Question

Consider the base plate of an 800-W household iron with a thickness of $L=0.6 \mathrm{~cm}\(, base area of \)A=160 \mathrm{~cm}^{2}$, and thermal conductivity of \(k=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. When steady operating conditions are reached, the outer surface temperature of the plate is measured to be \(85^{\circ} \mathrm{C}\). Disregarding any heat loss through the upper part of the iron, \((a)\) express the differential equation and the boundary conditions for steady one- dimensional heat conduction through the plate, \((b)\) obtain a relation for the variation of temperature in the base plate by solving the differential equation, and \((c)\) evaluate the inner surface temperature. Answer: (c) \(100^{\circ} \mathrm{C}\)

Short answer

Answer: The inner surface temperature of the iron's baseplate is 100°C.

Step-by-step solution

Writing the general heat conduction equation

The general heat conduction equation is given by the following formula: \(q = -k \frac{dT}{dx}\). Here, \(q\) is the heat flux, \(k\) is the thermal conductivity, \(T\) is the temperature at any point within the iron, and \(x\) is the location inside the iron.

Expressing the differential equation for steady one-dimensional heat conduction

In this problem, we have steady heat conduction, which means that the heat flow is constant through the iron, so: \(\frac{d^2T}{dx^2} = 0\).

Obtain the boundary conditions

Boundary condition 1: The outer surface temperature is given as 85°C (\(T_o = 85\)) at a position \(x = L\). Thus, \(T(L) = 85\). Boundary condition 2: Since there is no heat loss through the upper part of the iron, the heat flux (\(q\)) must be constant throughout the surface area A. Therefore, we have: \(-k \frac{dT}{dx}\Big|_{x=0} = -k \frac{dT}{dx}\Big|_{x=L} = q_0\), where \(q_0\) is the constant heat flux.

Solving the differential equation

Since the second derivative is zero for steady-state one-dimensional conduction, the temperature profile must be linear. Thus, integrating \(\frac{d^2T}{dx^2} = 0\), we obtain: \(\frac{dT}{dx} = C_1\). Integrating once again, we get \(T(x) = C_1x + C_2\). We will use the boundary conditions to find the constants \(C_1\) and \(C_2\).

Using boundary conditions to find the constants

Using boundary condition 1: \(T(L) = 85\), so \(C_1L + C_2 = 85\). Using boundary condition 2: \(-k \frac{dT}{dx}\Big|_{x=0} = -kC_1 = q_0\). Since total heat transfer \(Q = q_0A = -kA \frac{dT}{dx}\Big|_{x=0}\), and \(P = 800W\) (iron's power), we get \(800 = -kA(-C_1) \Rightarrow C_1 = -\frac{800}{kA}\).

Substituting \(C_1\) back into temperature profile

The temperature profile now becomes \(T(x) = -\frac{800}{kA}x + C_2\). Now, we can use the condition \(T(L) = 85\) to find \(C_2\). Substituting and solving for \(C_2\), we have: \(C_2 = 85 + \frac{800L}{kA}\).

Final temperature profile and inner surface temperature

The final temperature profile is: \(T(x) = -\frac{800}{kA}x + 85 + \frac{800L}{kA}\). To find the inner surface temperature, we need to find \(T(0)\). Substituting \(x = 0\) in our equation, we get \(T(0) = 85 + \frac{800L}{kA}\). Plugging in the values, we have \(T(0) = 85 + \frac{800(0.6\times10^{-2})}{20(160\times10^{-4})} = 85 + 15 = 100^\circ C\). Answer: (c) The inner surface temperature is \(100^\circ C\).