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Buildings 2020+. EnErgy sourcEs 5.3.4. Fourth stage – Isobaric evaporation p 4 q0 1 32 h Fig. 5.8. Heat pump evaporation in a p-h diagram (Source: own elaboration) The liquid refrigerant enters the evaporator, where it starts boiling (Fig. 5.8). The heat energy is taken from the surroundings or heat medium. The total output of the evaporator can be determined from the log p-h diagram and expressed by Eq. (5.6): Qo = (h1 – h4) · mR The specific refrigerating capacity is calculated by Eq. (5.7): qo = Qo/mR = h1 – h4 where: QO– evaporator output (kW), h2– refrigerant enthalpy before evaporator (kJ/kg), h3– refrigerant enthalpy after evaporator (kJ/kg), mR– refrigerant mass flow rate (kg/s), qo– specific evaporator output (kJ/kg). 5.3.5. Determining the output coefficient from a p-h diagram (5.6) (5.7) In order to compare different heat pumps, an output coefficient is used. It is consistent with the engine performance and is equal to the ratio between work and condenser output. The amounts of energy converted in the cyclic process can be taken directly as enthalpy differences from the p-h diagram. Thus, the output coefficient for the ideal process can be expressed by Eq. (5.8): ε=Qc =h2−h3 (5.8) 152 Ph−h in 2 1PDF Image | Heat Pumps 978-83-65596-73-4
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