Question

The outdoor and indoor temperature of a room are \(18^{\circ}\text{C}\) and \(22^{\circ}\text{C}\), respectively. There is also an internal heat gain of \(5\ \text{kW}\). Consider the volumetric heat capacity of air is \(1300\ \text{J/m}^3{\cdot}^{\circ}\text{C}\). Using the ventilation heat flow equation, the estimated rate of ventilation (in \(\text{m}^3\text{/s}\)) is (rounded off to two decimal places). Assume no other heat gains or losses and no net heat storage in the space.

Hint:

Use \(Q = \rho c_p V \Delta T\) with \(Q = 5000\ \text{W}\), \(\rho c_p = 1300\ \text{J/m}^3{\cdot}^{\circ}\text{C}\), and \(\Delta T = 4^{\circ}\text{C}\).

Answer 1

Correct Answer: 0.96

Step 1: Understanding the Question:
We are told the outdoor temperature is \(18^{\circ}\text{C}\), the indoor temperature is \(22^{\circ}\text{C}\), the internal heat gain is \(5\ \text{kW}\), and the volumetric heat capacity of air is \(1300\ \text{J/m}^3{\cdot}^{\circ}\text{C}\). We need to find the rate of ventilation, in \(\text{m}^3\text{/s}\), assuming steady state with no other heat gains or losses and no net heat storage.

Step 2: Key Formula or Approach:
In steady state, the heat gained inside the room must equal the heat carried away by the outgoing ventilation air, since the indoor air is warmer than the outdoor air and is steadily replaced by cooler outdoor air. The ventilation heat flow equation is
\[ Q = \rho c_p \, V \, \Delta T \]
where \(Q\) is the heat flow rate in watts, \(\rho c_p\) is the volumetric heat capacity of air in \(\text{J/m}^3{\cdot}^{\circ}\text{C}\), \(V\) is the ventilation rate in \(\text{m}^3\text{/s}\), and \(\Delta T\) is the temperature difference between indoor and outdoor air.

Step 3: Setting up the equation:
For the space to stay at steady state with no net heat storage, the internal heat gain has to be exactly balanced by the heat lost through ventilation, so \(Q\) equals the given internal heat gain of \(5\ \text{kW} = 5000\ \text{W}\). The temperature difference is
\[ \Delta T = 22^{\circ}\text{C} - 18^{\circ}\text{C} = 4^{\circ}\text{C} \]
so the equation becomes
\[ 5000 = 1300 \times V \times 4 \]

Step 4: Solving for the ventilation rate:
\[ 5000 = 5200 \, V \]
\[ V = \frac{5000}{5200} = 0.9615\ \text{m}^3\text{/s} \]
Rounding this to two decimal places gives \(V \approx 0.96\ \text{m}^3\text{/s}\).

Step 5: Final Answer:
The estimated rate of ventilation needed to remove the internal heat gain, given the indoor-outdoor temperature difference, is about \(0.96\ \text{m}^3\text{/s}\).
\[ \boxed{V \approx 0.96\ \text{m}^3\text{/s}} \]