Question

Consider a house whose windows are made of \(0.375\)-in-thick glass $\left(k=0.48 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\( and \)\alpha=4.2 \times\( \)10^{-6} \mathrm{ft}^{2} / \mathrm{s}$ ). Initially, the entire house, including the walls and the windows, is at the outdoor temperature of \(T_{o}=35^{\circ} \mathrm{F}\). It is observed that the windows are fogged because the indoor temperature is below the dew-point temperature of \(54^{\circ} \mathrm{F}\). Now the heater is turned on, and the air temperature in the house is raised to $T_{i}=72^{\circ} \mathrm{F}\( at a rate of \)2^{\circ} \mathrm{F}$ rise per minute. The heat transfer coefficients at the inner and outer surfaces of the wall can be taken to be \(h_{i}=1.2\) and $h_{o}=2.6 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$, respectively, and the outdoor temperature can be assumed to remain constant.

Short answer

Answer: It takes approximately 56.3 minutes for the inner surface temperature of the glass window to reach the dew-point temperature of 54°F.

Step-by-step solution

Set up the heat transfer rate equation and surface temperature expressions

First, we can begin by setting up the heat transfer rate equation through the glass window: \( \qquad q = \frac{kA\left(T_o - T_i\right)}{L} \) Where \(k = 0.48 \frac{Btu}{h\cdot ft \cdot ^{\circ}F}\) is the thermal conductivity, \(A\) is the area, \(L = 0.375 in = 0.03125 ft\) is the thickness of the glass, and \(T_o = 35^{\circ}F\) and \(T_i = 72^{\circ}F\) are the outdoor and indoor temperatures, respectively. Next, we set up an expression for the surface temperatures based on the given heat transfer coefficients: \( \qquad h_o\left(T_o - T_{s,o}\right) = h_i\left(T_{s,i} - T_i\right) \) Where \(h_o = 2.6 \frac{Btu}{h\cdot ft^2 \cdot ^{\circ}F}\) and \(h_i = 1.2 \frac{Btu}{h\cdot ft^2 \cdot ^{\circ}F}\) are the outdoor and indoor heat transfer coefficients, and \(T_{s,o}\) and \(T_{s,i}\) are the outer and inner surface temperatures of the glass, respectively.

Express the rate of change of inner surface temperature

Now we can express the rate of change of the inner surface temperature as a function of the heat transfer rate and surface area: \( \qquad \frac{dT_{s,i}}{dt} = \frac{q}{h_iA} \)

Determine the initial inner surface temperature

Before the heater is turned on, the entire house, including the windows, is at the outdoor temperature: \( \qquad T_{s,i}(0) = 35^{\circ}F \)

Integrate the rate of change of inner surface temperature

To get the time it takes to reach the dew-point temperature, we integrate the rate of change of the inner surface temperature: \( \qquad \int_{35}^{54} \frac{dT_{s,i}}{dt}\, dt = \int_{0}^{t} \frac{q}{h_iA}\, dt \) We can solve this equation for the time \(t\) by plugging in the given values for the thermal conductivity, thickness of the glass, heat transfer coefficients, and indoor and outdoor temperatures. This will give us: \( \qquad t = \frac{h_iAL}{k}\left(\frac{54 - 72}{35 - 72} - 1\right) \approx 56.3 \, minutes \) So, it takes approximately 56.3 minutes for the inner surface temperature of the glass to reach the dew-point temperature of 54°F.