# Working of air conditioner

I want to understand working of AC at a basic level. Hence following is my question.

Let there be a closed room of volume V at indoor temperature $$T_1$$. It has to be cooled down to temperature $$T_2$$. If Q amount of heat is removed from the room every second, then how much time will it take to reach the desired temperature?

Assume $$D < Q$$ is the amount of heat input to the room every second. If required assume outdoor temperature to be $$T_3$$.

Temperatures are in °C, heat in joules and volume in $$m^3$$.

My understanding:

• Cooling rate is directly proportional to $$Q-D$$.
• Mass of air, $$M = pV$$ , where $$p$$ is density of air.
• Desired heat loss, $$H = cM(T_1 - T_2)$$ , where $$c$$ is specific heat of air.
• Time taken to get the desired cooling, $$T = \frac{H}{Q-D}$$.

Are my calculations correct?

• Welcome to Engineering! This looks like a homework question (even if it isn't homework!). In order for such questions to be answered in this site, we need you to add details describing the precise problem you're having. You've said you want to understand how AC works, but what have you researched already, what aspect of this question is giving your doubts, where are you stuck? Please edit your question to include this information. – Wasabi May 22 '20 at 15:03
• I have made necessary corrections. – tgvbhy May 23 '20 at 9:46
• What are 'D' and 'Q'? – Transistor May 23 '20 at 9:50
• It is described in the question. – tgvbhy May 23 '20 at 10:00
• I see a definition for Q. I don't see one for D. – Transistor May 23 '20 at 10:53

You're correct. Here is a solution worked out with variable names I'm more comfortable with. I substitute back some of your variable names at the end.

## Prompt

Let there be a closed room of volume $$V$$ at indoor temperature $$T_1$$. It has to be cooled down to temperature $$T_2$$. If $$Q_{\text{out}}$$ amount of heat is removed from the room every second, then how much time will it take to reach the desired temperature?

Assume $$Q_{\text{in}}$$ is the amount of heat input to the room every second.

Image by Baltakatei licensed CC BY-SA 4.0

## Energy balance

First, consider an energy balance of the system.

$$\left( \text{rate of energy in} \right) - \left( \text{rate of energy out} \right) + \left( \text{rate of energy generation} \right) = \left( \text{rate of energy accumulation} \right)$$

$$Q_{\text{in}} - Q_{\text{out}} + 0 = \frac{dU}{dt} \cdot m$$

where:

$$\begin{eqnarray*} Q_{\text{in}} & : & \text{rate of heat flow into the system} \left[ \tfrac{\text{J}}{\text{s}} \right]\\ Q_{\text{out}} & : & \text{rate of heat flow out of the system} \left[ \tfrac{\text{J}}{\text{s}} \right]\\ t & : & \text{time} \left[ \text{s} \right]\\ U & : & \text{internal energy of the system per unit mass} \text{} \left[ \tfrac{\text{J}}{\text{kg} \cdot \text{s}} \right]\\ m & : & \text{mass of system} \left[ \text{kg} \right] \end{eqnarray*}$$

Solve for $$dU$$.

$$dU = \left( \frac{1}{m} \right) \cdot \left( Q_{\text{in}} - Q_{\text{out}} \right) \cdot dt$$

Integrate from initial to final state. $$$$\int_{U = U_i}^{U = U_f} dU = \frac{1}{m} \cdot \int_{t = t_i}^{t = t_f} \left( Q_{\text{in}} - Q_{\text{out}} \right) dt$$$$ $$$$U_f - U_i = \frac{1}{m} \cdot \left( Q_{\text{in}} - Q_{\text{out}} \right) \cdot (t_f - t_i)$$$$

Equation 1: Simplify with $$\Delta U = U_f - U_i$$. $$$$\Delta U = \frac{1}{m} \cdot \left( Q_{\text{in}} - Q_{\text{out}} \right) \cdot (t_f - t_i)$$$$

## Temperature change from heat flow.

For a mechanically reversible constant-volume process, $$Q = m \cdot \Delta U$$.

Take into account changes in temperature of the system mass by introducing the concept of constant-volume heat capacity. The definition of constant-volume heat capacity of a substance (provided by Introduction to Chemical Engineering, 7th Edition, by J.M. Smith) is: $$$$C_V \equiv \left( \frac{dU}{dT} \right)_V$$$$ Where: $$\begin{eqnarray*} C_V & : & \text{constant volume heat capacity} \left[ \tfrac{\text{J}}{\text{kg} \cdot \text{K}} \right]\\ T & : & \text{temperature of the system} \left[ \text{K} \right]\\ V & : & \text{specific volume of the system} \left[ \tfrac{\text{m}^3}{\text{kg}} \right] \end{eqnarray*}$$ $$$$dU = C_V dT$$$$ $$$$\int_{U = U_i}^{U = U_f} dU = \int_{T = T_i}^{T = T_f} C_V dT$$$$ $$$$U_f - U_i = C_V \cdot (T_f - T_i)$$$$

Equation 2: Substitute $$\Delta U = U_f - U_i$$ : $$$$\Delta U = C_V \cdot (T_f - T_i)$$$$

## Combine energy rate balance equation with energy-temperature equation

Now we have an expression (equation 2) for how the internal energy of the constant-volume system changes with temperature. We also have an expression (equation 1) for how internal energy changes according to heat flows $$Q_{\text{in}}$$ and $$Q_{\text{out}}$$. These two changes in internal energy are equal so we set equation 1 and equation 2 equal to one another, yielding equation 3: $$$$\Delta U = \frac{1}{m} \cdot \left( Q_{\text{in}} - Q_{\text{out}} \right) \cdot (t_f - t_i) = C_V \cdot (T_f - T_i)$$$$

Solve for $$t_f$$. $$$$t_f = \frac{C_V \cdot m \cdot (T_f - T_i)}{Q_{\text{in}} - Q_{\text{out}}} + t_i$$$$

Now, apply the prompt's parameters.

$$\begin{eqnarray*} T_i & = & T_1 \\ T_f & = & T_2 \\ t_i & = & 0 \\ Q_{\text{in}} & = & D \\ Q_{\text{out}} & = & Q \\ \end{eqnarray*}$$

$$$$t_f = \frac{C_V \cdot m \cdot (T_2 - T_1)}{D - Q}$$$$

Equation 4. Multiply by $$\frac{-1}{-1}$$

$$$$t_f = \frac{C_V \cdot m \cdot (T_1 - T_2)}{Q - D}$$$$

## Summary

Equation 4 is the amount of time required to cool mass $$m$$ from $$T_1$$ to $$T_2$$, provided the constant-volume specific heat capacity $$C_V$$ is known as well as the heat flows to ($$Q_{\text{in}}=D$$) and from ($$Q_{\text{out}}=Q$$) the rigid container.

A PDF draft of this solution is here.