In a dairy plant, whole milk at 160 degree F (specific volume, 0.01720 ft^3/Lb_m
ID: 1717916 • Letter: I
Question
In a dairy plant, whole milk at 160 degree F (specific volume, 0.01720 ft^3/Lb_m; specific enthalpy, 148.64 Btu/Ib_m; specific entropy, 0.2772 Btu/lb_m- degree R) enters the thermally insulated (adiabatic) stainless steel heat exchanger shown below at a flow rate of 200 gal./min. The milk is chilled and leaves the heat exchanger at 40 degree F (specific volume, 0.01682 ft^3/Ib_m; specific enthalpy, 37.16 Btu/lb_m; specific entropy, 0.0775 Btu/lb_m- degree R). The milk is cooled by R-134a (sec Table A-11 E), which enters the heat exchanger 35 degree F and saturated liquid conditions. The refrigerant leaves the heat exchanger at 40 psia and 100 degree F. The flow through the heat exchanger is steady, with negligible changes in kinetic and potential energies. Determine (a) the mass flow rate of refrigerant {Ib^m/sec.} and (b) the entropy generation rate {Btu/sec.- degree R} for the heat exchanger as the system.Explanation / Answer
Given data
Hot fluid enters at 160 °F & leaves at 40 °F
At 160 °F
h1 = 148.64 Btu/lbm
At 40 °F
h2 = 37.16 Btu/lbm
Mass flow rate mf = 200 gal/min = 1669.08 lbm/min = 27.818 lbm/sec
Similarly for refrigerant
At 35 °F
h3 = 86.95 Btu/lbm
At 40 psia & 100 °F
h4 = 185.95 Btu/lbm
We know that
Heat lost by hot fluid = Heat gained by cold fluid
mf (h1 - h2) = mr (h4 - h3)
27.818 (148.64 - 37.16) = mr (185.95 - 86.95)
mr = 31.324 lbm/sec
(b) Entropy generation rate
Sgen = mf (S1 - S2) + mr (S4 - S3)
= 27.818 (0.2772 - 0.0775) + 31.324 (0.4414 - 0.24075)
= 11.84 Btu/sec.R°