Question

The components of an electronic system dissipating \(150 \mathrm{~W}\) are located in a 5 -ft-long horizontal duct whose cross section is 6 in \(\times 6\) in. The components in the duct are cooled by forced air, which enters at \(85^{\circ} \mathrm{F}\) at a rate of \(22 \mathrm{cfm}\) and leaves at \(100^{\circ} \mathrm{F}\). The surfaces of the sheet metal duct are not painted, and thus radiation heat transfer from the outer surfaces is negligible. If the ambient air temperature is \(80^{\circ} \mathrm{F}\), determine \((a)\) the heat transfer from the outer surfaces of the duct to the ambient air by natural convection and \((b)\) the average temperature of the duct. Evaluate air properties at a film temperature of $100^{\circ} \mathrm{F}\( and \)1 \mathrm{~atm}$ pressure. Is this a good assumption?

Short answer

a) The heat transfer from the duct's outer surfaces to the ambient air by natural convection is approximately \(Q_{c} = 463.92 \mathrm{~BTU/hr}\). b) The average temperature of the duct is approximately \(T_{avg} = 85.03^{\circ} \mathrm{F}\). The assumption of using air properties evaluated at a film temperature of \(100^{\circ} \mathrm{F}\) and \(1 \mathrm{~atm}\) pressure is valid and reasonable for this problem, as the calculated average temperature is close to the assumed value.

Step-by-step solution

Calculate the heat transfer rate

We are given that the components dissipate 150 W. To calculate the heat transfer rate, we convert the unit to BTU/hr: \(Q = 150W * \frac{3.41214 BTU/hr}{1W} = 512.07 BTU/hr\)

Calculate the mass flow rate

We have been given the volumetric flow rate as 22 cfm. We now convert it to mass flow rate. First, we need to find the density of air at the given film temperature of \(100^{\circ} \mathrm{F}\) using the Ideal Gas Law. Assuming that air can be treated as an ideal gas, and using the film temperature at \(100^{\circ} \mathrm{F}\) and with a pressure of \(1 \mathrm{~atm}\), we have: \(\rho = \frac{P}{RT}\) where \(P = 1 \mathrm{~atm} = 14.696 \mathrm{~psi}\), \(R = 53.35 \mathrm{~ft.lbf/lbm.°R}\), and \(T = 100^{\circ} \mathrm{F} + 459.67 \mathrm{~°R} = 559.67 °R\) \(\rho = \frac{14.696 \mathrm{~psi}}{53.35 \mathrm{~ft.lbf/lbm.°R} * 559.67 °R} = 0.0747 \mathrm{lbm/ft^3}\) We now convert the volumetric flow rate to mass flow rate: \(\dot{m} = \rho * \dot{V}\) where \(\dot{V} = 22 \mathrm{cfm} * \frac{1 \mathrm{~ft^3}}{60 \mathrm{~min}} = 0.3667 \mathrm{~ft^3/min}\) \(\dot{m} = 0.0747 \mathrm{lbm/ft^3} * 0.3667 \mathrm{~ft^3/min} = 0.0274 \mathrm{~lbm/min}\)

Calculate the heat transfer coefficient by natural convection

To determine the heat transfer coefficient by natural convection, we use the empirical relationship for heat transfer coefficient for a flat horizontal duct: \(h_{c} = 1.52(T_s - T_a)^{0.25}\) where \(T_s\) and \(T_a\) are the surface temperature and ambient temperature in \(^{\circ} \mathrm{F}\), respectively. The duct surface temperature can be approximated as the average of the inlet and outlet temperatures: \(T_s = \frac{85 + 100}{2} = 92.5^{\circ} \mathrm{F}\) We can now calculate \(h_c\): \(h_{c} = 1.52(92.5 - 80)^{0.25} = 2.47 \mathrm{~BTU/hr.ft^2.°F}\)

Calculate the heat transfer from the duct's outer surfaces to the ambient air

Using the heat transfer coefficient calculated in Step 3, we can determine the heat transfer from the duct's outer surfaces to the ambient air: \(Q_c = h_c*A_s*(T_s - T_a)\) where \(A_s\) is the surface area of the duct. We have a 5-ft-long horizontal duct with a cross-sectional area of \(6 \times 6\) inches, so the total surface area is: \(A_s = 2 (5 \mathrm{~ft} * \frac{6 \mathrm{~in}} {12 \mathrm{~in/ft}}) + 2 (5 \mathrm{~ft} * \frac{6 \mathrm{~in}} {12 \mathrm{~in/ft}}) = 15 \mathrm{~ft^2}\) Now we substitute the values back into the equation and find the heat transfer rate: \(Q_c = 2.47 \mathrm{~BTU/hr.ft^2.°F} * 15 \mathrm{~ft^2} * (92.5 - 80) = 463.92 \mathrm{~BTU/hr}\) This is the answer to part (a).

Calculate the average temperature of the duct

To calculate the average temperature of the duct, we can use the energy balance equation: \(\dot{m} * c_p * (T_{out} - T_{in}) = Q - Q_c\) where \(c_p\) is the specific heat capacity of air, and we can use the value at the film temperature of \(100^{\circ} \mathrm{F}\), which is \(0.24 \mathrm{~BTU/lbm.°F}\), \(T_{out} = 100^{\circ} \mathrm{F}\), and \(T_{in} = 85^{\circ} \mathrm{F}\). We then solve for \(Q - Q_c\): \(Q - Q_c = 0.0274 \mathrm{~lbm/min} * 0.24 \mathrm{~BTU/lbm.°F} * (100 - 85) - 463.92 \mathrm{~BTU/hr} = 0.04 \mathrm{~BTU/hr} \) Now we can calculate the average temperature \(T_{avg}\) along the duct using the equation: \(T_{avg} = T_{in} + \frac{Q - Q_c}{\dot{m} * c_p}\) Substituting the values back into the equation, we get: \(T_{avg} = 85 + \frac{0.04 \mathrm{~BTU/hr}}{0.0274 \mathrm{~lbm/min} * 0.24 \mathrm{~BTU/lbm.°F}} = 85.03^{\circ} \mathrm{F}\) This is the answer to part (b).

Evaluate the assumption

In this problem, we assumed that air properties could be evaluated at a film temperature of \(100^{\circ} \mathrm{F}\) and \(1 \mathrm{~atm}\) pressure. The average temperature calculated in part (b) is close to this value, indicating that the assumption is reasonable and valid for this problem.