Tayters
18-10-2012, 11:28 PM
Customer wants to use an air to water to heat water in a tank. Time to heat up isn't too much of an issue as the water is used in a hostel so has all day or night to heat.
I don't want to get to bogged down in the whys/ do's and don'ts and specifics of the system, just need pointing in the right direction for the theory behind the sizing then I can work stuff out as required.
The tank holds 600lit.
Air to water capacity at ambient encountered 7.1kW
Nominal flow rate 28.7 lit/min.
I want to work out how long it will take to heat the water but can't quite fathom it,
Lets say I want to raise the water from 5*C to 55*C.
Q = mass x shc x dt
Q= 600 x 4.18 x 50
Q = 125,400kJ
Now how do I work out the time the heat pump will take to heat the water.
If the duty is 7.1kW then this is 7.1kJ/s, divide this into 125400 gives 17662 sec or 4.9hrs.
Or...
Lets assume I lose 5K between water into tank and water out.
Flow rate is 28.7lit/min = 0.48lit/sec = .48kg/sec
Q = 0.48 x 4.18 x 5
Q = 10kJ/s added to the tank.
Dividing the same way gives around 3.5 hrs to heat the water.
Second way seem wrong as total duty is 10kJ/s but unit is rated at 7.1kW (therefore 7.1kJ/s)but this way I get to use flow rate which must have some bearing. Sure, td will change, hey maybe 55*C is too ambitious but just looking for the method if anyone can help then check specifics of the air to water system where required.
Thanks for reading,
Andy.
I don't want to get to bogged down in the whys/ do's and don'ts and specifics of the system, just need pointing in the right direction for the theory behind the sizing then I can work stuff out as required.
The tank holds 600lit.
Air to water capacity at ambient encountered 7.1kW
Nominal flow rate 28.7 lit/min.
I want to work out how long it will take to heat the water but can't quite fathom it,
Lets say I want to raise the water from 5*C to 55*C.
Q = mass x shc x dt
Q= 600 x 4.18 x 50
Q = 125,400kJ
Now how do I work out the time the heat pump will take to heat the water.
If the duty is 7.1kW then this is 7.1kJ/s, divide this into 125400 gives 17662 sec or 4.9hrs.
Or...
Lets assume I lose 5K between water into tank and water out.
Flow rate is 28.7lit/min = 0.48lit/sec = .48kg/sec
Q = 0.48 x 4.18 x 5
Q = 10kJ/s added to the tank.
Dividing the same way gives around 3.5 hrs to heat the water.
Second way seem wrong as total duty is 10kJ/s but unit is rated at 7.1kW (therefore 7.1kJ/s)but this way I get to use flow rate which must have some bearing. Sure, td will change, hey maybe 55*C is too ambitious but just looking for the method if anyone can help then check specifics of the air to water system where required.
Thanks for reading,
Andy.