A counterflow heat exchanger is being used to cool a flow of

A counterflow heat exchanger is being used to cool a flow of hot water using a cold 20% ethylene glycol solution. The hot water enters at 80 degree F with a volumetric flow rate of 400 gpm. The ethylene glycol solution enters the heat exchanger at -10 degree F and leaves at 15 degree F. The ethylene glycol solution is flowing at a volumetric flow rate of 300 gpm. The overall heat transfer coefficient for the heat exchanger is found to be 155 Btu/h.ft^2.degree F. Determine the following: Outlet temperature of the water (degree F) Heat transfer rate between the water and ethylene glycol solution (Btu/h) Required heat transfer area of the heat exchanger (ft^2) Solve Problem 5.12 for a parallel-flow heat exchanger. Solve Problem 5.12 for a STHX with 1-shell pass and 2-tube passes. Problem 5.12 for a CFHX where both fluids are unmixed.

Solution

1)OUT LET TEMPERATURE OF WATER

ON CONVERSION GPM TO kg/Hr

MASS FLOW RATE OF HOT WATER=400GPM=90909Kg/Hr

1kg/hr=0.0044gpm

MASS FLOW RATE OF COLD ETHYLENE GLYCOL=300GPM=71748Kg/Hr

HOT WATER INLET TEMPERATURE=80 DEGREE FAHREIGHNHEAT=26.6DEGREE CENTIGRADE

ETHYLENE GLYCOL INLET TEMPERATURE=-10DEGREE FAHREIGHNHEAT=-23.5 DEGREECENTIGRADE

ETHYLENE GLYCOL OUTLET TEMPERATURE=15DEGREE FAHREIGHNHEAT=-9.5 DEGREECENTIGRADE

A)HOT WATER OUT LET TEMPERATURE=?

FROM ENERGY BALANCE EQUATION

M_H*C_H*(T_HI-T_HO)=M_C*C_C*(T_CO-T_CI)

M_C=MASS OF COLD FLUID=ETHYLENE GLYCOL

C_C=SPECIFIC HEAT OF COLD FLUID=202KJ/KG

90909*4.2*(26.06- T_HO) =1748*2.2*(-9.5-(-23.5))

T_{HO =} HOT WATER OUT LET TEMPERATURE=20.9 DEGREE CENTIGRADE

B)HEAT EXCHANGE BETWEEN WATER AND ETHYLENE GLYCOL

Q= M_H*C_H*(T_HI-T_HO) =90909*4.2*(26.06- 20.9 =604544.15watts

C)AREA OF HEAT EXCHANGER FOR SHELL AND TUBE WITH 1 SHELL PASS AND 2 TUBE PASS

Q=U*A*LMTD

LMTD=2- 1/ln(2/ 1)

1=T_Ho-Tc_O

2=T_Hi-Tc_O

LMTD=26.06-(-9.5)-(20.9-(-23.5))/ln(26.6-(-9.5))/20.9-(-23.5))

=401 degree centigrade =674 degree fahreighn heat

AREA OF HEAT EXCHANGER=Q/U*LMTD

U=OVER ALL HEAT TRANSFER=155BTU/h.FT^{2 0}F =1.92 W/M^{2 0}C

AREA=604544.15/674*1.92=467.17M²