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 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 pe = 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. 5–27). 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  \mathrm{kPa} \cdot \mathrm{m}^{3} / \mathrm{kg} \cdot \mathrm{K}\right)(283  \mathrm{~K})}{80  \mathrm{kPa}}=1.015  \mathrm{~m}^{3} / \mathrm{kg}

Then,

\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}\right)=78.8  \mathrm{~kg} / \mathrm{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

\begin{aligned}\underbrace{\dot{E}_{\mathrm{in}}-\dot{E}_{\mathrm{out}}}_{\begin{array}{c}\text { Rate of net energy transfer } \\\text { by heat, work, and mass }\end{array}} &=\underbrace{ dE_{\mathrm{system}}/dt^{\nearrow 0 (steady)}}_{\begin{array}{r}\text {Rate of change in internal, kinetic, } \\\text { potential, etc., energies }\end{array}}=0\\\dot{E}_{\mathrm{in}}&=\dot{E}_{\mathrm{out}}\\\dot{m}\left(h_{1}+\frac{V_{1}^{2}}{2}\right) &\left.=\dot{m}\left(h_{2}+\frac{V_{2}^{2}}{2}\right) \quad \text { (since } \dot{Q} \cong 0, \dot{W}=0, \text { and } \Delta \mathrm{pe} \cong 0\right) \\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 (V_2 << V_1) ; 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  \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}}\right) \\&=303.14  \mathrm{~kJ} / \mathrm{kg}\end{aligned}

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

T_{2} = 303  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.