A nozzle is required to produce a steady stream of R-134a at 790 ft/s at ambient conditions, 15 lbf/in.^2, 70 F. The isentropic efficiency may be assumed to be 90%. What pressure and temperature are required in the line upstream of the nozzle?

Question

A nozzle is required to produce a steady stream of R-134a at 790 ft/s at ambient conditions, 15 [latex]lbf / in ^{2}[/latex], 70 F. The isentropic efficiency may be assumed to be 90%. What pressure and temperature are required in the line upstream of the nozzle?

Accepted Answer

C.V. Nozzle, steady flow and no heat transfer.

Actual nozzle energy Eq.: [latex]h _{1}= h _{2}+ V _{2}^{2} / 2[/latex]

State 2 actual: Table F.10.2 [latex]h _{2}=180.975 Btu / lbm[/latex]

[latex]h _{1}= h _{2}+ V _{2}^{2} / 2=180.975+\frac{790^{2}}{2 imes 25037}=193.44 Btu / lbm[/latex]

Recall [latex]1 Btu / lbm =25037 ft ^{2} / s ^{2}[/latex] from Table A.1.

Ideal nozzle exit: [latex]h _{2 s }= h _{1}- KE _{ s }=193.44-\frac{790^{2}}{2 imes 25037} / 0.9=179.59 Btu / lbm[/latex]

Recall conversion [latex]1 Btu / lbm =25037 ft ^{2} / s ^{2}[/latex] from A.1

State 2s: [latex]\left( P _{2}, h _{2 s } ight) \Rightarrow T _{2 s }=63.16 F , \quad s _{2 s }=0.4481 Btu / lbm R[/latex]

Entropy Eq. ideal nozzle: [latex]s _{1}= s _{2 s }[/latex]

State 1: [latex]\left( h _{1}, s _{1}= s _{2 s } ight)[/latex] ⇒ Double interpolation or use software.

For 40 psia: given [latex]h _{1}[/latex] then s = 0.4544 Btu/lbm R, T = 134.47 F

For 60 psia: given [latex]h _{1}[/latex] then s = 0.4469 Btu/lbm R, T = 138.13 F

Now a linear interpolation to get P and T for proper s

[latex]P_{1}=40+20 \frac{0.4481-0.4544}{0.4469-0.4544}= 5 6 . 8 ext { psia }[/latex]

[latex]T _{1}=134.47+(138.13-134.47) \frac{0.4481-0.4544}{0.4469-0.4544}= 1 3 7 . 5 F[/latex]

Question 7.232E: A nozzle is required to produce a steady stream of R-134a at...

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