The roof of a house consists of a 15-cm-thick concrete slab \((k=2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that is \(15 \mathrm{~m}\) wide and \(20 \mathrm{~m}\) long. The convection heat transfer coefficients on the inner and outer surfaces of the roof are 5 and \(12 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\), respectively. On a clear winter night, the ambient air is reported to be at \(10^{\circ} \mathrm{C}\), while the night sky temperature is \(100 \mathrm{~K}\). The house and the interior surfaces of the wall are maintained at a constant temperature of \(20^{\circ} \mathrm{C}\). The emissivity of both surfaces of the concrete roof is \(0.9\). Considering both radiation and convection heat transfers, determine the rate of heat transfer through the roof, and the inner surface temperature of the roof. If the house is heated by a furnace burning natural gas with an efficiency of 80 percent, and the price of natural gas is \(\$ 1.20 /\) therm ( 1 therm \(=105,500 \mathrm{~kJ}\) of energy content), determine the money lost through the roof that night during a 14-h period.

Step by step solution

01

Calculate the heat transfer coefficient for the roof

To calculate the overall heat transfer coefficient for the roof, U, we need to consider both the conduction through the concrete slab and the convection at its inner and outer surfaces. $$ \frac{1}{U} = \frac{1}{h_{in}} + \frac{L}{k} + \frac{1}{h_{out}} $$ where \(h_{in}\) and \(h_{out}\) are the convective heat transfer coefficients on the inner and outer surfaces of the roof, respectively (given as 5 W/m²⋅K and 12 W/m²⋅K), L is the thickness of the concrete slab (15 cm or 0.15 m), and k is the thermal conductivity of the concrete slab (2 W/m⋅K). Plugging in the values and solving for U: $$ \frac{1}{U} = \frac{1}{5} + \frac{0.15}{2} + \frac{1}{12} $$ $$ U = \frac{1}{(1/5) + (0.15/2) + (1/12)} = 2.197 \text{ W/m}^2\text{K} $$

02

Calculate the heat transfer due to radiation

We will use the Stefan-Boltzmann law to calculate the heat transfer due to radiation: $$ q_{rad} = \sigma \varepsilon(T_{s}^4 - T_{sky}^4) $$ where \(\sigma\) is the Stefan-Boltzmann constant (equal to 5.67×10^(-8) W/m²⋅K⁴), \(\varepsilon\) is the emissivity of the roof surfaces (0.9), \(T_s\) is the roof surface temperature (assumed to be equal to the ambient temperature without loss of generality, \(283~\mathrm{K}\), since it lies between the inner surface temperature and the night sky temperature), and \(T_{sky}\) is the night sky temperature (100 K). Calculating the radiative heat transfer: $$ q_{rad} = (5.67 × 10^{-8}) × 0.9 × (283^4 - 100^4) = 457.58 \text{ W/m}^2 $$

03

Calculate the total heat transfer

Now we can calculate the total heat transfer by adding the heat transfer due to convection and conduction (calculated using the overall heat transfer coefficient, U) to the heat transfer due to radiation: $$ q_{total} = U \Delta T + q_{rad} $$ where \(\Delta T\) is the difference in temperature between the interior of the house and the ambient air (20 - 10 = 10 °C or 10 K). $$ q_{total} = 2.197 × 10 + 457.58 = 468.55 \text{ W/m}^2 $$

04

Calculate the rate of heat transfer through the roof

We can now calculate the rate of heat transfer through the entire roof by multiplying the total heat transfer per square meter by the area of the roof: $$ Q = q_{total} × A $$ where A is the area of the roof (15 m × 20 m = 300 m²). Calculating the rate of heat transfer: $$ Q = 468.55 × 300 = 140565 \text{ W} = 140.565 \text{ kW} $$

05

Calculate the inner surface temperature of the roof

We can determine the inner surface temperature by rearranging the heat transfer equation: $$ T_{in} = T_{house} - \frac{Q - q_{rad}×A}{U×A} $$ where \(T_{house}\) is the interior temperature of the house (20°C or 293 K). Calculate the inner surface temperature: $$ T_{in} = 293 - \frac{(140565 - 457.58 × 300)}{2.197 × 300} = 291.81 \text{ K} $$ $$ T_{in} = 291.81 – 273.15 = 18.66^{\circ} \mathrm{C} $$ The rate of heat transfer through the roof is 140.565 kW, and the inner surface temperature of the roof is 18.66 °C.

06

Calculate the money lost through the roof during the 14-hour period

The energy lost through the roof over the 14-hour period is given by: $$ E = Q × t $$ where t is the time in seconds (14 hours × 3600 s/hour = 50400 s). Calculate the energy lost: $$ E = 140.565 × 50400 = 7086471 \text{ kJ} $$ We also need to take into account the furnace efficiency (80%), so the actual energy consumption is: $$ E_{actual} = \frac{E}{\text{efficiency}} = \frac{7086471}{0.8} = 8858089 \text{ kJ} $$ Now, we can convert the energy lost to therms: $$ \text{Energy lost (therms)} = \frac{E_{actual}}{105,500} = 84 \text{ therms} $$ Finally, we can calculate the money lost: $$ \text{Money lost} = \text{Energy lost (therms)} × \text{Price per therm} = 84 × 1.20 = \$100.80 $$ The money lost through the roof during the 14-hour period is $100.80.

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