E5.11 Shell and Tube Heat Exchanger

Example 5.11: A shell and tube exchanger has one tube-side pass and one shell-side pass. Calculate the overall heat transfer coefficient under the following service conditions.

For the fluid flowing in the tube:

Specific heat, c = 0.5 Btu/(lb·°F), mass flow rate, w = 5´10⁵ lb/hr,

Absolute viscosity, μ = 1.21 lb/(ft·hr), thermal conductivity, k = 0.075 Btu/(hr·ft·°F)

For the fluid flowing in the shell:

Specific heat, C = 1.0 Btu/(lb·°F), mass flow rate, W = 2´10⁵ lb/hr

Absolute viscosity, μ = 2 lb/(ft·hr), thermal conductivity, k = 0.36 Btu/(hr·ft·°F)

Number of tubes, n = 413, inside diameter, d_i = 0.0516 ft, outside diameter, d_o = 0.0625 ft

Length of a tube, L = 16 ft, tubes arrangement = square, pitch, P_t = 0.083, internal diameter of shell, ID = 2.083 ft, baffle spacing, B = 0.7917 ft.

Solution: Let us first calculate the heat transfer coefficient for the fluid flowing through n tubes.

Mass velocity (mass flow rate per square foot of the flow area), G:

Reynolds number, Re:

Prandtl number, Pr = 8.067.

Film coefficient, h_i:

For the fluid flowing through shell side, flow area can be found using the following technique. The clearance, C, is

And the shell flow area a_s is found to be

The mass velocity through shell is calculated to be

The equivalent diameter for shell can be found by using the formula for a square pitch arrangement of the tubes as follows:

And the Reynolds number can be calculated as

And the Prandtl number is found to be 5.56. The heat transfer coefficient for the shell side can be found by using appropriate correlation and is found to be