Question

The exhaust gases of an automotive engine leave the combustion chamber and enter an 8 -ft-long and 3.5-in-diameter thin-walled steel exhaust pipe at \(800^{\circ} \mathrm{F}\) and \(15.5 \mathrm{psia}\) at a rate of $0.05 \mathrm{lbm} / \mathrm{s}$. The surrounding ambient air is at a temperature of \(80^{\circ} \mathrm{F}\), and the heat transfer coefficient on the outer surface of the exhaust pipe is $3 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2},{ }^{\circ} \mathrm{F}$. Assuming the exhaust gases to have the properties of air, determine \((a)\) the velocity of the exhaust gases at the inlet of the exhaust pipe and \((b)\) the temperature at which the exhaust gases will leave the pipe and enter the air.

Short answer

Answer: The initial velocity of the exhaust gases entering the pipe is 19.96 ft/s, and the temperature at which they leave the pipe is approximately 357.13°F.

Step-by-step solution

Calculate the initial velocity of the exhaust gases.

To calculate the velocity of the exhaust gases at the inlet of the exhaust pipe, we need to find the volume flow rate (\(Q_{in}\)) of the gases. The volume flow rate is equal to the mass flow rate (\(\dot{m}\)) divided by the density of the gas (\(\rho\)). Using the Ideal Gas Law, we can express the density as \(\rho = \frac{P M}{R T}\), where \(P\) is the pressure, \(M\) is the molar mass, \(R\) is the specific gas constant, and \(T\) is the temperature. For air, \(M = 28.97\mathrm{g/mol}\) and \(R = 1716\mathrm{ft.lb/slug}\). First, convert the temperature and pressure to Rankine and psf respectively: \(T_{1} = 800^{\circ} \mathrm{F} + 459.67 = 1259.67 \mathrm{R}\) \(P_{1} = 15.5 \mathrm{psia} \times 144 = 2232 \mathrm{psf}\) Now, we can calculate the density of the air: \(\rho_{1} = \frac{P_{1} M}{R T_{1}} = \frac{2232\mathrm{psf} \times 28.97\mathrm{g/mol}}{1716\mathrm{ft.lb/slug} \times 1259.67\mathrm{R}} = 0.0746 \mathrm{slug/ft^3}\) Then, we can calculate the volume flow rate (\(Q_{in}\)): \(Q_{in} = \frac{\dot{m}}{\rho_{1}} = \frac{0.05\mathrm{lbm/s}}{0.0746 \mathrm{slug/ft^3}}= 0.6701 \mathrm{ft^3/s}\) Finally, we can find the velocity (\(V_{1}\)) by dividing the volume flow rate by the cross-sectional area of the pipe (\(A\)): \(A = \pi (\frac{3.5}{24})^2 \mathrm{ft^2} = 0.0336 \mathrm{ft^2}\) \(V_{1} = \frac{Q_{in}}{A} = \frac{0.6701\mathrm{ft^3/s}}{0.0336 \mathrm{ft^2}} = 19.96 \mathrm{ft/s}\) (a) The velocity of the exhaust gases at the inlet of the exhaust pipe is \(19.96 \mathrm{ft/s}\).

Calculate the temperature at which the exhaust gases leave the pipe.

To determine the temperature at which the exhaust gases leave the pipe, we must take into account convective heat transfer from the gases to the surrounding ambient air. The heat transfer rate (\(Q\)) can be expressed as \(Q = h A_s (T_{2} - T_{ambient})\), where \(h\) is the heat transfer coefficient, \(A_s\) is the surface area of the pipe, \(T_{2}\) is the outlet temperature, and \(T_{ambient}\) is the temperature of the ambient air. The surface area of the pipe can be calculated as follows: \(A_s = 2 \pi R L = 2 \pi (\frac{3.5}{24}\mathrm{ft}) \times 8\mathrm{ft} = 5.82\mathrm{ft^2}\) By conservation of energy, the heat transfer rate can also be expressed as \(Q = \dot{m} c_p (T_{1} - T_{2})\), where \(c_p\) is the specific heat capacity of the gas at constant pressure. For air, we can assume \(c_p = 0.24\mathrm{Btu/lbm}^{\circ}\mathrm{F}\). Now, equating the heat transfer equations, we can find the outlet temperature \(T_2\): \(h A_s (T_{2} - T_{ambient}) = \dot{m} c_p (T_{1} - T_{2})\) Plugging in the given values, we can solve for \(T_2\): \(3\mathrm{Btu/h.ft^2.}^{\circ}\mathrm{F} \times 5.82\mathrm{ft^2} (T_{2} - 80^{\circ}\mathrm{F}) = 0.05\mathrm{lbm/s} \times 0.24\mathrm{Btu/lbm}^{\circ}\mathrm{F} \times (1259.67^{\circ}\mathrm{F}- T_{2})\) Solving the equation, we find \(T_{2} = 357.13^{\circ}\mathrm{F}\). (b) The temperature at which the exhaust gases will leave the pipe and enter the air is approximately \(357.13^{\circ} \mathrm{F}\).