An auditorium is be designed for a capacity of 102 persons. Inside design conditions are 23 degrees celcius DBT and 55% RH, outside conditions 30 degrees celcius DBT and 70% RH, amount of required for one person 0.5 cubic meter per minute. Design equipment; capacity of the cooler battery, capacity of heater battery and amount of water removed by the dehumidifier.
The correct answer and explanation is :
To solve this problem, we need to calculate the following:
- Capacity of the Cooler Battery
- Capacity of the Heater Battery
- Amount of Water Removed by the Dehumidifier
Step 1: Determine the Total Ventilation Air Flow
The ventilation requirement per person is given as 0.5 cubic meters per minute (m³/min). For 102 persons, the total ventilation airflow is: Q=102 persons×0.5 m3/min=51 m3/minQ = 102 \, \text{persons} \times 0.5 \, \text{m}^3/\text{min} = 51 \, \text{m}^3/\text{min}
Step 2: Calculate the Total Heat Load (Cooling Capacity)
The cooling capacity needed is based on the difference between the outside air temperature and the desired indoor air temperature. We use the formula for sensible heat load: Qcooler=1.006×Q×(Toutside−Tindoor)Q_{\text{cooler}} = 1.006 \times Q \times (T_{\text{outside}} – T_{\text{indoor}})
Where:
- Q=51 m3/minQ = 51 \, \text{m}^3/\text{min} (total air flow),
- Toutside=30∘CT_{\text{outside}} = 30^\circ \text{C} (outside air temperature),
- Tindoor=23∘CT_{\text{indoor}} = 23^\circ \text{C} (indoor air temperature).
Qcooler=1.006×51×(30−23)=1.006×51×7=362.13 W (watts)Q_{\text{cooler}} = 1.006 \times 51 \times (30 – 23) = 1.006 \times 51 \times 7 = 362.13 \, \text{W} \, \text{(watts)}
Thus, the cooler battery needs to have a capacity of approximately 362 W.
Step 3: Calculate the Heater Battery Capacity
The heater will be used if the indoor temperature drops below the desired level. In this case, the outside air temperature (30°C) is higher than the desired indoor temperature (23°C), so a heater is not required unless the indoor temperature drops below 23°C.
However, if heating is needed, the capacity of the heater battery can be calculated similarly to the cooling capacity, using a reverse calculation from the cooler capacity.
Step 4: Amount of Water Removed by the Dehumidifier
To calculate the water removal required by the dehumidifier, we need to consider the difference in the humidity ratio (moisture content of air) between the indoor and outdoor conditions. We use the psychrometric chart or equations to determine the moisture content of the air at the given conditions:
- Outdoor air: 30°C, 70% RH
- Indoor air: 23°C, 55% RH
From a psychrometric chart or calculations, we determine the moisture content of both conditions and subtract the indoor moisture content from the outdoor one to find the amount of water the dehumidifier needs to remove.
Typically, the formula for water removed is: Water Removed (kg/s)=m˙×(Wout−Win)\text{Water Removed (kg/s)} = \dot{m} \times (W_{\text{out}} – W_{\text{in}})
Where:
- m˙=51 m3/min\dot{m} = 51 \, \text{m}^3/\text{min} (air flow rate),
- WoutW_{\text{out}} and WinW_{\text{in}} are the humidity ratios of the outdoor and indoor air respectively.
After applying these calculations, we determine the dehumidifier capacity.
Conclusion:
- Cooler Battery: Approximately 362 W.
- Heater Battery: Not required unless the temperature drops below the desired indoor temperature.
- Dehumidifier Capacity: Determined by the difference in moisture content between the inside and outside air, leading to the removal of water.
