Question

Warm air is blown over the inner surface of an automobile windshield to defrost ice accumulated on the outer surface of the windshield. Consider an automobile windshield $\left(k_{w}=0.8 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot \mathrm{R}\right)$ with an overall height of 20 in and thickness of \(0.2\) in. The outside air \((1 \mathrm{~atm})\) ambient temperature is \(8^{\circ} \mathrm{F}\), and the average airflow velocity over the outer windshield surface is \(50 \mathrm{mph}\), while the ambient temperature inside the automobile is \(77^{\circ} \mathrm{F}\). Determine the value of the convection heat transfer coefficient for the warm air blowing over the inner surface of the windshield that is needed to cause the accumulated ice to begin melting. Assume the windshield surface can be treated as a flat-plate surface.

Short answer

The convection heat transfer coefficient for the warm air blowing over the inner surface of the windshield needed to cause the accumulated ice to begin melting is approximately \(7.73\, \mathrm{Btu/h} \cdot \mathrm{ft}^2 \cdot \mathrm{R}^{-1}\).

Step-by-step solution

Calculate heat transfer through the windshield by conduction

We can use the formula for steady-state heat conduction to determine the rate of heat transfer through the windshield: \(q = k_w \cdot A \cdot \frac{\Delta T}{L}\) Where \(q\) is the rate of heat transfer, \(k_w\) is the thermal conductivity of the windshield material, \(A\) is the area of the windshield, \(\Delta T = (T_{\text{inside}} - T_{\text{outside}})\) is the temperature difference between the inside and outside surfaces of the windshield, and \(L\) is the thickness of the windshield. Given values: \(k_w = 0.8\, \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot \mathrm{R}\), \(L = 0.2\, \text{in} = 1/60\, \text{ft}\), \(T_{\text{inside}} = 77^{\circ} \mathrm{F}\), \(T_{\text{outside}} = 8^{\circ} \mathrm{F}\). In order to calculate the area \(A\), we assume that the windshield is a rectangular surface with height \(h = 20\, \text{in} = 5/3\, \text{ft}\). Assuming a standard windshield width (let's say \(5\, \text{ft}\)), we compute the area: \(A = 5\, \text{ft} \cdot 5/3\, \text{ft} = 25/3\, \text{ft}^2\) Now we can calculate the rate of heat transfer \(q\): \(q = 0.8\, \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot \mathrm{R} \cdot (25/3\, \mathrm{ft}^2) \cdot \frac{(77\, ^{\circ} \mathrm{F} - 8\, ^{\circ} \mathrm{F})}{(1/60\, \mathrm{ft})} \) \(q \approx 5133.3\, \mathrm{Btu/h}\)

Determine the heat transfer coefficient required to start melting ice

Now that we have the rate of heat transfer through the windshield by conduction, we can use Newton's law of cooling to determine the convection heat transfer coefficient \(h_c\) for the warm air blowing over the inner surface of the windshield required to maintain an equilibrium temperature that will start melting the ice: \(q = h_c \cdot A \cdot \Delta T_c\) Where \(\Delta T_c = (T_{\text{inside}} - T_{\text{ice}})\), and the ice melting point is \(T_{\text{ice}} = 32^\circ\, \mathrm{F}\). We can rearrange this equation to obtain the convection heat transfer coefficient \(h_c\): \(h_c = \frac{q}{A \cdot \Delta T_c}\) Now, we can compute the required convection heat transfer coefficient: \(h_c = \frac{5133.3\, \mathrm{Btu/h}}{(25/3\, \mathrm{ft}^2) \cdot (77^\circ\, \mathrm{F} - 32^\circ\, \mathrm{F})}\) \(h_c \approx 7.73\, \mathrm{Btu/h} \cdot \mathrm{ft}^2 \cdot \mathrm{R}^{-1}\). The convection heat transfer coefficient for the warm air blowing over the inner surface of the windshield needed to cause the accumulated ice to begin melting is approximately \(7.73\, \mathrm{Btu/h} \cdot \mathrm{ft}^2 \cdot \mathrm{R}^{-1}\).