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.

Question

Auditorium capacity of 1200 with 1/3 females & 2/3 Male. WC requirement in integer is
Female: 3 per 100 Upto 200- 2 per 100 for 200 onwards
Male: 2 per 100 upto 400. I per 250 for 400 onwards

Answer 1

This is a straightforward heat-balance problem: whatever heat is made inside the room has to leave through the ventilation air, since we are told nothing is stored and there are no other losses.

Start by writing down what is known, with units attached so nothing gets lost in the arithmetic:

Indoor temperature: $22^{\circ}\text{C}$
Outdoor temperature: $18^{\circ}\text{C}$
Internal heat gain: $5\ \text{kW} = 5000\ \text{W} = 5000\ \text{J/s}$
Volumetric heat capacity of air: $1300\ \text{J/m}^3\text{-}^{\circ}\text{C}$

Every second, the ventilation system swaps out some of the warm indoor air for cooler outdoor air. Each cubic metre of that swapped air carries away heat equal to $1300\ \text{J/m}^3\text{-}^{\circ}\text{C}$ times the temperature drop it goes through, which is $22 - 18 = 4^{\circ}\text{C}$.

So each cubic metre of ventilated air removes:

\[ 1300 \times 4 = 5200\ \text{J per cubic metre} \]

Since the room must shed exactly $5000$ joules every second to stay at steady state, the volume of air that needs to pass through per second is the required heat loss divided by the heat removed per cubic metre:

\[ V = \frac{5000\ \text{J/s}}{5200\ \text{J/m}^3} = 0.9615\ \text{m}^3\text{/s} \]

Rounded to two decimal places, this gives $V = 0.96\ \text{m}^3\text{/s}$.

So the estimated rate of ventilation is 0.96 cubic metres per second.

Answer 2

This is a straightforward heat-balance problem: whatever heat is made inside the room has to leave through the ventilation air, since we are told nothing is stored and there are no other losses.

Start by writing down what is known, with units attached so nothing gets lost in the arithmetic:

Indoor temperature: $22^{\circ}\text{C}$
Outdoor temperature: $18^{\circ}\text{C}$
Internal heat gain: $5\ \text{kW} = 5000\ \text{W} = 5000\ \text{J/s}$
Volumetric heat capacity of air: $1300\ \text{J/m}^3\text{-}^{\circ}\text{C}$

Every second, the ventilation system swaps out some of the warm indoor air for cooler outdoor air. Each cubic metre of that swapped air carries away heat equal to $1300\ \text{J/m}^3\text{-}^{\circ}\text{C}$ times the temperature drop it goes through, which is $22 - 18 = 4^{\circ}\text{C}$.

So each cubic metre of ventilated air removes:

\[ 1300 \times 4 = 5200\ \text{J per cubic metre} \]

Since the room must shed exactly $5000$ joules every second to stay at steady state, the volume of air that needs to pass through per second is the required heat loss divided by the heat removed per cubic metre:

\[ V = \frac{5000\ \text{J/s}}{5200\ \text{J/m}^3} = 0.9615\ \text{m}^3\text{/s} \]

Rounded to two decimal places, this gives $V = 0.96\ \text{m}^3\text{/s}$.

So the estimated rate of ventilation is 0.96 cubic metres per second.

Show Hint

Correct Answer: 0.96

Solution and Explanation

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