How to calculate Heat Dissipation of PLC ?

[naishadgosai]

Dear All,

Back to college days,
can anyone give me a simple formula to calculatr a Heat dissipation of the PLC / DI / Do / AI cards / modules.

All specifications gives rated voltage, inrush current, i2t, current consumption, power loss. from this how we can calculate?????or need some other data.

Regards,
naishad

 

[Bob O]

I "think" the power loss is the heat rejected into the enclosure by the device.

 

[naishadgosai]

Power consumption = heat dissipation.......true?

 

[TConnolly]

100% of the power consumed is converted to heat at 100% efficiency. Electrical watts consumed equals heat watts produced. A power supply using 25 watts produces 25 watts of heat. An input point that consumes .005 watts produces .005 watts of heat. A 2 amp triac output that consumes 3.5 watts produces 3.5 watts of heat.

 

[panic mode]

what exatly you need to find out? my guess is that you are estimating whaty kind of cooling is needed for enclosure.

there is no perfect answer since some of the power will be disipated outside enclosure or converted to other form of energy (light for example etc). the best you can do without going to extreme (lab conditions) is estimate.

another thing that is hard to get straight answer would be ambinet temperature in target area (even if it's indoors, many plants don't have airconditioning for the shop floor etc.).

one thing mentioned above is efficiency. if you have power supply with 80% efficiency and 100W load then this will need 100W/0.8=125W of power. this means that this power supply alone will produce 125W-100W=25W of heat. depending where is the load, some or all of the 100W can also contribute to total amount of heat disipated.

in case of output card you will have two factors - disipation of the card itself plus disipation of each load (and duty cycle of each output). if the loads are ouside enclosure, no problem.

to get the toal for one panel, you simply add all heat losses...
some components will not even list efficiency or loses (like some transforers etc).

 

[Bob O]

Here is a link that may help for sizing the cooling if this is what you're after...
http://www.hoffmanonline.com/produc...t_1=34&cat_2=2383&SelectCatID=2383&CatID=2383

 

[David Reyes]

Reference

Ia have the same question, looking in the Net I foun this link.
http://www.cabinetcooler.info/heatload.html
You need to find the power in watts convert to BTU/H, so, check the link...

Dear All,

Back to college days,
can anyone give me a simple formula to calculatr a Heat dissipation of the PLC / DI / Do / AI cards / modules.

All specifications gives rated voltage, inrush current, i2t, current consumption, power loss. from this how we can calculate?????or need some other data.

Regards,
naishad

 

[leitmotif]

Power consumption = heat dissipation.......true?

Power in = power out + inefficiency

the power lost in inefficiency can be quantified in terms of watts or BTU (British Thermal Unit)
1 Kw = 3412 BTU

So if a PLC is rated at drawing 100 watt and has an efficiency of 95% then heat lost is 5 watt or approx 17 BTU.

IF the PLC can dissipate the heat ie shed it it will stay cool if not it overheats and burns up. Normally the PLC will shed the heat to the enclosure - if enclosure cannot shed PLC heat plus the heat from other devices in it then it overheats.

SO total heat load of enclosure is the sum of all its items and is equal to the amount of cooling needed. You should allow for solar heating if outside and for other ambient conditions.

Hoffman has some infor on this.

Dan Bentler

 

[allentcm]

You can get the heat dissipation information from the data sheet of the PLC and the I/O modules.

For example for all Allen-Bradley I/O modules you just need to refer to the specification:

 

[shooter]

measure the amps when nothing is connected , put all outputs, inputs on for even better result.
The specs only give maximum, however that is also a good way to find the amount of cooling you will need.
Do not forget power supply and all cablings in the cabinet, and displays, lights, relays etc.

 

[Jeff23spl]

100% of the power consumed is converted to heat at 100% efficiency. Electrical watts consumed equals heat watts produced. A power supply using 25 watts produces 25 watts of heat. An input point that consumes .005 watts produces .005 watts of heat. A 2 amp triac output that consumes 3.5 watts produces 3.5 watts of heat.

Misunderstanding here, power can be modified but never be created.....

Heat generating is the reverse of efficiency x capacity.

for a theoric device of 100% efficiency, no heat is generated....

For a 4amps 24vdc 100 watts power supply and 80% efficiency:

The output = 98w (100w for easiest calcul) given to others device (Maybe a 100w heater but for the power supply generation this portion = 0w) To give this power with 80% eff power supply need to suck 125w from the line so 25w is lost in heat production.....

 

[Jeff23spl]

Do someone has the exact control panel heat dissipation formula:

Including:
Heat generation
Differential temperature from inside to outside
Panel surface

that give inside temperature rise ?

I found calculation diagram at Hoffman site but i need to show the proof to &?%$#@ inginneer

 

[peterjung]

Do someone has the exact control panel heat dissipation formula:

Including:
Heat generation
Differential temperature from inside to outside
Panel surface

that give inside temperature rise ?

I found calculation diagram at Hoffman site but i need to show the proof to &?%$#@ inginneer

Have a look in the Appendix of this document:

http://www.rittal-corp.com/literature/literature_view.cfm?id=622&sbu=03&action=view

This gives two formulas:

1. Qe = Qv - Qs

or:

Required Cooling Power = Heat Generated - Heat Dissipated by enclosure.

and:

2. Qs = kA(Ti - Tu)

k = heat transfer coefficient (5.5W/m2 for sheet steel)
A = Surface area of enclosure.
Ti = internal temperature.
Tu = external ambient temperature.

An important point to remember is that if the ambient temperature is higher than the internal temperature the heat dissipation of the enclosure becomes negative - i.e. changes to become heat absorption.

The example in the Appendix is used to calculate how much cooling is required to maintain the internal temperature at a specific value. However this can also be used to calculate temperature rise based on known ambient temperature and heat output of the panel components.

To calculate the temperature rise with no cooling:

assume:
A = 0.4m2
Tu = 60 deg C
Qv = 14W

These are all values that you should know or can estimate with some accuracy.

With no cooling then Qe = 0

Qe = Qv - Qs = 0

Qv = Qs = kA(Ti - Tu)

14 = 5.5 x 0.4 (Ti - 60)

14 = 2.2 (Ti - 60)

Ti = 60 + 6.4

Ti = 66.4

Therefore in this example at an ambient temperature of 60 degree C the temperature rise inside the enclosure for the given conditions will be 6.4 deg C.

 

[Ahmedsafaa]

how can i size the Fan

this is great, but is there a link between Fan size and heat dissipation? and how can i find the proper fan after calculate heat dissipation?

Have a look in the Appendix of this document:

http://www.rittal-corp.com/literature/literature_view.cfm?id=622&sbu=03&action=view

This gives two formulas:

1. Qe = Qv - Qs

or:

Required Cooling Power = Heat Generated - Heat Dissipated by enclosure.

and:

2. Qs = kA(Ti - Tu)

k = heat transfer coefficient (5.5W/m2 for sheet steel)
A = Surface area of enclosure.
Ti = internal temperature.
Tu = external ambient temperature.

An important point to remember is that if the ambient temperature is higher than the internal temperature the heat dissipation of the enclosure becomes negative - i.e. changes to become heat absorption.

The example in the Appendix is used to calculate how much cooling is required to maintain the internal temperature at a specific value. However this can also be used to calculate temperature rise based on known ambient temperature and heat output of the panel components.

To calculate the temperature rise with no cooling:

assume:
A = 0.4m2
Tu = 60 deg C
Qv = 14W

These are all values that you should know or can estimate with some accuracy.

With no cooling then Qe = 0

Qe = Qv - Qs = 0

Qv = Qs = kA(Ti - Tu)

14 = 5.5 x 0.4 (Ti - 60)

14 = 2.2 (Ti - 60)

Ti = 60 + 6.4

Ti = 66.4

Therefore in this example at an ambient temperature of 60 degree C the temperature rise inside the enclosure for the given conditions will be 6.4 deg C.

 

[Ahmedsafaa]

can i know what fan size i need after calculate heat dissipation?

this is great, but is there a link between Fan size and heat dissipation? and how can i find the proper fan after calculate heat dissipation?