Heat Exchangers - 1

# Heat Exchangers - 1 Notes | Study Heat Transfer - Mechanical Engineering

## Document Description: Heat Exchangers - 1 for Mechanical Engineering 2022 is part of Heat Exchangers for Heat Transfer preparation. The notes and questions for Heat Exchangers - 1 have been prepared according to the Mechanical Engineering exam syllabus. Information about Heat Exchangers - 1 covers topics like and Heat Exchangers - 1 Example, for Mechanical Engineering 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Heat Exchangers - 1.

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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––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

Do = 1.66 in. = 0.138 ft.

S= πD2 /4 = 0.0104 ft (cross-sectional area for flow)

Di = 1.38 in = 0.115 ft.

the outside pipe :

Schedule 40 2” pipe

Di = 2.07 in = 0.115 ft.

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

The Design Equation is

The overall heat transfer coefficient, Uo , 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

Tb = 140˚F,  ρ= 52.3 lbm/ft3 , Cp = 0.45 BTU/lb-°F

kb = 0.085 Btu / h · ft ·°F

μb = 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

hi = 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

Doi = 0.138 ft and Dio = 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

Wwater = 9300 lbm/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 Tb 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 Re0.8Pr0.4 = 127

so that the external film coefficient, ho , is

ho = 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 Qh = 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 ΔTlm are different.

Counter current

T(water) = 99

T1 (benzene) = 180

T(water) = 70

T2 (benzene) = 100

ΔT= 81

ΔT2 = 30

ΔTlm = 51

L = 74

Co-current

T(water) = 70

T1 (benzene) = 180

T(water) = 99

T2 (benzene) = 100

ΔT= 110

ΔT2 = 1

ΔTlm = 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.

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