Question

It is proposed to air-cool the cylinders of a combustion chamber by joining an aluminum casing with annular fins (k = 240 W/m K) to the cylinder wall (k = 50 W/m K). The air is at 320 K and the corresponding convection coefficient is 100 W/m2 K. Although heating at inner surface is periodic, it is reasonable to assume steady-state conductions with a time-averaged heat flux of = 105 W/m2. The contact resistance between the wall and casing is = 3 times 10-4 m2 K/W. Determine the wall inner temperature Ti, the interface temperatures at the cylinder side and the casing side, and the base temperature Tb. What would be the inner wall temperature without the casing? Hint: The expression for the efficiency of rectangular annular fin is given by Eq. (3.96), which can be found in Table 3.5. Alternatively, the efficiency of this type of fins can be evaluated graphically

Answer 1

To determine the wall inner temperature Ti, the interface temperatures at the cylinder side and the casing side, and the base temperature Tb, we need to consider heat conduction, convection, and contact resistance.

Step-by-step explanation:

To determine the wall inner temperature Ti, the interface temperatures at the cylinder side and the casing side, and the base temperature Tb, we need to consider heat conduction, convection, and contact resistance.

The heat flux through the aluminum casing is given by Q = k1 * A * (T1 - T2) / d1, where k1 is the thermal conductivity of aluminum, A is the surface area of the casing, T1 is the inner temperature of the casing, T2 is the outer temperature of the casing, and d1 is the thickness of the casing.

The heat flux through the cylinder wall is given by Q = k2 * A * (T2 - T3) / d2, where k2 is the thermal conductivity of the cylinder wall, T3 is the temperature at the inner wall of the cylinder, and d2 is the thickness of the cylinder wall.

The heat flux from the cylinder to the air is given by Q = h * A * (T3 - T4), where h is the convection coefficient, A is the surface area of the cylinder, and T4 is the temperature of the air.

The total heat flux through the system is given by Q = Q1 + Q2 + Q3, where Q1 is the heat flux through the casing, Q2 is the heat flux through the cylinder wall, and Q3 is the heat flux from the cylinder to the air.

Equating the heat flux through the system to the given value of the heat flux, we can solve for the temperatures Ti, T1, T2, T3, and T4.

Answer 2

The student must use Fourier's law of conduction and principles of convective heat transfer to determine the various temperatures in the system, taking into account the fin efficiency and contact resistance.

Step-by-step explanation:

The problem at hand involves finding the inner wall temperature (Ti), interface temperatures at the cylinder side and the casing side, and the base temperature (Tb) of a cylinder with an aluminum casing cooled by air with a given convection coefficient and contact resistance. Additionally, it asks for the inner wall temperature without the casing as well. The process involves steady-state heat conduction, and the effects of contact resistance and fin efficiency must be considered.

To calculate the temperatures, one must apply the principles of thermal resistance analogous to electrical resistance. Given the steady-state assumption and time-averaged heat flux (Q/t), we can use Fourier's law of conduction (Q/t = kA(T₂ - T₁)/d) in combination with the concept of convective heat transfer (Q = hA(T_{surface} - T_{fluid})) and the efficiency of fins to solve for the unknown temperatures.

It is important to recognize that these calculations would be complex and require numerous steps, including considering the thermal contact resistance and fin efficiency, which might involve iterative calculations or tabulated data.