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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––114 inch steel."
Pipe "schedules" are simply agreedupon 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 }= πD^{2} /4 = 0.0104 ft^{2 } (crosssectional 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/ft^{3} , 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·ft^{2}·˚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/hft°F , ρ = 62.4 lb/ft^{3}
so that the Reynolds number can be calculated.
From the DittusBoelter equation, the Nusselt number is given as:
Nu = 0.023 Re^{0.8}Pr^{0.4} = 127
so that the external film coefficient, ho , is
h_{o} = 1270 Btu/h·ft^{2}·˚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
LogMean 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 cocurrent flow we could be more fluid. Now is the pipe longer or shorter ?
A Cocurrent 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
T_{1 }(water) = 99
T_{1} (benzene) = 180
T_{2 }(water) = 70
T_{2} (benzene) = 100
ΔT_{1 }= 81
ΔT_{2} = 30
ΔT_{lm }= 51
L = 74
Cocurrent
T_{1 }(water) = 70
T_{1} (benzene) = 180
T_{2 }(water) = 99
T_{2} (benzene) = 100
ΔT_{1 }= 110
ΔT_{2} = 1
ΔT_{lm }= 23.2
L = 163 ft
There are two observations to be made. First that the tube length required for cocurrent flow is more than twice as long. Secondly that the approach temperature for cocurrent flow becomes diminishingly small.
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