Standardista 14511-3
Heating capacityThe heating capacity of air-to-water, water-to-water heat pumps and liquid chilling packages shall be determined
in accordance with the direct method at the water or brine heat exchanger, by determination of the volume flow of
the heat transfer medium, and the inlet and outlet temperatures, taking into consideration the specific heat
capacity and density of the heat transfer medium.
For steady state operation, the heating capacity shall be determined using the following formula:
P q c t H = × ρ ×
p∆ (1)
where
PH is the heat capacity, in Watts;
q is the volume flow rate, in cubic metres per second;
ρ is the density, in kilograms per cubic metre;
cp is the specific heat at constant pressure, in joules per kilogram and kelvin;
∆t is the difference between inlet and outlet temperatures, in kelvin.
The heating capacity shall be corrected for the heat from the fan or pump:
− if the fan or pump at the indoor heat exchanger is an integral part of the unit, the same power (calculated
in 4.1.5.1 or 4.1.6.1) which is excluded from the total power input shall be also subtracted from the
heating capacity;
− If the fan or pump at the indoor heat exchanger is not an integral part of the unit, the same power
(calculated in 4.1.5.2 or 4.1.6.2) which is included in the effective power input shall be also added to the
heating capacity.Eli yritetään tätä uudestaan.
Eli jos pumppu on osa konetta ja tehoa ei ole huomioitu ottotehossa tulee (4.1.6.1):ssä laskettu teho vähentää lämmityskapasiteetista.
Jos pumppu ei ole osa konetta ja teho on huomioitu ottotehossa tulee 4.1.6.2:ssa laskettu teho lisätä lämpökapasiteettiin.
Menikö nyt oikein? pahis että on meikäläisen englannille hankalaa tekstiä vai mikä tässä mättää? Kai tuon nyt yksinkertaisemminkin voisi todeta.
Sitten 4.1.6 Power input of liquid pumps
4.1.6.1 If a liquid pump is an integral part of the unit, only a fraction of the input to the pump motor shall
be included in the effective power absorbed by the unit. The fraction which is to be excluded from the total
power absorbed by the unit shall be calculated using the following formula:
4.1.6.2 If no liquid pump is provided with the unit, the proportional power input which is to be included in
the effective power absorbed by the unit, shall be calculated using the following formula:
η
i q × ∆p [W]
η is 0,3 by convention;
∆pe is the measured available external static pressure difference, in Pascals;
q is the nominal water flow rate, in cubic meters per second.
Sama kaava selityksillä
The shaft power - the power required transferred from the motor to the shaft of the pump - depends on the efficiency of the pump and can be calculated as
Ps = Ph / η (2)
where
Ps = shaft power (kW)
η = pump efficiency
Ja tuossa sitten taidetaan laskea pumpun teho olettaen että sen hyötösuhde on 0,3.
