Question 5.9: Consider a diffuser as shown in Figure 5.11a with inlet and ...
Consider a diffuser as shown in Figure 5.11a with inlet and exit pressures of90 and 94 kPa, respectively. The inlet air temperature, the entrance area and the inlet air velocity are 10 °C, 6.4 m^2 and 94 m/s. Treat air as an ideal gas and use its properties. (a) Write the mass, energy, entropy, and exergy balance equations, (b) find the mass flow rate and exit temperature of the air leaving the diffuser, and (c) calculate the entropy generationrate and exergy destruction rate.
In addition, there is a nozzle, as shown in Figure 5.11b, with helium flowing through an entrance area of 0.6 m^2 at a velocity and temperature of 24 m/s and 25 °C. The exit area is measured to be 0.1 m^2 and the pressure is measured to be 90 kPa. The exit density is given as 0.142 kg/m^3. (a) Write the mass, energy, entropy, and exergy balance equations, (b) find the mass flow rate and exit velocity of helium, and (c) calculate the entropy generation rate and exergy destruction rate. Both of these units can be treated as adiabatic devices.
Solution for the Diffuser
a) Write the balance equations for mass, energy, entropy, and exergy.
MBE : \dot{m}_{1}=\dot{m}_{2}=\dot{m}EBE : \dot{m}_{1} h_{1}+m \frac{V_{1}^{2}}{2}=\dot{m}_{2} h_{2}+m \frac{V_{2}^{2}}{2}
EnBE : \dot{m}_{1} s_{1}+\dot{S}_{g e n}=\dot{m}_{2} s_{2}
\operatorname{ExBE}: \dot{m}_{1} e x_{1}+m \frac{V_{1}^{2}}{2}=\dot{m}_{2} e x_{2}+m \frac{V_{2}^{2}}{2}+\dot{E} x_{d}
b) Calculate the mass flow rate and exit temperature of the air leaving the diffuser. The mass flow rate of the air can be calculated using the ideal gas equation. As the gas can be assumed to be an ideal gas:
Pv=RT
We can rearrange the specific ideal gas equation, to find the specific volume:
v_{1}=\frac{R T_{1}}{P_{1}}v_{1}=\frac{\left(0.287 \frac{ kPa m 3}{ kg K}\right)(283 K )}{90 kPa }=0.902 \frac{ m ^{3}}{ kg }
\dot{m}=\frac{1}{v_{1}} V_{1} A_{1}
\dot{m}=\frac{1}{0.902 \frac{m^{3}}{k g}}\left(94 \frac{m}{s}\right)\left(6.4 m^{2}\right)
\dot{m}=666.9 kg / s
We now need to calculate the exit temperature of the air leaving the diffuser by using the EBE:
\dot{m}_{1} h_{1}+\frac{m V_{1}^{2}}{2}=\dot{m}_{2} h_{2}+\frac{m V_{2}^{2}}{2}\dot{m}_{1}\left(h_{1}+\frac{V_{1}^{2}}{2}\right)=\dot{m}_{2}\left(h_{2}+\frac{V_{2}^{2}}{2}\right)
As the mass flow rate is constant we can cancel out the inlet and outlet values, leaving us with just the enthalpy and inlet and exit velocities:
h_{2}=h_{1}-\left(\frac{V_{2}^{2}-V_{1}^{2}}{2}\right)As the diffuser is a pressure increasing device, the velocity at the exit of the diffuser can be assumed to be zero V_{1} \gg V_{2}:
h_{2}=283.14 \frac{ kJ }{ kg }-\left(\frac{0-\left(94 \frac{ m }{ s }\right)^{2}}{2}\left(\frac{1 \frac{ kJ }{ kg }}{1000 \frac{ m ^{2}}{ s ^{2}}}\right)\right)
h_{2}=287.56 kJ / kg
Using the property tables (such as Appendix E-1) or EES, the enthalpy can then be used to find the temperature of the air at the exit, which in this case is:
T_{2}=14.02^{\circ} Cc) Calculate the entropy generation rate and exergy destruction rate. The entropy generation can be calculated by isolating \dot{S}_{\text {gen }}:
\dot{m}_{1} s_{1}+\dot{S}_{g e n}=\dot{m}_{2} s_{2}
The entropy change of the system when rearranged is equal to \dot{S}_{\text {gen }} as follows:
\dot{S}_{g e n}=\dot{m}_{2} S_{2}-\dot{m}_{1} S_{1}\dot{S}_{\text {gen }}=\dot{m}\left(s_{2}-s_{1}\right)
As we do not have the final entropy value, it can be assumed that under ideal gas relations that the specific heat remains constant. The relationship is:
s_{2}-s_{1}=C_{p} \ln \left(\frac{T_{2}}{T_{1}}\right)-R \ln \left(\frac{P_{2}}{P_{1}}\right)s_{2}-s_{1}=1.004 kJ / kgK \ln \left(\frac{287}{283}\right)-0.287 \frac{ kPa m ^{3}}{ kg K } \ln \left(\frac{94 kPa }{90 kPa }\right)
s_{2}-s_{1}=0.00161 kJ / kgK
\dot{S}_{\text {gen }}=666.9 kg / s \times 0.00161 kJ / kgK
\dot{S}_{g e n}=1.073 k W / K
After the entropy generation is calculated one can proceed to calculate the exergy destruction rate using the following relationship:
\dot{E} x_{d}=\dot{S}_{g e n} T_{0}=1.073 kW / K \times 298 K\dot{E x}_{d}=319.77 k W
Solution for the Nozzle
a) Write the balance equations for mass, energy, entropy, and exergy:
MBE : \dot{m}_{1}=\dot{m}_{2}=\dot{m}EBE : \dot{m}_{1} h_{1}+\frac{m V_{1}^{2}}{2}=\dot{m}_{2} h_{2}+\frac{m V_{2}^{2}}{2}
\operatorname{EnBE}: \dot{m}_{1} s_{1}+\dot{S}_{\text {gen }}=\dot{m}_{2} s_{2}
\operatorname{ExBE}: \dot{m}_{1} e x_{1}+\frac{m V_{1}^{2}}{2}=\dot{m}_{2} e x_{2}+\frac{m V_{2}^{2}}{2}+\dot{E} x_{d}
b) Find the mass flow rate and exit velocity of helium.
Helium flows through a nozzle through an entrance area of 0.6 m^2 at a speed and temperature of 24 m/s and 25 °C. The exit area is measured to be 0.1 m^2, the pressure is measured at 90 kPa. The exit density is 0.142 kg/m^3.
The mass flow rate can be defined with the following relationship:
\dot{m}=\rho A Vwhere ρ is the density, A is the area, and V is the velocity of the fluid.
\dot{m}=0.1615 \frac{ kg }{ m ^{3}}\left(0.6 m ^{2}\right)\left(24 \frac{ m }{ s }\right)\dot{m}=2.33 kg / s
The final velocity at the exit of the nozzle can be calculated based on the mass balance equation, as follows:
\dot{m}=\rho_{1} A_{1} V_{1}=\rho_{2} A_{2} V_{2}One can then isolate for the exit velocity as the mass flow rate stays constant.
2.33 \frac{ kg }{ s }=0.142 \frac{ kg }{ m ^{3}} \times 0.1 m ^{2} \times V_{2}V_{2}=164.08 m / s
Using the property tables or EES one can find the entropy values for the inlet and exit flows:
s_{1}=31.58 \frac{ kJ }{ kgK } \text { and } s_{2}=31.76 \frac{ kJ }{ kgK }One may obtain the following equation from the EnBE for calculating the entropy generation rate:
\dot{S}_{\text {gen }}=\dot{m}\left(s_{2}-s_{1}\right)\dot{S}_{g e n}=2.33 \frac{k g}{s}\left(31.76 \frac{k J}{k g K}-31.58 \frac{k J}{k g K}\right)
\dot{S}_{g e n}=0.4194 k W / K
After the entropy generation is calculated, one can proceed to calculate the exergy destruction rate using the following relationship:
\dot{E} x_{d}=\dot{S}_{\text {gen }} T_{0}=0.4194 kW / K \times 298 K\dot{E x}_{d}=124.98 k W