4.1813 actually
Probably an arbitrary maximum acceptable flow speed was multiplied up (or down) by the cross-section of the pipe. Looks like 1 m/s was chosen for this example.I understand the formula, but where do you get the flow rate of 0.14 litres/sec from?
I understand the formula, ?
15mm tube will carry a 6kw load @ 10c dt. (0.14 x 4.2 x 10c = 5.8 kW)
15 mm tube @ 20 dt (0.14 x 4.2 x 20c = 11.6 kW)
That was obvious.Probably an arbitrary maximum acceptable flow speed was multiplied up (or down) by the cross-section of the pipe. Looks like 1 m/s was chosen for this example.
A load of cock and bull.
The same flow rate can't deliver double the energy.
With double the temperature drop then double the power can be achieved
It can, if you double the temperature differential.The same flow rate can't deliver double the energy.
That assumes you are delivering the same amount of heat.Doubling the temperature drop will have a relatively small effect on the heating surface required but will reduce the flow rate by half.
It can, if you double the temperature differential.
I suggest you further educate yourself , I'm not handing it to you on a plate...A load of cock and bull.
The same flow rate can't deliver double the energy.
Doubling the temperature drop will have a relatively small effect on the heating surface required but will reduce the flow rate by half.
CIBSE
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