Question 13.5: 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—34°C DBT and 70% R.H.
- Required air inlet conditions—25°C DBT and 65% R.H.
- Amount of free air circulated—5 m³/s
- Coil dew temperature—15°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. 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 34°C. Move a pencil up this line to meet the intersection with the exponential line 70% 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 15°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 = 95.3 kJ/kg, H_3 = 48.27 kJ/kg, H_2 = 58.143 kJ/kg, H_4 = 42.11 kJ/kg
W_1 = 0.023 kg/kg dry air, W_2 = 0.012 kg/kg dry air, V_1 = 0.9022 m^3 /kg.
Step 6: Determine the mass of air per minute
G = \frac{m}{W} = \frac{5}{0.902} = 5.54 kg/sec
Step 7: Determine the cooling coil capacity
Cooling coil capacity = G (H_1 – H_3) = 95.3 – 48.27 = 47 kJ/s
\frac{5.54 \times 47 \times 3600}{14000} = 66.95 ton
Step 8: Determine the capacity heating coil
Cooling coil capacity = G (H_2 – H_3) = 58.143 – 48.27 = 9.877 kJ/s
9.877 × 5.44 = 54.71 kW.
Step 9: Determine the mass of water vapour removed
Cooling coil capacity = G(W_1 – W_3) = 0.023 – 0.012 = 0.011
0.011 × 5.54 = 0.06 kg/s.
