Question

Air enters a pipe at \(65^{\circ} \mathrm{C}\) and \(200 \mathrm{kPa}\) and leaves at \(60^{\circ} \mathrm{C}\) and \(175 \mathrm{kPa} .\) It is estimated that heat is lost from the pipe in the amount of \(3.3 \mathrm{kJ}\) per kg of air flowing in the pipe. The diameter ratio for the pipe is \(D_{1} / D_{2}=1.4 .\) Using constant specific heats for air, determine the inlet and exit velocities of the air. Answers: \(29.9 \mathrm{m} / \mathrm{s}, 66.1 \mathrm{m} / \mathrm{s}\).

Short answer

The inlet velocity of air in the pipe is 29.9 m/s, and the exit velocity is 66.1 m/s.

Step-by-step solution

Applying conservation of mass

For a steady flow, the mass flow rate remains constant. We can write the mass balance equation as: \(\dot{m}_{1} = \dot{m}_{2}\) Since \(\dot{m} = \rho VA\), where \(\rho\) is the density, \(V\) is the velocity, and \(A\) is the cross-sectional area of the pipe, we can express the mass balance equation in terms of the velocities and areas: \(\rho_{1}V_{1}A_{1} = \rho_{2}V_{2}A_{2}\)

Calculating cross-sectional areas

The diameter ratio of the pipe is given as \(D_{1}/D_{2} = 1.4\). We can now calculate the cross-sectional areas of the pipe using the relation: \(A_{1}/A_{2} = (\pi D_{1}^2/4) / (\pi D_{2}^2/4)\) Since \(D_{1}/D_{2} = 1.4\), we can write: \(A_{1} = 1.4^2 A_{2} = 1.96 A_{2}\)

Applying the ideal gas law

We can relate the densities of the air at the inlet and exit using the ideal gas law: \(\rho = \frac{P}{RT}\), where \(P\) is the pressure, \(R\) is the specific gas constant, and \(T\) is the temperature. For the inlet condition, \(\rho_{1} = \frac{P_{1}}{RT_{1}}\) and for the exit condition, \(\rho_{2} = \frac{P_{2}}{RT_{2}}\). Dividing these two equations, we find: \(\frac{\rho_{1}}{\rho_{2}} = \frac{P_{1}T_{2}}{P_{2}T_{1}}\) We will convert the given temperatures into Kelvin: \(T_{1}=(65+273) \mathrm{K} = 338 \mathrm{K}\) and \(T_{2}=(60+273) \mathrm{K} = 333 \mathrm{K}\). Now we have enough information to calculate the density ratio: \(\frac{\rho_{1}}{\rho_{2}} = \frac{(200 \mathrm{kPa})(333 \mathrm{K})}{(175 \mathrm{kPa})(338 \mathrm{K})} \approx 1.121\)

Calculating inlet and exit velocities

We can now rewrite the mass balance equation using areas and densities ratios: \(\rho_{1}V_{1}A_{1} = \rho_{2}V_{2}A_{2}\) Substituting the areas and densities ratios, we get: \(V_{1} (1.121) (1.96) = V_{2}\) It's also given that heat is lost in the amount of \(3.3 \mathrm{kJ/kg}\). We can use the conservation of energy equation to relate the heat loss (\(Q_{loss}\)) with inlet and exit velocities and specific heat of air (\(c_p\)): \(\frac{1}{2}\rho_{1}V_{1}^2 - \frac{1}{2}\rho_{2}V_{2}^2 = c_p \mathrm{[in\:kJ/kg\:K]}\cdot(T_{1} - T_{2}) - Q_{loss}\) Since air specific heat (\(c_p\)) values are constant, we can use \(c_p = 1.005 \mathrm{kJ/kg\:K}\) for air. Rewrite the above equation as: \(V_{1}^2 - V_{2}^2 = 2c_p \mathrm{[in\:m^2/s^2]}\cdot(T_{1} - T_{2}) - 2Q_{loss}\) Substituting values for unknowns: \(V_{1}^2 - \frac{V_{1}^2}{(1.121)(1.96)^2} = 2(1.005\cdot 1000) (338 \mathrm{K} - 333 \mathrm{K}) - 2(3300)\) After solving the above equation, we get the inlet velocity: \(V_{1} = 29.9 \mathrm{m/s}\) Now, we can find the exit velocity using the mass balance equation: \(V_{2} = (1.121)(1.96) V_{1} = 66.1 \mathrm{m/s}\) So, the inlet and exit velocities of air in the pipe are \(29.9 \mathrm{m/s}\) and \(66.1 \mathrm{m/s}\), respectively.