Question 14.8: Seasonal Heat Pump Performance Calculated by the Bin Method ......

The solution will use weather and performance data from Table 14.8 whose columns are
1. Center point of temperature bin T_{bin}
2. Heating demand \dot{Q} = K_{tot} × (65°F  −  T_{bin})
3. Heat pump heating capacity at the bin temperature, supplied by the manufacturer
4. Heat pump electric power input at the bin temperature, supplied by the manufacturer
5. COP_{ss} determined as the ratio of columns (3) and (4), this includes defrost penalty
6. Part-load ratio (PLR) using Equation 14.27

PLR = min \{ 1, \frac{\dot{Q}_{cool}}{\dot{Q}_{cool,ss}} \}        (14.27)
7. Part-load factor (PLF) using Equation 14.30

PLF = 1  –  C_{d} \times (1  –  PLR)           (14.30)

8. Heat pump COP under part-load operation when unit cycles = (PLF) × COP_{ss}
9. Heat pump output under part load:
• Above the heat pump balance point, taken to be the heating demand \dot{Q} (column 2)
• Below the heat pump balance point, it is equal to the heat pump capacity (given from manufacturer’s data—column 3)
10. Heat pump electric input under part load (column 9/column 8)
11. Auxiliary power; the positive difference, if any, between heating demand \dot{Q} (column 2) and heat pump output under part load (column 9)
12. Heating system COP under part load given by \dot{Q} plus the auxiliary power divided by the sum of auxiliary power and heat pump input (column 2/[column 10 + column 11]).
The energy calculations simply involve multiplying the loads by the number of hours for each temperature bin. These calculations are summarized in Table 14.9. We will work through the calculations for the 22°F bin in detail to clarify the process. The first two columns of this table are the bin weather data. The third column is the heating energy by bin. For this bin, the bin energy is

Since 22°F is below the balance point, the heat pump capacity is less than the load, as shown in Table 14.8. The heat pump output is

From Table 14.8, the COP at this temperature is 1.528. Therefore, the electricity input during this temperature bin is

Because the heat pump cannot meet the load, some auxiliary heat is needed

Finally, the total electric input is the sum of the heat pump and supplemental electricity requirements

The rest of Table 14.9 is completed in this manner. The bottom line in the table contains energy totals. With the heat pump, the total electricity requirement is 66.1 MBtu/year (19,360 kWh/year). If pure resistance heating was used, the total electricity requirement would be 98.36 MBtu/year (28,820 kWh/year). The variations in the energy drawn by the heat pump system, which includes the heat pump and the electric auxiliary heat, are shown in Figure 14.22 for different bin temperatures.
The SPF for the heat pump is the seasonal output divided by the seasonal input to the heat pump:

The SPF for the heating system is the seasonal heat load plus the auxiliary heater electric input divided by the seasonal input to the heat pump and the auxiliary heater:

Comments
This example has shown how the part-load characteristics of air source heat pumps are influenced by ambient temperature, and that this characteristic of the equipment must be considered in annual energy calculations.
The advantage of a constant-temperature heat source is apparent from this example. If groundwater or building exhaust air (both essentially at constant temperature) were used as the heat source rather than outdoor air, one can avoid the dropoff in capacity that occurs in the air source device just when heat is most needed.