In an air conditioning plant, airflow rate of 2 kg/s passes through a coil. The dry-bulb temperature decreases from 24 C to 12 C. The moisture content of the air decreases from 0.010 to 0.008. Determine the load on the coil, contact factor of the coil and apparatus dew-point temperature, ADP.
| Inputs |
Units |
|
|
on-coil air, DBT |
24,00 |
C |
|
on-coil air, moisture content |
0,010 |
. |
|
on-coil air, mass flow rate |
2,00 |
kg/s |
|
off-coil air, DBT |
12,00 |
C |
|
off-coil air, moisture content |
0,008 |
. |
|
Atmospheric pressure |
1,01 |
bar |
The contact factor of a coil is defined as the efficiency for dehumidification. A 100% efficient coil will bring the moisture content of the air to the saturation moisture content at the apparatus dew-point, mcC. The contact factor of the coil can be defined by moisture content differences:
cf = (mcA - mcB)/(mcA - mcC)
or
cf = (hA - hB)/(hA - hC)
| Output | |||
|
Specific enthalpy of the on-coil air |
49,4733 |
kJ/kg |
|
|
Specific enthalpy of the off-coil air |
32,2385 |
kJ/kg |
|
|
Load on the coil |
34,4695 |
kW |
|
|
Apparatus dew-point moisture content |
0,0077 |
. |
Assumed |
|
Apparatus dew-point temperature |
10,4 |
C |
|
|
Apparatus dew-point enthalpy |
29,9 |
kJ/kg |
|
|
Coil contact factor Eq1 |
0,88 |
. |
|
|
Coil contact factor Eq2 |
0,88 |
. |
|
|
Diff = Eq1-Eq2 = 0 |
0,0003 |
. |
Goal |
|
Use the solver to reach the goal |
. |
. |