A shell-and-tube heat exchanger is to be used to cool 25.2 kg/s of water from 38°C to 32°C. The exchanger has one shell-side pass and two tube side passes. The hot water flows through the tubes and the cooling water flows through the shell. The cooling water enters at 24°C and leaves at 32°C. The shell-side (outside) heat transfer coefficient is estimated to be 5678W/(m 2 K). Design specifications require that the pressure drop through the tubes be as close to 13.8 kPa as possible and that the tubes be 18 BWG copper tubing (1.24 mm wall thickness), and each pass is 4.9 m long. Assume that the pressure losses at the inlet and outlet are equal to one and one half of a velocity heat (?V2/gc), respectively. For these specifications, what tube diameter and how many tubes are needed?

Question

GIVEN

A water-to-water shell-and-tube exchanger, hot water in tubes, cooling water in shell One shell and two tube passes Hot water flow rate m

h= 25.2 kg/s Water temperatures

? Hot: Th,in = 38°C Th,out = 32°C

? Cold: Tc,in = 24°C Tc,out = 32°C Shell-side transfer coefficient h o= 5678 W/(m2 K) Pressure drop (?p) = 13.8 kPa Tube wall thickness (t) = 1.24 mm = 0.00124 m Tube length per pass (Lp) = 4.9 m

FIND

The tube diameter (Do) and number of tubes (N)

ASSUMPTIONS

Pressure losses at inlet and outlet (?pii) = 1.5 (? V2/gc) Variation of thermal properties with temperature is negligible Fouling resistance is negligible Thermal resistance of the tube walls is negligible

SKETCH



PROPERTIES AND CONSTANTS

for water at 30°C Density (?) = 995.7 kg/m3

Specific heat (cp) = 4176 J/(kg K)

Thermal conductivity (k) = 0.615 W/(m K)

Kinematic viscosity (?) = 0.805 × 10–6 m2/s

Prandtl number (Pr) = 5.4

Answer 1

This must be corrected





An iterative solution is required. For a first guess, let the tubing be 1-in.-OD for 1-in. BWG 18 tubing: Di = 0.902 in. = 0.0229 m; Do = 0.0254 m that the overall heat transfer coefficient Uo = 1700 W/(m2 K). The rate of heat transfer is



The water velocity in the tubes for 86 tubes is



The Reynolds number is





The overall heat transfer coefficient, neglecting fouling and tube wall thermal resistance,



The pressure drop through the tube is obtained by adding the inlet and outlet pressure drops to



The friction factor f, is for turbulent flow



There is about half of the required pressure drop. Therefore, smaller tubes should be used. For a second iteration, let the tubes be 3/4 in. 18 BWG tubes



Following the same procedure shown above but using Uo = 2000 W/(m2 K) yields



Performing the procedure for the same tubes but using the Uo derived above 2502W/(m 2 K) yields



This will give an even higher pressure drop, therefore, use the 1 in. 18 BWG tubes.