Procedure of Heat Exchanger Design and Sizing Calculation Simplified with Example

The heat exchanger (HX) finds numerous applications throughout the wide spectrum of industry.

 

Three main types of HX are:

 

1.Plate heat exchanger

2.Tube heat exchanger

3.Fin and tube heat exchanger

 

Although, the detail design methodologies are different for the different types of HX but the basic steps required for sizing any HX or coolers can be sumerised as below:

 

a. Design thermal load calculation through energy balance equation: The heat balance calculation equation is used here to find out the design heat load or cooling load value, this is the amount of heat to be removed (or added) by the heat exchanger. Typically the design heating (or cooling) load is expressed in terms of kilowatts (KW).

 

The required equation here is:

 

H=M*Cv*dt……………………………….eq.1

Where,

H – Heat load to be removed (or added) (KW)

M – Mass of the fluids (flowing per unit time ) from where the heat to be removed (or added) (Kg/Sec)

Cv- Specifc heat capacity of the fluid (from where the heat to be removed or added) (KJ/Kg-C)

 

 

b. LMTD calculation: The log mean temperature difference (LMTD) is calculated based on the required inlet and outlet temperatures to/from the heat exchanger.

 

The required equation here is:

 

LMTD=[(T1-t2)-(T2-t1)]/ln[(T1-t2)/(T2-t1)]…………………..eq.2

Where,

T1,T2 – Inlet and outlet temperature of primary fluid

t1,t2 – inlet and outlet temperature of secondary fluid

 

 

 

c. Required heat transfer area calculation: The design heat load value and the LMTD value is used in the final step of the heat exchanger sizing calculation t find out the minimum required surface area.

 

The required equation here is:

 

A=H/(U*LMTD)…………………………………eq.3

 

 

Where,

U – Heat transfer coefficient (W/m2-C)

 

Heat Exchanger Design / Sizing Calculation Example:

Calculate the minimum heat transfer surface area required for the following heat exchanger:

heat exchanger design example

Given:

Specif heat capacity of water, Cv = 4.2 KJ/Kg-degreeC

Heat transfer coefficient for the heat exchanger, U= 2000 W/m2-degreeC

Flow rate of water, M = 20 kg/s

 

Solution:

 

Use the energy balance calculation equation eq.1 to find out the design heat load:

 

H=[20*4.2*(45-40)]=420 KW

 

 

Now, use the LMTD calculation equation eq.2:

 

LMTD=[(70-45)-(60-40)]/ln[(70-45)/(60-40)]=0.223

 

 

Further use the eq.3 to find out the minimum required heat transfer area for the heat exchanger design calculation example:

 

A=420/(2000*0.223)=0.94 sq.m

 

Shibashis Ghosh

Hi, I am Shibashis, a blogger by passion and an engineer by profession. I have written most of the articles for mechGuru.com. For more than a decades i am closely associated with the engineering design/manufacturing simulation technologies. I am a self taught code hobbyist, presently in love with Python (Open CV / ML / Data Science /AWS -3000+ lines, 400+ hrs. )

This Post Has 4 Comments

  1. Name

    If
    T1 is 40 deg C (Hot fluid)
    T2 is 35.5 deg C (Hot fluid)
    t1 is 32.5 deg C (Cold fluid)
    t2 is 37 deg C (Cold fluid)

    what would be the LMTD?

    1. MechGuru

      Use the LMTD equation discussed in the article (eq.2), you will be able to solve it by yourself.

  2. Barry Johnston @GreenGrounded

    Very useful thanks!

  3. James Sugden

    I don’t follow your LMTD. The result of 0.223 doesn’t equal the left hand side of the equation but only the denominator in the example above. Also for the SI unit of power shouldn’t it be in Watts? Therefore multiplying 420 by 1000. Giving final answer of 9.36m^2.

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.