Question

Four power transistors, each dissipating \(10 \mathrm{~W}\), are mounted on a thin vertical aluminum plate $(k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( \)22 \mathrm{~cm} \times 22 \mathrm{~cm}$ in size. The heat generated by the transistors is to be dissipated by both surfaces of the plate to the surrounding air at \(20^{\circ} \mathrm{C}\), which is blown over the plate by a fan at a velocity of \(5 \mathrm{~m} / \mathrm{s}\). The entire plate can be assumed to be nearly isothermal, and the exposed surface area of the transistor can be taken to be equal to its base area. Determine the temperature of the aluminum plate. Evaluate the air properties at a film temperature of \(40^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Short answer

Answer: The temperature of the aluminum plate is approximately 30.76°C.

Step-by-step solution

Calculate the total heat dissipation

The total heat dissipation of the four power transistors can be calculated as follows: $$Q_{total} = N \cdot Q_{transistor}$$ Where, \(Q_{total}\) is the total heat dissipation, \(N\) is the number of transistors, and \(Q_{transistor}\) is the heat dissipation per transistor. Plugging in the given values: $$Q_{total} = 4 \cdot 10 \mathrm{~W} = 40 \mathrm{~W}$$

Determine the air properties

Evaluate the air properties at a film temperature of \(40^{\circ} \mathrm{C}\) and a pressure of \(1 \mathrm{~atm}\). For the given conditions, the following properties can be obtained from the air property tables: - Density, \(\rho = 1.127 \mathrm{~kg} / \mathrm{m}^{3}\) - Specific heat capacity, \(c_p = 1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) - Thermal conductivity, \(k_a = 0.0271 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) - Dynamic viscosity, \(\mu = 1.966 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\)

Calculate the convective heat transfer coefficient

We will first calculate the Reynolds number \(Re\) to determine the flow regime, and then use an appropriate correlation to find the convective heat transfer coefficient \(h\). Since the airflow is forced convection, we will use the Dittus-Boelter correlation for forced convections. Reynolds number is given by: $$Re = \frac{\rho V d_h}{\mu}$$ Where, - \(d_h\) is characteristic length for the forced convection over a flat plate, which can be taken as the length of the plate, given here as \(0.22 \mathrm{~m}\). - \(V\) is the air velocity of \(5 \mathrm{~m}/\mathrm{s}\). Plugging in the values, we calculate \(Re\): $$Re = \frac{1.127 \cdot 5 \cdot 0.22}{1.966 \times 10^{-5}} = 63357$$ Since \(Re > 4000\), the flow is turbulent. Next, we'll calculate the Nusselt number, which is given by the Dittus-Boelter correlation: $$Nu = 0.027 \cdot Re^{4/5} \cdot Pr^{1/3}$$ Where, \(Pr\) is the Prandtl number, which can be calculated as: $$Pr = \frac{c_p \mu}{k_a}$$ $$Pr = \frac{1007 \cdot 1.966 \times 10^{-5}}{0.0271} = 0.727$$ Plugging in the values obtained, we calculate \(Nu\): $$Nu = 0.027 \cdot 63357^{4/5} \cdot 0.727^{1/3} = 324.5$$ Now, the convective heat transfer coefficient, \(h\) can be found as follows: $$h = \frac{k_a \cdot Nu}{d_h}$$ $$h = \frac{0.0271 \cdot 324.5}{0.22} = 40.3 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}$$

Calculate the temperature of the aluminum plate

We are now able to calculate the temperature of the aluminum plate, knowing the heat transfer rate and convective heat transfer coefficient. The heat transfer rate can be found by using the heat balance equation: $$Q_{total} = h \cdot A_p \cdot (T_p - T_\infty)$$ Where, \(A_p\) is the plate area, which can be calculated as \(A_p = 2 \times (0.22 \mathrm{~m} \times 0.22 \mathrm{~m}) = 0.0968 \mathrm{~m}^2\) (we consider both sides of the plate), \(T_p\) is the temperature of the plate, and \(T_\infty\) is the air temperature, given as \(20^{\circ} \mathrm{C}\). Now, solve for \(T_p\): $$T_p = T_\infty + \frac{Q_{total}}{h \cdot A_p}$$ $$T_p = 20 + \frac{40}{40.3 \cdot 0.0968}$$ $$T_p = 20 + 10.76 = 30.76^{\circ} \mathrm{C}$$ The aluminum plate's temperature is approximately \(30.76^{\circ} \mathrm{C}\).