{3} = 0.78 ft^{3}/lbm and v_{b} = 0.81 ft^{3}/lbm. Assuming that n = k = 1.16, the volumetric efficiency may be computed from Eq. 15-8 as
\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 + 0.03 – (0.03) \left\lgroup\frac{300}{68} \right\rgroup^{1/1.16} \right] \frac{0.78}{0.81} = 0.89
For the evaporator
\dot{q}_{e} = \dot{m} (i_{3^{′}} – i_{2}) = \dot{m} (i_{3^{′}} – i_{1})
or
\dot{m} = \frac{\dot{q}_{e}}{i_{3^{′}} – i_{1}}
From Chart 5 in Appendix E, i_{3^{′}} = 109 Btu/lbm, i_{1} = 41 Btu/lbm, and
\dot{m} = \frac{10 (12,000)}{109 – 41} = 1765 lbm/hr
From Eq. 15-10
\eta _{v} = \frac{\dot{m} v_{3}}{PD} (15-10)
\eta _{v} = \frac{\dot{m} v_{3}}{PD^{'}} or PD = \frac{\dot{m} v_{3}}{\eta _{v}}
whence
PD = \frac{1765 (0.78)}{0.89} = 1547 ft^{3}/hr
so that
\frac{PD}{cycle} = 25.8 in.^{3}
The work per pound of refrigerant may be computed from Eq. 15-12:
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)
w = \frac{1.16}{1.16 – 1} 68 (144) (0.81) \left[\left\lgroup\frac{300}{68} \right\rgroup^{(1.16 – 1)/1.16} – 1\right]
w = 13,000 (ft-lbf)/lbm
\dot{W} = \dot{m} w = \frac{13,000 (1765)}{778} = 29,500 Btu/hr
\dot{W}_{sh} = \frac{\dot{W}}{\eta _{m}} = \frac{29,500}{0.75} = 39,300 Btu/hr = 15.4 hp
The heat rejected in the condenser is given by
\dot{q}_{c} = \dot{m} (i_{1} – i_{4^{′}})
However, state 4′ has not been completely determined. Since the polytropic exponent was assumed equal to k, the isentropic compression process follows a line of constant entropy on Chart 4. Then t_{c} = t_{d} = 188 F, t_{4} ≈ 185 F, and t_{4^{′}} = t_{4} – 10 = 175 F. Then i_{4^{′}} = 125 Btu/lbm, so
\dot{q}_{c} = 1765 (41 – 125) = -148,000 Btu/hr
The total power input to the compressor per ton of cooling effect is
\frac{hp}{ton} = \frac{15.4}{10} = 1.54
