Question

To defrost ice accumulated on the outer surface of an automobile windshield, warm air is blown over the inner surface of the windshield. Consider an automobile windshield with thickness of \(5 \mathrm{~mm}\) and thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The outside ambient temperature is \(-10^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the ambient temperature inside the automobile is $25^{\circ} \mathrm{C}$. Determine the value of the convection heat transfer coefficient for the warm air blowing over the inner surface of the windshield necessary to cause the accumulated ice to begin melting.

Short answer

Short Answer: The convection heat transfer coefficient for the warm air blowing over the inner surface of the windshield necessary to cause the accumulated ice to begin melting is \(0.007 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Step-by-step solution

Calculate the heat transfer through the windshield

First, let's determine the heat transfer through the windshield using the equation mentioned above. We are given \(Q_{conv}\) and we have provided the values for thermal conductivity \((k)\), the density difference \((T_{hot} - T_{cold})\), and thickness \((L)\). Therefore, we can set up the following equation: \(Q_{conv} = k\frac{A(T_{hot} - T_{cold})}{L}\)

Substitute the values

Plug in the given values: $$Q_{conv} = 1.4\frac{A(25-(-10))}{0.005}$$ $$Q_{conv} = 9.8A$$

Calculate the heat transfer at the outer surface of the windshield

Now, let's find \(Q_{conv}\) from the outer surface's convection heat transfer coefficient \((h_{out})\) and temperature difference: $$Q_{conv} = 200A(-10-T_0)$$ Divide \(9.8A\) by \(200A\) $$\frac{9.8}{200} = \frac{Q_{conv}}{200A} = \frac{-10-T_0}{-10}$$ Now we can find \(T_0\): $$T_0 = -10 - \frac{9.8}{200}(10)$$

Calculate the temperature at the outer surface of the windshield, \(T_0\)

Solve for \(T_0\) $$T_0 = -10 - \frac{49}{10} = -14.9^{\circ}C$$

Set up the equation for the heat transfer rate by convection inside the car

Now, we will find the heat transfer rate by convection inside the car using the following relationship: \(Q_{conv_{in}} = h_{in}A(T_{in} - T_1)\) Since we have determined \(Q_{conv}\) (the heat transfer through the windshield), we can equate it with \(Q_{conv_{in}}\): $$9.8A = h_{in}A(25 - T_1)$$

Calculate the temperature at the inner surface of the windshield, \(T_1\)

From the equation in Step 1, we have: $$9.8A = 1.4\frac{A(25-T_1)}{0.005}$$ Divide \(1.4A\) by \(0.005A\): $$\frac{9.8}{0.007} = 25 - T_1$$ Now we can find \(T_1\): $$T_1 = 25 - \frac{9.8}{0.007}$$

Solve for \(T_1\)#

Calculate \(T_1\): $$T_1 = 25 - 1400 = -1375^{\circ}C$$

Calculate the convection heat transfer coefficient for the warm air, \(h_{in}\)

Now, we can find \(h_{in}\) by plugging in the values of \(T_{in}\) and \(T_1\) in the equation from Step 5: $$9.8A = h_{in}A(25 - (-1375))$$ Divide \(9.8A\) by \(1400A\): $$h_{in} = \frac{9.8A}{1400A}$$

Solve for \(h_{in}\)

Calculate the convection heat transfer coefficient, \(h_{in}\): $$h_{in} = \frac{9.8}{1400} = 0.007 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$$ The convection heat transfer coefficient for the warm air blowing over the inner surface of the windshield necessary to cause the accumulated ice to begin melting is \(0.007 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).