Heat Exchangers - 1 | Heat Transfer - Mechanical Engineering PDF Download

The Exchanger

The Design Equation for a Heat Exchanger

Problem : Find the Required Length of a Heat Exchanger with Specified Flows: Turbulent Flow in Both Streams 

The design constraints are given in the schematic above. We show this as a countercurrent configuration, but we will examine the cocurrent case as well. The benzene flow is specified as a mass flow rate (in pound mass units), and the water flow is given as a linear velocity. Heat transfer coefficients are not provided; we will have to calculate them based on our earlier discussions and the correlations presented in earlier lectures. The inside tube is specified as "Schedule 40––1-14 inch steel."

Pipe "schedules" are simply agreed-upon standards for pipe construction that specify the wall thickness of the pipe. Perry’s Handbook specifies the following dimensions for

the inside pipe :

Schedule 40 1 1/4” pipe

D_o = 1.66 in. = 0.138 ft.

S_c = πD² /4 = 0.0104 ft²  (cross-sectional area for flow)

D_i = 1.38 in = 0.115 ft. 

the outside pipe : 

Schedule 40 2” pipe

D_i = 2.07 in = 0.115 ft.

To calculate the heat exchanger area, we must find A_o = πDL. We know the diameter; what is the length ?

The Design Equation is 

The overall heat transfer coefficient, U_o , is given by

We can write it as:

To evaluate the parameters of the problem, we need the physical and thermal properties and conditions for flow in the system

T_b = 140˚F,  ρ_b = 52.3 lbm/ft³ , C_p = 0.45 BTU/lb-°F

k_b = 0.085 Btu / h · ft ·°F

μ_b = 0.39 Cp  

Internal Film Resistance

The Nusselt number on the inside of the inner pipe is given by the DittusBoelter equation

so that the film heat transfer coefficient

h_i = 249 Btu/h·ft²·˚F

The heat transfer area per unit length is

so that the inner film resistance is

The other tube dimensions are

D_oi = 0.138 ft and D_io = 0.172 ft

Calculation of the Water Flow Rate 

The hydraulic diameter is

Given the water velocity of 5 ft/s, we can solve for the water flow rate

W_water = 9300 lb_m/h

The Overall Energy balance

(wCp ΔT)_benz = (wCp ΔT)_water

Solving for the outlet water temperature:

7500 (0.45) (100 – 180) = 9300 (1) (70 – Tout)

gives the exit temperature as: Tout = 99˚F

External Film Resistance 

The physical properties of the water must be estimated in order to determine the film heat transfer coefficient in the annular shell. The average water temperature T_b is calculated as 84.7 °F

μ = 0.8 cp, k = 0.34 BTU/h-ft-°F , ρ = 62.4 lb/ft³

so that the Reynolds number can be calculated.

From the Dittus-Boelter equation, the Nusselt number is given as:

Nu = 0.023 Re^0.8Pr^0.4 = 127

so that the external film coefficient, ho , is

h_o = 1270 Btu/h·ft²·˚F

The external area/length is

so that the external film resistance is

Conduction Resistance

The last term in the equation for the overall heat transfer coefficient is

Overall Heat Transfer Coefficient

The overall resistance is

Log-Mean ΔT

Heat Load Q_h = wCpΔT = 7500 (0.45) (180 - 100) = 2.7 x 105 Btu/h

Heating Rate/unit Length

Given the heat load, we can calculate the length of tubing so that

The case we considered was countercurrent flow, but we noted in an earlier example that in co-current flow we could be more fluid. Now is the pipe longer or shorter ?

A Co-current Flow Heat Exchanger

The Design Equation for a Heat Exchanger

The heat loads are identical, the Overall Resistances to heat transfer (UA)^-1 are no different since the film coefficients do not change, but the ΔT_lm are different.

Counter current

T₁ (water) = 99

T₁ (benzene) = 180

T₂ (water) = 70

T₂ (benzene) = 100

ΔT₁ = 81

ΔT₂ = 30

ΔT_lm = 51

L = 74

Co-current

T₁ (water) = 70

T₁ (benzene) = 180

T₂ (water) = 99

T₂ (benzene) = 100

ΔT₁ = 110

ΔT₂ = 1

ΔT_lm = 23.2

L = 163 ft

There are two observations to be made. First that the tube length required for co-current flow is more than twice as long. Secondly that the approach temperature for co-current flow becomes diminishingly small.