This post was updated on 8/29/2008 to clarify a few things
This is a heavily viewed page. Please leave me a comment if this was helpful or if something is missing or confusing. I would like it to be more comprehensive.
I posted a simplified Watt-to-BTU post if you are not looking for server-specific information
Shouts to Lavrik for getting this started.
To calculate heat output (BTU) of your systems, you need to remember (or read below) your high school electrical class. There are three questions to answer.
- How much power am I going to draw?
- How much heat will that power useage generate?
- How much cooling do I need for that heat?
How much power am I going to draw?
Power is measured in Watts and used in math as the letter ‘P’. The formula for calculating power requires Volts (How much juice you can provide) and Amperes (How hard you suck on the straw).
Power in Watts (P) = Volts (V) * Current in Amperes (I) or:
P = V * I
If you know how many Watts your power supplies can feed and you know how much voltage you have, you can figure out your Amperes.
Take a simple server:
– A server with a 1000 Watt power supply.
– 208 Volt service.
– Running at theoretical maximum load.
P = 1000
V = 208
I = ?
I = P/V or 1000/208 = 4.8 Amperes
So now that you know how much you’re tugging on the grid, find out how much heat that will produce.
How much heat will that power usage generate?
BTUs or British Thermal Units are heat measurements. Generally measured in per-hour increments. In common usage 12,000 BTU/h is considered a Ton. This is important because we cool and heat over time so we need to know how much cooling to throw in to the room per hour.
1 Watt will produce 3.412 BTU per hour
To calculate the BTU/heat output of a system, substitute BTUs for Watts and use the following equation:
BTU/h = (V * I) * 3.412
Using the example server above running full blast in a sealed room with no air moving in and out and no heat loss through the walls you can do the math:
BTU/h = 3.412 x 1000 = 3412BTU/h
How much cooling do I need for that heat?
About 1/3 Ton of cooling is required to cool this system. However, your actual requirements will probably be lower because we can assume you won’t run the equipment at full capacity and there is some energy loss through the walls, ceiling and floors.
Using the techniques above, you can make a simple worksheet to calculate your total theoretical maximum cooling requirements. Don’t forget to take in to account other sources of heat. Ambient building temperature (do you have hot walls from the boiler in the next room?), lights, cooling compressor (it should list it’s heat output) and other equipment.
You can probably undercut your requirements if your equipment does not run at full capacity, but be sure you leave enough room to grow and to keep up in case a cooling unit fails or you experience a period of heavy load and you do use your gear at capacity.
Did you find this post useful or have questions or comments? Please let me know!
I have a question
Im trying to calculate the true BTU/Hour output of a geothermal heatpump. Since the compressors energy is being used to move heat and not generate it ( allthough the heat generated by it is added to the net output) how can I calculate BTU output using cfm , and delta T?
Thank yuo
This is a simplistic idea, what also needs to be looked at is the amount of time the unit will be working at or near full load, i.e, diversity factor 1.0 operates all the time at 100% load, diversity factor 0.25 = 25% of the time at 100% load. When you are wrong in estimating the unit run time you are causing heat to build up in the space. Additionally how many pieces of heat producing equipment will be put in the server rack or in the equipment room that also have to be kept cool. Then you start getting into server fan noise getting out of the server room into surrounding areas and how do you provide varible cooling to a variable heat load.
If heat variations occur over the time inside the telecom center,then how do u calculate the BTU? For Example, 1.5 ton A/C is using inside the server room which is not providing necessary cooling due to the the temperature increases 10 degree celsius/ hour then how can u calculate the additional (Btu)cooling capacity required?
Cooling requirements are defined as a function of time. “1.5 ton cooling” would refer to 1.5 ton/hour. If you have an increase in heat over time, (10 degrees per your example) you need to find how many BTU/hour produce 10 degree/hour change in temp and add equivalent cooling. I don’t have the math in front of me, but it shouldn’t be too difficult to find those values. Since my other examples use Watts, you could in theory work backward to determine how much more wattage you are using then your current cooling system can accommodate.
After a little research, there is no conversion of BTU to degree since BTU is a unit of energy and degree is a unit of temperature which depends on the properties of the medium. For example, air will heat at a different rate than water given the same amount of energy put in. What we need to do is use a conversion for for the space being heated. The heat entropy formulas I found here: http://www.themeter.net/conv20_e.htm con convert BTUs for a given space to watts for the same space. Specifically look at “FROM Britsh thermal unitIT foot per hour square foot degree Fahrenheit [BtuIT*ft/(h*ft2*°F)] TO watt per meter kelvin [W/(m*K)]”
The math will get a little tricky, but if you know the size of your room, you can likely figure out the difference in BTUs.
I have 1.5 ton split Aircondition, Now i want to check my Air condition wether is working as 1.5 ton or not?
@Anbu you need a professional to check the unit.
In the example of the server with a 1000 W power supply, even if the sever is working at full load, all of that 1000 W would not be converted into heat. Most of the power would be used by the server actually to do its work. How would you factor this for your cooling capacity calculation?
@sunil
The first law of thermodynamics essentially states that energy cannot be created or destroyed (conservation of energy).
Computer work is an abstract term to describe the switching on and off of transistors and logic controllers within the chips.
Energy enters the server and so it has to leave the server either in the form of heat, motion or potential energy (electricity). Since very little electricity leaves the server down the Ethernet cable and none at all in the case of fiber optics, 100% of the energy is eventually converted to heat or motion (sound). If the room is sound proof, even the sound will eventually become heat. Spinning hard drives will store some energy as momentum, but the bearings are very efficient so once the energy is spent to spin them, very little is required to maintain the rotation as some is lost to heat and friction in the bearings.
i have a question
how much amount of heat is released from the server(computer)and how much amount of eletricity can be produced from it
how much amount of heat is generated from a single server(computer) and how much amount of heat can be generated from it…?
I want to calculate the amount of heat dissipated by a 1 ton Air conditioner per hour ? how to calculate it ?
Praveen,
In common usage 12,000 BTU/h is considered a Ton.
If in case i dont the instrument watt ratings. and i have to find out the dissipation of that instrument.is there any practical method to find out the dissipation if we dont the waatage rating of that instrument.plz help me out by replying on above mail addresss
i want to know how i can calculate 3 phase 378000 cooling capacity screw compressor btu value….
The calculation assume all power convert to heat, which is wrong. A server system draws most of the power to circuitry and fan. That cannot be all considered as heat. Watt to BTU can only be directly converted for device exclusively for heating.
Shing L,
As stated in a previous response:
Energy enters the server and so it has to leave the server either in the form of heat, motion or potential energy (electricity). Since very little electricity leaves the server down the Ethernet cable and none at all in the case of fiber optics, 100% of the energy is eventually converted to heat or motion. In the case of motion, this is wind (fans used to dissapate heat) or sound. Even the sound will eventually become heat as the energy is absorbed by surrounding materials.. It is true that some energy may escape in the form of sound or air motion, but this amount is minimal. Spinning hard drives will store some energy as momentum, but the bearings are very efficient so once the energy is spent to spin them, very little is required to maintain the rotation as some is lost to heat and friction in the bearings.