In an industiial plant a shell-and-tube heat exchanger is heating water initiaPy

Question

In an industiial plant a shell-and-tube heat exchanger is heating water initiaPy at 120°F by means of steam condensing at 280°F on the outside of the tubes. The heat exchanger has 50 steel tubes (ID = 0.052 ft and OD= 9.070 ft) in a tube bundle which is 10 ft long. The water flows through the tubes, while the steam condenses in the shell. Devise a MathCad solution to calculate the water outlet temperature and the tube-side pressure drop as the water dow rate is varied from 0 1bm/hi to 600,000 1bm/In. Use ‹r — 0.4 for the pressure-drop calcula- tions, A reasonable transition Reynolds number is 1,500. Plot the results in terms of the water mass flow rate; clearly show the change in performance and parameters at transition. (Con- sider mass flow rates that yield Reynolds numbers just above and just below the transition limit.) Useful parameters are (1) the water outlet temperature, (2) the rating, (3) the water- side connective heat transfer coefficient, and (4) the water-side pressure drop. Discuss the op- eration of the heat exchanger as the mass flow rate is varied. Be sure to state all assumptions.

The properties of water can be considered as constant, and any corrections due to property variations can be neglected.

Explanation / Answer

AS INFORMATION GIVEN IN QUESTION WE HAVE

MASS FLOW RATE OF HOT WATER = mh=100000 lb/hr

MASS FLOW RATE OF COLD WATER= mc= 200000 lb/hr

ENTERING TEMPERATURE OF HOT WATER = Th1= 160 degree F

ENTERIMG TEMPERATURE OF COLD WATER= Tc1= 90 degree F

EXITING TEMPERATURE OF COLD WATER= Tc2=120 degree F

LENGTH OF STEEL TUBE =20ft

OD OF TUBE= .75in = .0625 ft , {Ro=(.75)/2 = .375 in =.03125ft}

ID of TUBE= .62in =.051666ft , { Ri=(.62)/2 = .31 in =.025833ft}

THERMAL CONDUCTIVITY K= 25.9 btu/(ft*hr* F)

ho= 200 btu/(ft2*hr*F)

hi =200 btu/(ft2*hr*F)

FIRST ANSWER

AS WE KNOW THAT HEAT FLOWS ON THIS LAW

HEAT TAKEN = HEAT LOST

SO WE FIND EXITING TEMPERATURE OF HOT WATER AS SHOWN IN FIGURE BELOW

NEXT WE FIND OUT THE OVERALL HEAT TRANSFER COEFFICIENT FOR HEAT EXCHANGER

NEXT WE FIND OUT THE LMTD FOR COUNTER FLOW HEAT EXCHANGER AS GIVEN BELOW

and WE KNOW AREA FOR HEAT TRANSFER IN COUNTER SHELL HEAT EXCHANGER

WHERE Q =

wher Cp=1.799 BTU/lb

and mh=200000lb/hr =55.55 lb/sec

Th1= 160 degree F

Th2= 100 degree F

the Q=5989.6343 BTU/sec

and U = OVERALL HEAT TRANSFER COEFFICIENT=17.714

PUT THESE VALUES IN THIS FORMULA =

AREA FOR HEAT TRANSFER =

=15.57

second part

no of tubes required   

n=

=

=3.8

4 (approx.)

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