calc Q at given Pout and dT
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Assumed or measured voltage110Voltfill out
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Measured current RMS20Ampsfill out
Updated:
11/11/2018 13:30:19
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Calculated input power2200Watt
Result: Fill out in 'Input power measured', please be aware that this is not an accurate measurment
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Input power measured2200Wattfill outJoules
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specific heat water (SH)4187 J/kgKWatt
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If quantity of water unknown:
Watt
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delta T40degree K/Cfill out
If you do know the difference between inlet and outlet temperature,
kg or litre for water
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Claimed Output power (Pout) 20,000Wattfill out
and the reactor is claimed to produce 20,000 W,
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time (t)60 secondsfill out
and during 60 seconds you have collected the water,
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calculated quantity (Q) during 60 sec
7.17 kgResult
than that amount of water should be 7.17 kg or litres and
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COP9Result
if the input power is 2200 W, then the COP is 9.
Used formulae:
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If output power is unknown:
Energy (E)=SH*dT*Q
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measured quantity (Q) during 100 sec
10kgfill out
If you have collected 10 litres or kg of water
Pout=E/t
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time (t)100secfill out
during 100 seconds,
Pout= SH*dT*Q/t
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delta T30degree K/Cfill out
and the difference between outlet en inlet temperature is 30 degrees,
Q=Pout*t/(SH*dT)
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Calculated output power12,561WattResult
than the reactor produces 12,561 watts and
dT=Pout*t/(SH*Q)
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COP6Result
with a COP of 6, if the input power is 2200 Watt.
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If output delta T is unknown:
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measured quantity (Q) during 75 sec
5kgfill out
If you have collected 5 kg of water
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time (t)75secfill out
during 75 seconds,
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Claimed Output power (Pout) 20,000Wattfill out
and the claimed output power is 20,000 Watt,
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delta T72
degree K/C
Result
then the difference between the out- and inlet temperatures should be 71.7 degrees,
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COP9Result
and the COP 9 with an input power of 2200 watts.
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Below you will find similar calculations assuming air is used. I believe the required measurements are not easily done. Maybe use a hot air balloon? ;-)
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Air pressure 1020mBarfill out
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Calclated Ro @ 1020 mBar and 500C
0.457kg/m3
The specific mass of air is strongly temperture dependent so must be calculated.
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Input inlet temperature air20Cfill out293K
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Input outlet temperature air500Cfill out773K
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calculated specific heat air (SH)1088 J/kgKCp
Air has not a constant specific heat factor so it must be calculated.
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If quantity of air unknown:
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calculated delta T480degree K/C
If you do know the difference between inlet and outlet temperature,
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Claimed Output power (Pout) 20,000Wattfill out
and the reactor is claimed to produce 20,000 W,
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time (t)60 secondsfill out
and during 60 seconds you should produce,
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calculated quantity (Q) during 60 sec
2.30 kg airResult
2.30 kg air, which equates to about 5.03 m3 @ 500 C
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COP9Result
if the input power is 2200 W, then the COP is 9.
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If output power is unknown:
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measured quantity (Q) during 100 sec
3kgfill out
If you have collected 3 kg or 6.57 m3 of air of 500 C
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time (t)100secfill out
during 100 seconds,
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calculated delta T480degree K/C
and the difference between outlet en inlet temperature is 480 degrees,
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Calculated output power15,674WattResult
than the reactor produces 15,674 watts and
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COP7Result
with a COP of 7, if the input power is 2200 Watt.
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If output delta T is unknown:
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measured quantity (Q) during 75 sec
5kgfill out
If you have collected 5 kg or 10.95 m3 of air
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time (t)75secfill out
during 75 seconds,
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Claimed Output power (Pout) 20,000Wattfill out
and the claimed output power is 20,000 Watt,
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delta T276
degree K/C
Result
then the difference between the out- and inlet temperatures should be 275.6 degrees,
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COP9Result
and the COP 9 with an input power of 2200 watts.
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