Question 19.7: Dual-Duct System Part-Load Operation A dual-duct conditioned......

The solution involves two steps. First, the airflow at peak load is determined. Since the dualduct system is a constant-volume system, this same flow will apply to part-load conditions.
Second, solve for the heating and cooling rates at part load.

The airflow rate needed for peak heating is found from Equation 19.12

$\dot{V}_{a,heat} = \frac{\dot{Q}_{space,heat}}{\rho_{a} c_{a} (T_{db,supply}  –  T_{db,space})}$               (19.12)

and for peak cooling, the airflow rate is

The higher flow rate (0.59 kg/s) will be used for both heating and cooling since the constant-volume dual-duct system uses a single constant-volume fan.
The supply air temperature entering the zone at part load (3.5 kW cooling) is found from a sensible heat balance

from which the supply air temperature is found to be $T_{5} = 19.1°C$.
At the part-load condition, we use Equation 19.18

$\dot{m}_{a,cc,B} = \dot{m}_{a,B} \frac{(T_{db,4}  –  T_{db,5B})}{(T_{db,4}  –  T_{db,3})}$   and    $\dot{m}_{a,hc,B} = \dot{m}_{a,B}  –  \dot{m}_{a,cc,B}$             (19.18)

and $\dot{m}_{a,hc} = 0.59  –  0.475 = 0.115  kg/s$
Finally, the cooling and heating coil loads are

= 5.23  kW (17,850  Btu/h)

= 1.73  kW (5900  Btu/h)
The difference between these two coil loads is the 3.5 kW cooling rate, as required by the problem statement.
Comments
This example illustrates the key problem with dual-duct systems from an energy viewpoint. The heating energy in this example is not needed at all.
However, it is present due to the system’s basic configuration requiring constant airflow. To compound the problem, not 3.50 kW but 5.23 kW of cooling is needed, since the unwanted 1.7 kW of heat must be extracted by the cooling system in addition to the zone’s load. The energy waste is obvious.