14.6 Convection

Calculating Heat Transfer by Convection: Convection of Air Through the Walls of a House

Most houses are not airtight: air goes in and out around doors and windows, through cracks and crevices, following wiring to switches and outlets, and so on. The air in a typical house is completely replaced in less than an hour. Suppose that a moderately-sized house has inside dimensions $12.0\text{m}\times 18.0\text{m}\times 3.00\text{m}$ high, and that all air is replaced in 30.0 min. Calculate the heat transfer per unit time in watts needed to warm the incoming cold air by $\text{10}\text{.}\mathrm{0º}\text{C}$, thus replacing the heat transferred by convection alone.

Strategy

Heat is used to raise the temperature of air so that $Q=\text{mc}\text{Δ}T$. The rate of heat transfer is then $Q/t$, where $t$ is the time for air turnover. We are given that $\text{Δ}T$ is $\text{10}\text{.}\mathrm{0º}\text{C}$, but we must still find values for the mass of air and its specific heat before we can calculate $Q$. The specific heat of air is a weighted average of the specific heats of nitrogen and oxygen, which gives $c=c_{\text{p}}\cong \text{1000}\text{ J/kg}\cdot º\text{C}$ from Table 14.4 (note that the specific heat at constant pressure must be used for this process).

Solution

1. Determine the mass of air from its density and the given volume of the house. The density is given from the density $\rho$ and the volume
   $$m=\text{ρV}=\left(1\text{.}\text{29}{\text{kg/m}}^3\right)\left(\text{12}\text{.}0\text{m}\times \text{18}\text{.}0\text{m}\times 3.00\text{m}\right)=\text{836}\text{kg.}$$

   14.37
2. Calculate the heat transferred from the change in air temperature: $Q=\text{mc}\text{Δ}T$ so that
   $$Q=\left(\text{836}\text{kg}\right)\left(\text{1000 J/kg}\cdot º\text{C}\right)\left(\text{10.0º}\text{C}\right)\text{= 8}\text{.}\text{36}\times {\text{10}}^6\text{J.}$$

   14.38
3. Calculate the heat transfer from the heat $Q$ and the turnover time $t$. Since air is turned over in $t=0\text{.}500\text{h}=\text{1800}\text{s}$, the heat transferred per unit time is
   $$\frac{Q}{t}=\frac{8\text{.}\text{36}\times {\text{10}}^{\mathrm{6 }}\text{J}}{\text{1800 }\text{s}}=4\text{.}64\text{kW.}$$

   14.39

Discussion

This rate of heat transfer is equal to the power consumed by about forty-six 100-W light bulbs. Newly constructed homes are designed for a turnover time of 2 hours or more, rather than 30 minutes for the house of this example. Weather stripping, caulking, and improved window seals are commonly employed. More extreme measures are sometimes taken in very cold (or hot) climates to achieve a tight standard of more than 6 hours for one air turnover. Still longer turnover times are unhealthy, because a minimum amount of fresh air is necessary to supply oxygen for breathing and to dilute household pollutants. The term used for the process by which outside air leaks into the house from cracks around windows, doors, and the foundation is called “air infiltration.”

Calculate the Flow of Mass during Convection: Sweat-Heat Transfer away from the Body

The average person produces heat at the rate of about 120 W when at rest. At what rate must water evaporate from the body to get rid of all this energy? (This evaporation might occur when a person is sitting in the shade and surrounding temperatures are the same as skin temperature, eliminating heat transfer by other methods.)

Strategy

Energy is needed for a phase change ($Q={\text{mL}}_{\text{v}}$). Thus, the energy loss per unit time is

We divide both sides of the equation by $L_{\text{v}}$ to find that the mass evaporated per unit time is

Solution

(1) Insert the value of the latent heat from Table 14.2, $L_{\text{v}}=\text{2430}\text{ kJ/kg}=\text{2430}\text{ J/g}$. This yields

Discussion

Evaporating about 3 g/min seems reasonable. This would be about 180 g (about 7 oz) per hour. If the air is very dry, the sweat may evaporate without even being noticed. A significant amount of evaporation also takes place in the lungs and breathing passages.