Deceleration of Air in a Diffuser Air at 10°C and 8 0 kPa enters the diffuser of a jet engine steadily with a velocity of 200 m/s. The inlet area of the diffuser is 0 .4 m^2. The air leaves the diffuser with a velocity that is very small compared with the inlet velocity. Determine (a) the mass

Question

Deceleration of Air in a Diffuser

Air at 10°C and 8 0 kPa enters the diffuser of a jet engine steadily with a velocity of 200 m/s. The inlet area of the diffuser is 0.4m[latex]^2[/latex]. The air leaves the diffuser with a velocity that is very small compared with the inlet velocity. Determine (a) the mass flow rate of the air and (b) the temperature of the air leaving the diffuser.

Accepted Answer

Air enters the diffuser of a jet engine steadily at a specified velocity. The mass flow rate of air and the temperature at the diffuser exit are to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time at any point and thus [latex]\Delta m_{\mathrm{CV}}=0[/latex] and [latex]\Delta E_{\mathrm{CV}}=0[/latex]. 2 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical-point values. 3 The potential energy change is zero, [latex]\Delta p e=0[/latex]. 4 Heat transfer is negligible. 5 Kinetic energy at the diffuser exit is negligible. 6 There are no work interactions.

Analysis We take the diffuser as the system (Fig. 5-26). This is a control volume since mass crosses the system boundary during the process. We observe that there is only one inlet and one exit and thus [latex]\dot m_{1}=\dot{m}_{2}=\dot{m}[/latex].
(a) To determine the mass flow rate, we need to find the specific volume of the air first. This is determined from the ideal-gas relation at the inlet conditions:

[latex]v_{1}=\frac{R T_{1}}{P_{1}}=\frac{\left.0.287 \mathrm{kPa} \cdot \mathrm{m}^{3} / \mathrm{kg} \cdot \mathrm{K} ight)(283 \mathrm{~K})}{80 \mathrm{kPa}}=1.015 \mathrm{~m}^{3} / \mathrm{kg}[/latex]

Then,

[latex]\dot{m}=\frac{1}{v_{1}} V_{1} A_{1}=\frac{1}{1.015 \mathrm{~m}^{3} / \mathrm{kg}}(200 \mathrm{~m} / \mathrm{s})\left(0.4 \mathrm{~m}^{2} ight)=78.8 \mathrm{~kg} / \mathrm{s}[/latex]

Since the flow is steady, the mass flow rate through the entire diffuser remains constant at this value.

(b) Under stated assumptions and observations, the energy balance for this steady-flow system can be expressed in the rate form as

[latex]\underbrace{E_{ ext {in }}-E_{ ext {out}}}_{\begin{array}{c} ext { Rate of Net energy transfer } \ ext { by heat, work, and mass }\end{array}} =\underbrace{{d E_{ ext {system }}}/dt earrow ^{0( ext{steady})}}_{\begin{array}{c} ext { Rate of Change in internal, kinetic, } \ ext { potential, etc., energies }\end{array}}=0[/latex]

[latex]\dot{E}_{ ext {in }} =\dot{E}_{ ext {out }} [/latex]

[latex]\dot{m}\left(h_{1}+\frac{V_{1}^{2}}{2} ight) =\dot{m}\left(h_{2}+\frac{V_{2}^{2}}{2} ight) \quad( ext { since } \dot{Q} \cong 0, \dot{W}=0, ext { and } \Delta ext { pe } \cong 0) [/latex]

[latex]h_{2} =h_{1}-\frac{V_{2}^{2}-V_{1}^{2}}{2}[/latex]

The exit velocity of a diffuser is usually small compared with the inlet velocity [latex]\left(V_{2} \ll V_{1} ight)[/latex]; thus, the kinetic energy at the exit can be neglected. The enthalpy of air at the diffuser inlet is determined from the air table (Table A-17) to be

[latex]h_{1}=h_{@ 283 \mathrm{~K}}=283.14 \mathrm{~kJ} / \mathrm{kg}[/latex]

Substituting, we get

[latex]h_{2} =283.14 \mathrm{~kJ} / \mathrm{kg}-\frac{0-(200 \mathrm{~m} / \mathrm{s})^{2}}{2}\left(\frac{1 \mathrm{~kJ} / \mathrm{kg}}{1000 \mathrm{~m}^{2} / \mathrm{s}^{2}} ight) [/latex]

[latex]=303.14 \mathrm{~kJ} / \mathrm{kg}[/latex]

From Table A-17, the temperature corresponding to this enthalpy value is

[latex]T_{2}=303 \mathrm{~K}[/latex]

Discussion This result shows that the temperature of the air increases by about 20°C as it is slowed down in the diffuser. The temperature rise of the air is mainly due to the conversion of kinetic energy to internal energy.

Question 5.4: Deceleration of Air in a Diffuser Air at 10°C and 8 0 kPa en...

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