Question 5.4: Deceleration of Air in a Diffuser Air at 10°C and 80 kPa ent...

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 \Delta m_{ CV }=0 \text { and } \Delta E_{ CV }=0 . 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, \Delta p e=0 . 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 \dot{m}_{1}=\dot{m}_{2}=\dot{m}
(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:

v_{1}=\frac{R T_{1}}{P_{1}}=\frac{\left.0.287 kPa \cdot m ^{3} / kg \cdot K \right)(283 K )}{80 kPa }=1.015 m ^{3} / kg

Then,

\dot{m}=\frac{1}{v_{1}} V_{1} A_{1}=\frac{1}{1.015 m ^{3} / kg }(200 m / s )\left(0.4 m ^{2}\right)=78.8 kg / s

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

\underbrace{\dot{E}_{\text {in }}-\dot{E}_{\text {out }}}_{\begin{array}{c}\text { Rate of net energy transfer } \\\text { by heat, work, and mass }\end{array}}=\underbrace{d E_{\text {system }} / {d t}^{\nearrow ^{0(steady)}} }_{\begin{array}{c}\text { Rate of change in internal, kinetic, } \\\text { potential, etc., energies }\end{array}}=0

\begin{aligned}\dot{E}_{\text {in }} &=\dot{E}_{\text {out }} \\\dot{m}\left(h_{1}+\frac{V_{1}^{2}}{2}\right)&=\dot{m}\left(h_{2}+\frac{V_{2}^{2}}{2}\right) \quad(\text { since } \dot{Q} \cong 0, \dot{W}=0, \text { and } \Delta \text { pe } \cong 0) \\h_{2} &=h_{1}-\frac{V_{2}^{2}-V_{1}^{2}}{2}\end{aligned}

The exit velocity of a diffuser is usually small compared with the inlet velocity \left(V_{2} \ll V_{1}\right) ; 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

h_{1}=h_{@ 283 K }=283.14 kJ / kg

Substituting, we get

\begin{aligned} h_{2} &=283.14 kJ / kg -\frac{0-(200 m / s )^{2}}{2}\left(\frac{1 kJ / kg }{1000 m ^{2} / s ^{2}}\right) \\ &=303.14 kJ / kg \end{aligned}

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

T_{2}= 3 0 3 K

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.