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Calculating efficiency of a boiler 
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#1
Jun1009, 12:34 PM

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Hi, I've been attempting this problem all week now and I can't seem to even begin to obtain an answer which is highly frustrating, so here's the problem:
An electric boiler is required to raise the temperature of 90 litres of water per hour by 35oC. If the input required to affect this change is 4.3235kW, calculate the efficiency of the boiler. Find the cost of operating the boiler for a period of 5 hours if electrical energy costs 13p/kWhr. 2. Relevant equations I have some equations which could be useful, but I can't seem to apply them to this particular problem, the equations I have looked at are: n (efficiency) = Useful energy output/Total energy input and Maximum theoretical thermal efficiency = (Temperature Hot  Temperature Cold)/Temperature Hot Unfortunately I have no attempted solution to show as I haven't been able to get anywhere with this. I do know however that the temperature will have to be converted into Kelvin (the easy bit). I have thought about considering the Specific Heat Capacity of water but don't know how it would apply to this problem. Any help is greatly appriciated. 


#2
Jun1009, 12:59 PM

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P: 5,196

The specific heat capacity of water tells you how much energy input (in the form of heat transfer) is *required* in order to raise the temperature of a unit mass of water by a unit (e.g. by 1 degree celsius).
Knowing that, and knowing how much water you have to heat by what temperature in a time of one hour, can you see that it is possible to deduce how much energy per unit time (i.e. power) is going into the water to heat it? Can you also see that comparing this to how much energy per unit time (power) is being used by the boiler will give you the efficiency. If the former power level is less than the latter, obviously some power is being lost due to inefficiency. 


#3
Jun1009, 01:50 PM

P: 2

Thanks for that, think it's done the job. My solution to the problem is now this:
From my notes the s.h.c of water is: 4200J/Kg K Using the formula: Q = mc(t1t2) In this case m=90 (as far as i know 1 litre of water has a mass of 1kg) c=4200 and t1t2=35oC (the change in temperature) so.... Q = 90 x 4200 x 35 = 13,230,000 J and 1W = 1J/s So for the amount of energy used per hour, the amount of seconds in one hour must be obtained, which is 3600 seconds. so, 13,230,000/3600 = 3675W/hr Now the effiency can be calculated, using n (efficiency) = useful energy output/Total energy input so, n (efficiency) = 3675/4323.5 = 0.8500057... Multiply by 100 to get percentage efficiency: 0.8500057... x 100 = 85.00057% Does this look about right? 


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