Heat Pump Energy Analysis

In a heat pump cycle, obtaining Q0 kcal/h of heat from a low-temperature heat source (outdoor air or circulating water, both of which are higher than the evaporation temperature t0) consumes mechanical work ALkcal/h, while supplying Q1 kcal/h of heat to a high-temperature heat source (indoor heating system) follows the first law of thermodynamics, that is, Q1=Q0+AL kcal/h. If a heat pump device is not used, The heat converted from mechanical work (or directly heated by electrical energy to a high-temperature heat source) yields ALkcal/h of heat, while using a heat pump device, the high-temperature heat source (heating system) gains more heat: Q1-AL=Q0 kcal/h. This heat is obtained from a low-temperature heat source, and it cannot be obtained without a heat pump device. Therefore, using a heat pump device can save fuel and utilize waste heat. The working cycle of a heat pump is exactly opposite to that of a heat engine. A heat engine uses the energy of a high-temperature heat source to generate mechanical work, while a heat pump transfers the heat from a low-temperature heat source to a high-temperature object by consuming mechanical work. If a heat pump has two identical heat source temperatures, the relationship between them is: φ= Q1/AL=(Q0+AL)/AL= ε+ 1, ε It is the cooling coefficient of the refrigerator. From this, it can be seen that the minimum value of the heat conversion coefficient is φ= 1. In this extreme case ε= 0, Q0=0, that is, no heat is absorbed from the low-temperature heat source.