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Thread ID: 98535	2009-03-28 09:13:00	Hot Water Calculation Question	murphfan51 (14762)	PC World Chat

Post ID	Timestamp	Content	User
760186	2009-03-28 09:13:00	Hi All, Have looked everywhere for the answer to this question regarding calculating Hot Water Cylinder power and time . I found this forum while trying to find the answer through Google . Maybe someone here can help . Im an electrician and would like to find out how much energy (kWh) is to be consumed to increase the water temperature up to a certain level . Say if you had a 300 litre cylinder, with a 3kW element, specific type of thermal insulation, and had to increase it from 30 degrees celcius to 60 degrees . Surely there must be a formula on calculating this value? As we know the volume of a Cylinder (in meter cubed) is = pi * diameter squared * height / 4 . . . Then by including the losses (efficiency), element power rating, temperature increase required and maybe some other factors will give us the answer??? Theres also the time, how long would it take to heat the water to 60 degrees from 30 degrees? Thanks for your help, murphfan51	murphfan51 (14762)
760187	2009-03-28 09:15:00	Welcome to Press F1. I know some manufacturers have a "recovery time" specification on their cylinders, which tells you how long it takes to heat a full tank of cold water up to a certain temperature (55 degrees I think). It might be worth finding a few of these figures, as it's likely to be quite an unexact science.	somebody (208)
760188	2009-03-28 10:03:00	Hi All, Have looked everywhere for the answer to this question regarding calculating Hot Water Cylinder power and time . Say if you had a 300 litre cylinder, with a 3kW element, specific type of thermal insulation, and had to increase it from 30 degrees celcius to 60 degrees . Surely there must be a formula on calculating this value? As we know the volume of a Cylinder (in meter cubed) is = pi * diameter squared * height / 4 . . . Then by including the losses (efficiency), element power rating, temperature increase required and maybe some other factors will give us the answer??? Theres also the time, how long would it take to heat the water to 60 degrees from 30 degrees? Thanks for your help, murphfan51 physics 101 was a few years ago but . . . . . . . The specific heat of water is 4 . 186 joules/gram-C Q = mc(DeltaT) T1 = (30c) start temperature T2 = (60c) finish temperature Q = m ( 4 . 186 joules/gram-C ) (T2-T1) Q = 300 litres * 1000 grams * (4 . 186j/g-C)(30C) Q = 300,000*4 . 186*30 = 37674000 Q = 37674000 joules for a 30 degree C rise in temperature A joule is a watt-second so: 37674000 watts will heat 300 litres up by 30C in a second . or 627900 watts will heat 300 litres up by 30C in 60 seconds . or 10465 watts will heat 300 litres up by 30C in 1 hour . or 3488 watts will heat 300 litres up by 30C in 3 hours . I think that's right . . . . .	robsonde (120)
760189	2009-03-30 07:31:00	Ah, thank you, that looks about right. What about efficiency? I will show my boss tomorrow. Thanks!!!:thanks	murphfan51 (14762)
760190	2009-03-30 21:08:00	This is probably one of the few situations where using American numbers is easier. A calorie is defined as the amount of energy required to raise the temperature of one gram of water by one degree Celsius. You've got 300 litres of water, which is close enough to 300kgs. So you need 300,000 calories for each degree change. And you need to raise the temperature by 30 degrees. Oh, and 1 calorie = 4.184 J. Energy = 30*4.184*300,000 = same answer as rob. Efficiency: Think about where the energy goes when you have an inefficient light bulb!	shermo (12739)
760191	2009-03-31 02:08:00	It's not so much 'efficiency' that's involved in the calculation, as the rate of heat loss from the tank. This will deduct a little from the electrical heat input and so the actual heating time would in practice be a little bit longer than the calculated time ignoring heat loss. Not enough to worry about I'd guess, but if the insulation was poor the time difference may be significant.. Only the tank manufacturer could give you this figure, though catalogues may quote heat loss values. The way to measure the loss at any temperature (the higher the temperature, the greater the loss) would be to take a cooling curve, measure average tank temperature against time after the power is turned off. (Not so easy as you cant get into the tank easily, maybe a thermocouple inserted through a special pipe fitting) From the curve a value of dQ\dT can be found for any temperature, and the rate of heat loss is mass x specific heat x dQ\Dt.	Terry Porritt (14)
760192	2009-03-31 18:07:00	Whoops, no one spotted the undeliberate mistake or typo. The slope of the cooling curve is dT\dt, so the rate of heat loss from the tank is: mass(of water) x specific heat x dT\dt	Terry Porritt (14)
760193	2009-04-08 10:12:00	Thanks for that guys. I have showed my boss (as he was asking me if I knew it) but I told him I got it from a person on this site. Thanks:thumbs:	murphfan51 (14762)
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