Question 13.6: It is required to design an air-conditioning system for an i......
It is required to design an air-conditioning system for an industrial process for the following hot and wet summer conditions:
- Outdoor conditions—32°C DBT and 65% RH.
- Required air inlet conditions—25°C DBT and 60% RH.
- Amount of free air circulated—250 m³/min
- Coil dew temperature—13°C.
The required condition is achieved by first cooling and dehumidifying and then by heating.
Calculate the following:
(a) The cooling capacity of the cooling coil in tonnes and its by-pass factor.
(b) Heating capacity of the heating coil in kW
(c) The mass of water vapour removed per hour. Solve this problem with the use of psychrometric chart. (AU, 2014)
Step 1: Draw the psychrometric chart.
Step 2: By using psychrometric chart, take a look at horizontal axis to find DBT of 32°C. Move a pencil up this line to meet the intersection with the exponential line 65% relative humidity. Mark it as 1.
Step 3: Similarly take a look at horizontal axis to find DBT of 25°C. Move a pencil up this line to meet the intersection with the exponential line 65% relative humidity. Mark it as 2.
Step 4: Similarly take a look at horizontal axis to find dew point of 13°C. Move a pencil up this line to meet the intersection with the exponential line 100% relative humidity. Mark it as 4. The relative humidity is 100% as the dew point is in saturation.
Step 5: Join the line 1–4. Draw the constant specific humidity line through 2 which cuts the line 1–4 at point 3.
By this way the point 3 is located for which we can determine the enthalpy.
The enthalpy and specific humidity from each point is determined from the psychrometric chart.
From psychrometric chart, we calculate the values:
H_1 = 82.5 kJ/kg, H_3 = 47.5 kJ/kg, H_2 = 55.7 kJ/kg, H_4 = 36.6 kJ/kg
W_1 = 19.6 g/kg, W_2 = 11.8 g/kg, V_1 = 0.892 m^3/kg
Step 6: Determine the mass of air per minute
G = \frac{m}{W} = \frac{250}{0.892} = 280.26 kg/min
Step 7: Determine the cooling coil capacity
Cooling coil capacity = G (H_1 – H_3) = 82.5 – 47.5 = 35 kJ/s
\frac{280.26 \times 35 \times 3600}{14000} = 42.039 ton
Step 8: Determine the capacity heating coil
Cooling coil capacity = G (h_2 – h_3) = 55.7 – 47.5 = 8.2 kJ/s
8.2 × 280.26 = 2,298.132 kJ/min
\frac{2,298.132}{60} = 38.3 kW
Step 9: Determine the mass of water vapour removed
Cooling coil capacity = G (W_1 – W_3) = 19.6 – 11.8 = 7.8
\frac{280.26 \times 7.8 \times 60}{1000} = 131.16 kg/h
