Inst ToolsInst ToolsInst Tools
  • Courses
  • Automation
    • PLC
    • Control System
    • Safety System
    • Communication
    • Fire & Gas System
  • Instrumentation
    • Design
    • Pressure
    • Temperature
    • Flow
    • Level
    • Vibration
    • Analyzer
    • Control Valve
    • Switch
    • Calibration
    • Erection & Commissioning
  • Interview
    • Instrumentation
    • Electrical
    • Electronics
    • Practical
  • Q&A
    • Instrumentation
    • Control System
    • Electrical
    • Electronics
    • Analog Electronics
    • Digital Electronics
    • Power Electronics
    • Microprocessor
  • Request
Search
  • Books
  • Software
  • Projects
  • Process
  • Tools
  • Basics
  • Formula
  • Power Plant
  • Root Cause Analysis
  • Electrical Basics
  • Animation
  • Standards
  • 4-20 mA Course
  • Siemens PLC Course
Reading: How to Calculate Heat Load in Electrical/Electronic Panel Enclosure
Share
Font ResizerAa
Inst ToolsInst Tools
Font ResizerAa
  • Courses
  • Design
  • PLC
  • Interview
  • Control System
Search
  • Courses
  • Automation
    • PLC
    • Control System
    • Safety System
    • Communication
    • Fire & Gas System
  • Instrumentation
    • Design
    • Pressure
    • Temperature
    • Flow
    • Level
    • Vibration
    • Analyzer
    • Control Valve
    • Switch
    • Calibration
    • Erection & Commissioning
  • Interview
    • Instrumentation
    • Electrical
    • Electronics
    • Practical
  • Q&A
    • Instrumentation
    • Control System
    • Electrical
    • Electronics
    • Analog Electronics
    • Digital Electronics
    • Power Electronics
    • Microprocessor
  • Request
Follow US
All rights reserved. Reproduction in whole or in part without written permission is prohibited.
Inst Tools > Blog > Instrumentation Design > How to Calculate Heat Load in Electrical/Electronic Panel Enclosure

How to Calculate Heat Load in Electrical/Electronic Panel Enclosure

Last updated: August 5, 2019 3:07 pm
Editorial Staff
Instrumentation Design
5 Comments
Share
4 Min Read
SHARE

Total heat load consists of the heat transfer from outside your panel and from the heat dissipated inside the control unit.

Useful terms & Conversions:

1 BTU/hr = 0.293 watts
1 BTU/hr – 0.000393 horsepower
1 Watt = 3.415 BTU/hr
1 horsepower = 2544 BTU/hr
1 Watt = 0.00134 horsepower
1 Square Foot = 0.0929 square meters
1 Square Meter = 10.76 square foot

Typical fan capacity:

4″ fan: 100 CFM (2832 LPM)
6″ fan: 220 CFM (6230 LPM)
8″ fan: 340 CFM (9628 LPM)
10″ fan 550 CFM (15574 LPM)

BTU/hr. cooling effect from fan 1.08 x (temp. inside panel in ºF – temp. outside panel in degrees F) x CFM

Watts cooling effect from fan: 0.16 x (temp. inside panel in ºC – temp. outside panel in degrees C) x LPM

Calculating BTU/hr. or Watts:

  1. Determine the heat generated inside the enclosure. Approximations may be necessary. For example, if you know the power generated inside the unit, assume 10% of the energy is dissipated as heat.
  2. For heat transfer from the outside, calculate the area exposed to the atmosphere except for the top of the control panel.
  3. Choose the internal temperature you wish to have, and choose the temperature difference between it and the maximum external temperature expected.
  4. From the conversion table that follows, determine the BTU/hr. per square foot (or watts per square meter) for the temperature difference.
  5. Multiply the panel surface area times the BTU/hr. per square foot (or watts per square meter) to get the external heat transfer in BTU/hr or in watts.
  6. Sum the internal and external heat loads calculated.
  7. If you do not know the power used in the enclosure but you can measure temperatures, then measure the temperature difference between the outside at current temperature, and the present internal cabinet temperature.
  8. Note the size and number of any external fans. Provide this information to Nex FlowT to assist in sizing the appropriate cooling system.
Temperature Difference in Deg F BTU/hr./sq. ft. Temperature Difference in Deg C Watts/sq.m
5 1.5 3 5.2
10 3.3 6 11.3
15 5.1 9 17.6
20 7.1 12 24.4
25 9.1 15 31.4
30 11.3 18 39.5
35 13.8 21 47.7
40 16.2 24 55.6

Example:

The control panel has two frequency drives totaling 10 horsepower and one module rated at 100 watts. The maximum outside temperature expected is ºC. The area of the control panel exposed sides, except for the top is 42 square feet or 3.9 square meters. We want the internal temperature to be ºC.

Total internal power is 10 hp x 746 watts/hp – 7460 plus 100 watts = 7560 watts.
Assume 10% forms heat = an internal heat load of 756 watts.

Or

Total internal power is 10 hp x 2544 BTU/hp = 25440 BTU/hr plus 100 watts x 3.415 BTU/hr/watt = 25782 BTU/hr.

Assume 10% forms heat = an internal heat load of 2578 BTU/hr.

External heat load: The temperature difference between the desired temperature and the outside is ºC. Using the conversions (and interpolating where necessary) we multiply the area by the conversion factor:

42 sq. ft x 3.3 – 139 BTU/hr or 3.9 sq. m x 10.3 = 40 watts

Total Heat Load: 756 + 40 – 796 watts or 2578 + 139 – 2717 BTU/hr.

Don't Miss Our Updates
Be the first to get exclusive content straight to your email.
We promise not to spam you. You can unsubscribe at any time.
Invalid email address
You've successfully subscribed !

Continue Reading

Redundant Power Supply for Critical Field Panels
Draw a ladder logic circuit for the electric motor of an air compressor ?
Package System Architecture – Control & Instrumentation
Turbine-Compressor System Architecture
Voting Concept in Package Safety System
Level Instruments Design Rules
Share This Article
Facebook Whatsapp Whatsapp LinkedIn Copy Link
Share
5 Comments
  • mayank says:
    July 22, 2019 at 12:40 pm

    I’m was searching Heat Load in Electrical Enclosure box!
    And I’m here ?!

    Reply
    • reality says:
      September 1, 2021 at 8:09 pm

      Gee, you don’t sound like the author supporting their own article, at all….. LOL

      Reply
  • FAHMI QASEM says:
    September 14, 2022 at 5:33 pm

    BTU/hr. cooling effect from fan = 1.08 x (temp. inside panel in ºF – temp. outside panel in degrees F) x CFM
    the other formula
    Watts cooling effect from fan= 0.16 x (temp. inside panel in ºC – temp. outside panel in degrees C) x LPM, these two fomulas are not matching, let say that the temperature difference between inside & outside a panel is 10 c, & an 4” fan(100CFM) accordingly
    BTU/h= 1.08*18*100= 1944 which is =570 w

    when apply the other formula in watt, the result will be
    Watt =0.16*10*2832= 4531
    to get the same result the constant 0.16 should be 0.02
    please check

    Reply
  • Ryan says:
    November 14, 2024 at 5:55 pm

    All of the temperatures in the example are blank. Can you please revise the article to include these values?

    Reply
  • Luqman says:
    December 16, 2024 at 9:18 am

    may i know the reference to consider 10% of internal power consumption as heat load? why not more or less and why 10% only.

    Reply

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Stay Connected

128.3kFollowersLike
69.1kFollowersFollow
210kSubscribersSubscribe
38kFollowersFollow

Categories

Explore More

What is a Mimic Panel ?
Explain Operation of the Lamp Circuit ?
Temperature Gauges and Elements : Detailed Specifications
Instrumentation Books Download
Instrumentation Air System Design
How to Fill up Instrument Datasheet? – Pressure Gauge Specifications & Standards
Magnetic Flow Meter Corrosion Effects
System Architecture and Process Control Systems Philosophy

Keep Learning

Instrumentation Role during Steam Blowing

Instrumentation Role during Steam Blowing

Control System Vendor Internal Test Procedures

Vendor View of Supplying Industrial Process Control & Safety Systems

Loop Diagram

15 Loop Diagram Questions

Pneumatic Piping Design and Specification

Pneumatic Piping Design and Specification

Field Instruments

Is Vacuum affects Field Instruments in Industries?

Instrumentation and Control Engineer Subject

Competency Factors of Instrument & Control Engineer (Design Engineering)

Diaphragm seal for pressure measurement device

When to use a Diaphragm Seal?

Flow Control Valves

Difference Between a Flow Control & a Needle Valve

Learn More

Optical Torsion Meter Principle

Optical Torsion Meter Principle

Submersible Pressure Transmitter

What is a Submersible Pressure Transmitter? Principle, Advantages

Multiplexers and Demultiplexers Objective Questions

Multiplexers and Demultiplexers Objective Questions

flow-meter-accuracy-calculation

Impact of Flow Meter Accuracy

Instrument Zero and Span Calibration

Instrument Zero and Span Calibration

PLC Instructions

Program Flow Control Instructions in PLC Programming

Studio 5000 Logix Emulate

Simulation of Studio 5000 and FactoryTalk View Studio

Testing of Pneumatic Systems

Panel and Field Pneumatic Controller’ Bumpless Transfer Issues

Menu

  • About
  • Privacy Policy
  • Copyright

Quick Links

  • Learn PLC
  • Helping Hand
  • Part Time Job

YouTube Subscribe

Follow US
All rights reserved. Reproduction in whole or in part without written permission is prohibited.
Welcome Back!

Sign in to your account

Username or Email Address
Password

Lost your password?