Cooling with dehumidification, Using Thermo Utilities, MS Excel Add-ins

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

.

.


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