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:

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

Disclaimer: I work for Altair. mechGuru.com is my personal blog. Although i have tried to put my neutral opinion while writing about different competitor's technologies, still i would like you to read the articles by keeping my background in mind.

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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?

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

Very useful thanks!

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.