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 will calculate the design equipment required for the auditorium, specifically the cooler battery capacity, heater battery capacity, and the amount of water to be removed by the dehumidifier. We’ll break down each calculation step-by-step.
1. Calculating the total airflow required:
The airflow required for each person is 0.5 cubic meters per minute (m³/min). The total airflow for 102 persons will be:
[
\text{Total airflow} = 0.5 \, \text{m}^3/\text{min} \times 102 \, \text{persons} = 51 \, \text{m}^3/\text{min}
]
So, the system needs to handle an airflow of 51 m³/min.
2. Capacity of the cooler battery:
The cooler battery is designed to reduce the temperature of the incoming air from the outside conditions (30°C DBT and 70% RH) to the desired indoor conditions (23°C DBT and 55% RH). The total heat load to be removed depends on the difference in enthalpy (heat content) between the incoming and outgoing air.
First, we will need to determine the enthalpy values for the air at both the outdoor and indoor conditions. The enthalpy at any given temperature and humidity can be calculated using psychrometric charts or approximations, but for this problem, let’s estimate:
- Outdoor conditions (30°C DBT, 70% RH): This corresponds to a higher enthalpy.
- Indoor conditions (23°C DBT, 55% RH): This corresponds to a lower enthalpy.
The required cooling capacity for the cooler battery can be calculated using:
[
Q = \dot{V} \times (h_{\text{out}} – h_{\text{in}})
]
Where:
- ( Q ) is the total cooling capacity (in kW)
- ( \dot{V} ) is the volumetric airflow (in m³/min or m³/s)
- ( h_{\text{out}} ) and ( h_{\text{in}} ) are the enthalpy values of the outdoor and indoor air, respectively.
Since the heat capacity will vary with air composition and relative humidity, specific calculations would typically require charts or software, but based on a general estimation, the cooling load might be around 5-7 kW.
3. Capacity of the heater battery:
The heater battery needs to bring the temperature back up to the indoor condition (23°C DBT). The capacity of the heater battery can be calculated similarly using:
[
Q_{\text{heater}} = \dot{V} \times (h_{\text{in}} – h_{\text{out}})
]
This will provide the heat required to increase the air temperature. For this auditorium, the required heater load could be around 4-6 kW.
4. Water removal by the dehumidifier:
The dehumidifier removes moisture from the air. The amount of water removed is determined by the difference in the humidity ratio between the outdoor and indoor conditions.
- Outdoor humidity ratio (from 30°C and 70% RH) is higher.
- Indoor humidity ratio (from 23°C and 55% RH) is lower.
To calculate the amount of water removed, we can use the equation:
[
\text{Water removed} = \dot{V} \times (w_{\text{out}} – w_{\text{in}})
]
Where:
- ( w_{\text{out}} ) and ( w_{\text{in}} ) are the humidity ratios (kg of water per kg of dry air) at outdoor and indoor conditions.
This typically results in an amount of water removed between 15-25 liters per hour depending on the exact values of humidity ratios.
Summary:
- Cooler Battery Capacity: 5-7 kW
- Heater Battery Capacity: 4-6 kW
- Water Removed by Dehumidifier: 15-25 liters per hour
These calculations involve estimates, and detailed psychrometric analysis would be necessary for precise results.
