Question 4.21: Problem: Air flowing through an adiabatic diffuser enters at...
Problem: Air flowing through an adiabatic diffuser enters at a pressure of 100 kPa and 10 °C with a velocity of 200 m / s and leaves with low velocity at a pressure of 150 kPa. The inlet diameter is 2 cm and the exit diameter is 5 cm. Find the exit temperature and velocity.
Find: Exit temperature T_2 and exit velocity \pmb{V}_2 .
Known: Inlet pressure P_1 = 100 kPa, inlet temperature T_1 = 10 °C, inlet velocity \pmb{ V}_{11} = 200 m / s, exit Pressure P_{22} = 150 kPa, inlet diameter D_1 = 2 cm, exit diameter D_2 = 5 cm.
Assumptions: Negligible kinetic energy at exit, negligible heat and work transfer in diffuser, negligible change in potential energy, air is an ideal gas with constant specific heat.
Governing Equations:
Inlet velocity for diffusers \pmb{V}_1=\sqrt{2(h_2-h_1)}
Mass flow rate \dot{m} =\frac{A\pmb{V}}{\nu }
Properties: For air at 300 K specific heat capacity c_p = 1.005 kJ / kg°C (Appendix 4).
\pmb{V}_1=\sqrt{2(h_2-h_1)} =\sqrt{2c_p(T_2-T_1)}
T_2-T_1=\frac{\pmb{V}_1^2}{2c_p} =\frac{(200 \ m/s)^2}{2 \times 1.005 \times 10^3 \ J/kg^{\circ} C} =19.900 \ ^\circ C
T_2 = 19.9 \ °C + 10 \ °C = 29.9 \ °C
To find the exit velocity, since mass flow rate into an out of the diffuser are equal:
\frac{\pmb{V}_2}{\pmb{V}_1} =\frac{A_1}{A_2} \frac{\nu _2}{\nu _1} .
For an ideal gas, \nu = RT / P, so
\frac{\pmb{V}_2}{\pmb{V}_1} =\frac{A_1}{A_2} \frac{T_2}{T_1} \frac{P_1}{P_2} =\left\lgroup\frac{D_1}{D_2} \right\rgroup ^2\frac{T_2}{T_1} \frac{P_1}{P_2} =\left\lgroup\frac{2 \ cm}{5 \ cm} \right\rgroup ^2\left\lgroup\frac{303.15 \ K}{283.15 \ K} \right\rgroup \left\lgroup\frac{100 \ kPa}{150 \ kPa} \right\rgroup =0.11420,
\pmb{V}_2 = 0.1142 \times 200 \ m/s=22.84 \ m/s.
Answer: The exit temperature is 30 °C and the exit velocity is 22.8 m / s. The ratio of the exit kinetic energy to the inlet kinetic energy is ( \pmb{V}_2 / \pmb{V}_1 )^2 ≈ 0.01. Our assumption of negligible exit kinetic energy is therefore valid.