# UY1: Convective Heat Transfer

In this part, we will look at an example to facilitate understanding of convective heat transfer.

Consider a situation where the air in a room is at a temperature of $25^{\circ}\text{C}$, and the outside air is at $-15^{\circ}\text{C}$. What is the rate of heat transfer per unit area through a 2 mm thick glass window pane of thermal conductivity $k = 2.5 \times 10^{-3} \, \text{cal} .\text{s}^{-1} .\text{cm}^{-2} .^{\circ}C^{-1}$?

Analysis of problem:

At steady state, the rates of heat transfer by convection in the room, conduction through the glass and convection in the outside air are all the same.

Let $\text{T}_{1}$ be the temperature of the surface of the glass window pane facing outside and $\text{T}_{2}$ be the temperature of the surface of the glass window pane facing the room.

Hence,

$$\Delta T_{2} = 25 – T_{2}$$

$$\Delta T_{1} = T_{1} + 15$$

Note: $h_{1}$ and $h_{2}$ is the convection coefficient of the glass window pane facing outside and the room respectively. L is the thickness of the glass window pane.

Rate of heat transfer by convection in the room = $h_{2}A \Delta T_{2}$

Rate of heat transfer by convection outside = $h_{1} A \Delta T_{1}$

Rate of heat transfer by conduction through the glass:

\begin{aligned} &= \frac{kA \left( T_{2} – T_{1} \right)}{L} \\ &= kA \frac{\left(40 – \Delta T_{2} – \Delta T_{1} \right)}{L} \end{aligned}

At steady state, the three rates are all equal.

$$h_{2}A \Delta T_{2} = h_{1} A \Delta T_{1} = kA \frac{\left(40 – \Delta T_{2} – \Delta T_{1} \right)}{L}$$

This 3 way equality in the steady state gives 2 independent equations, with which one can solve for the two unknowns $\Delta T_{1}$ and $\Delta T_{2}$ in principle. Once $\Delta T_{1}$ and $\Delta T_{2}$ have been determined, the rate of energy transfer can be calculated.

Typically, the convection coefficient are proportional to $\left(\Delta T \right)^{\frac{1}{2}}$. This means that $h_{1}$ and $h_{2}$ are proportional to $\left(\Delta T_{1} \right)^{\frac{1}{2}}$ and $\left(\Delta T_{2} \right)^{\frac{1}{2}}$ respectively. Therefore, the solutions can only be determined numerically.

As a first approximation in the solution of the problem, we can assume that the $\Delta T_{1} = \Delta T_{2} = 20^{\circ}C$ (In other words, $T_{2} = T_{1} = 5^{\circ}C$). This means that conduction is much more efficient than convection for transferring heat. (Which seems like a reasonable approximation) The convection coefficient on the inside and outside of the window pane are then given by: (Using $h = 0.0424 \times 10^{-4} \left(\Delta T \right)^{\frac{1}{4}} \, \text{cal}. \text{s}^{-1} .\text{cm}^{-1} .^{\circ}C^{-1}$)

\begin{aligned} h &= 0.0424 \times 10^{-4} \left( 20 \right)^{\frac{1}{4}} \\ &= 0.897 \times 10^{-5} \, \text{cal} .\text{s}^{-1} .\text{cm}^{-2} .^{\circ}C^{-1} \end{aligned}

The rate of heat transfer per unit area is thus:

\begin{aligned} h \Delta T_{1} &= 0.897 \times 10^{-5} \times 20 \\ &= 17.9 \times 10^{-5} \, \text{cal} .\text{s}^{-1} .\text{cm}^{-2} \end{aligned}

Heat transfer is typically limited by convection which is far less efficient than conduction.

Previous: Convection

Back To Thermodynamics (UY1)

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