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Isentropic Flow In A Converging Nozzle

Isentropic Flow In A Converging Nozzle. Converging Nozzle. M = 0. x = 0. M can equal 1 only where dA = 0 Can one find M > 1 upstream of exit?. M can equal 1 only where dA = 0 Can one find M > 1 upstream of exit?. Converging Nozzle. M = 0. x = 0.

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Isentropic Flow In A Converging Nozzle

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  1. Isentropic Flow In A Converging Nozzle

  2. Converging Nozzle M = 0 x = 0 M can equal 1 only where dA = 0 Can one find M > 1 upstream of exit?

  3. M can equal 1 only where dA = 0 Can one find M > 1 upstream of exit? Converging Nozzle M = 0 x = 0 No, since M = 0 at x = 0, can not increase to > 1 without at some x =1 which is not possible because dA  0 anywhere but at exit.

  4. Now that we have isentropic equations can explore the following problems (assuming they are isentropic) ~ Converging Nozzle pb may or may not equal pe How the mass flow changes with decreasing back pressure in a converging nozzle?

  5. pb = pe if Me<1 pb=po;V=0 pb=0; V=max pb pe Converging nozzle operating at various back pressures. Plotting mass flow as a function of pb/po as pb decreases from vacuum (pb/po = 0) to when the nozzle is closed (pb/po = 1; then po everywhere and flow = 0).

  6. Me = 1 Me < 1 =tVtAt 0 0.528pop0 Mass Flow Rate vs Back Pressure po/p = [1 + {(k-1)/2}M2]k/(k-1) pe*/po = {2/(k+1)}k/(k-1); k=1.4, pe*/po =0.528

  7. po/pe = [1 + {(k-1)/2}M2]k/(k-1) po/pe* = [1 + {(k-1)/2}]k/(k-1) po/pe* = [(k + 1)/2}]k/(k-1) pe*/po = [2/(k+1)] k/(k-1) k=1.4, pe*/po =0.528 p* = pe pb = pe

  8. What is the maximum mass flux through nozzle? dm/dtmax = dm/dtchoked = *V*A* = f(Ae, To, po, k, R) A* = Ae V* = c*= [kRT*]1/2 = [{2k/(k+1)}RTo]1/2 Eq. 11.22 * = p*/(RT*) p* = po[2/(k+1)]k/(k-1) Eq. 11.21a T* = To[2/(k+1)] Eq. 11.21b

  9. dm/dtmax = dm/dtchoked = *V*A* *V*A* ={p*/(RT*)}{[{2k/(k+1)}RTo]1/2}Ae *V*A* = {po[2/(k+1)]k/(k-1)}/(R{To[2/(k+1)]){[{2k/(k+1)}RTo]1/2}Ae *V*A* = Aepo(RTo)(-1+1/2) [k1/2] [2/(k+1)](k/(k-1) –1 +1/2) (k/(k-1) –1 +1/2) = [2k-2(k-1)+(k-1)]/(2[k-1]) = [k+1]/(2[k-1]) *V*A* = Aepo (RTo)-1/2k1/2[2/(k+1)][k+1]/(2[k-1]) *V*A* = Aepo (k/[RTo]-1/2)k1/2[2/(k+1)][k+1]/(2[k-1])

  10. dm/dtmax = dm/dtchoked = *V*A* *V*A* ={p*/(RT*)}{[{2k/(k+1)}RTo]1/2}Ae *V*A* = {po[2/(k+1)]k/(k-1)}/(R{To[2/(k+1)]){[{2k/(k+1)}RTo]1/2}Ae *V*A* = Aepo(RTo)(-1+1/2) [k1/2] [2/(k+1)](k/(k-1) –1 +1/2) (k/(k-1) –1 +1/2) = [2k-2(k-1)+(k-1)]/(2[k-1]) = [k+1]/(2[k-1]) *V*A* = Aepo (RTo)-1/2k1/2[2/(k+1)][k+1]/(2[k-1]) *V*A* = Aepo (k/[RTo]-1/2)k1/2[2/(k+1)][k+1]/(2[k-1])

  11. dm/dtmax = dm/dtchoked = *V*A* *V*A* ={p*/(RT*)}{[{2k/(k+1)}RTo]1/2}Ae *V*A* = {po[2/(k+1)]k/(k-1)}/(R{To[2/(k+1)]){[{2k/(k+1)}RTo]1/2}Ae *V*A* = Aepo(RTo)(-1+1/2) [k1/2] [2/(k+1)](k/(k-1) –1 +1/2) (k/(k-1) –1 +1/2) = [2k-2(k-1)+(k-1)]/(2[k-1]) = [k+1]/(2[k-1]) *V*A* = Aepo (RTo)-1/2k1/2[2/(k+1)][k+1]/(2[k-1]) *V*A* = Aepo (k/[RTo]-1/2)k1/2[2/(k+1)][k+1]/(2[k-1])

  12. k = 1.4 R = 287 mks = 0.04Aepo/To1/2 Note! Maximum mass flow rate will depend on: the exit area Ae, properties of the gas, k and R, conditions in the reservoir, po and To but not the pressure at the exit, pe.

  13. (Air so k = 1) As long as pb/po > 0.528 then pe = pb and M at throat is < 1. pe If pb/po is = 0.528 then = pb and M at throat is = 1. If pb/po is < 0.528 then pe < pb and M at throat is = 1.

  14. pb = po; Me < 1 pb < po ; Me < 1 s = 0 pb < po ; Me < 1 Me = 1 Me = 1 s > 0

  15. Once pe=p*, even if pe is continually lowered nothing happens upstream of the throat. Disturbances traveling at the speed of sound can not pass throat and propagate upstream.

  16. True or false? If pe > p* then pe = pb; but if pe = p* then as pb decreases pe = p* and pb< pe.

  17. True or false? If pe > p* then pe = pb; but if pe = p* then as pb decreases pe = p* and pb< pe.

  18. True or false? If pe > p* then pe = pb; but if pe = p* then as pb decreases pe = p* and pb< pe. TRUE

  19. Should be able to be able to explain why pe can never be lower than p*.

  20. Regime II: pb/po < p*/po Isentropic to throat Me=1 pe = p* > pb nonisentropic expansion occurs after leaving throat Regime I: 1 pb/po  p*/po isentropic and pe=pb

  21. Label a, b, c, d, e from above on this graph.

  22. Given: Ae = 6 cm2; po = 120 kPa; To = 400K; pb = 90kPa Find: pe; dm/dt

  23. Check for choking: p*/po = 0.528 p* = 0.528 po = 0.528 (120 kPa) = 63.4 kPa If back pressure is less than this than flow is choked. pb = 90 kPa > p* so flow is not choked and pe = pb = 90 kPa Given: Ae = 6 cm2; po = 120 kPa; To = 400K; pb = 90kPa Find: pe; dm/dt

  24. dm/dt = VA = e VeAe Isentropic relations: po/pe = [1+(k-1)Me2/2]k/(k-1); To/Te = 1 + (k-1)Me2/2; o/e = [1 + (k-1)Me2/2]1/(k-1) pe = eRTe Me = Ve/ce = Ve/ (kRTe)1/2 Ae/A* = (1/Me){[(1+(k-1)Me2]/(k+1)/2}(k+1)/2(k-1) Given: Ae = 6 cm2; po = 120 kPa; To = 400K; pb = pe= 90kPa Find: pe; dm/dt

  25. dm/dt = VA = e VeAe po/pe = [1+(k-1)Me2/2}k/(k-1) Me2 = 5[(po/pe)2/7 –1] = 0.428 Me= 0.654 To/Te = 1 + (k-1)Me2/2 Te = To / [1 + 0.2 Me2] = 400/{1 + 0.2(0.654)2} = 368K Given: Ae = 6 cm2; po = 120 kPa; To = 400K; pb = pe= 90kPa Find: pe; dm/dt

  26. dm/dt = VA = e VeAe Me = 0.654 kPa; Te = 368K Me = Ve/ce = Ve/ (kRTe)1/2 Ve = Me (kRTe)1/2 = 0.654[(1.4)(287)(268)]1/2 Ve = 251 m/s e = pe/(RTe) = 90,000/{(287)(368)} = 0.851 kg/m3 e VeAe = 0.128 kg/s Given: Ae = 6 cm2; po = 120 kPa; To = 400K; pb = pe= 90kPa Find: pe; dm/dt

  27. Given: Ae = 6 cm2; po = 120 kPa; To = 400K; pb = 90kPa = 45 kPa Find: pe; dm/dt

  28. Check for choking: p*/po = 0.528 p* = 0.528 po = 0.528 (120 kPa) = 63.4 kPa If back pressure is less than this than flow is choked. pb = 45 kPa < p* so flow is choked and pe pb; pe = p* = 63.4 kPa Given: Ae = 6 cm2; po = 120 kPa; To = 400K; pb = 45kPa Find: pe; dm/dt

  29. dm/dtchoked = 0.04Aepo/To1/2 k = 1.4 R = 287 mks = 0.04Aepo/To1/2 e VeAe = 0.04(0.0006)120000/4001/2 = 0.144 kg/s Given: Ae = 6 cm2; po = 120 kPa; To = 400K; pb = pe= 90kPa Find: pe; dm/dt

  30. THE END … until next time

  31. (A/A*)=(1/M2){(2/[k+1])(1+(k-1)M2/2}(k+1)/(k-1)

  32. CONVERGING-DIVERGING CONVERGING I S E N T R O P I C

  33. i – if flow is slow enough, V<0.3M, then  incompressible so B. Eq. holds. ii – still subsonic but compressibility effects more apparent, B.Eq. Not good. iii – highest pb where flow is choked; Mt=1 i, ii and iii are all isentropic flows But replace Ae by At.

  34. Note: diverging section decelerates subsonic flow, but accelerates supersonic flow. What is does for sonic flow depends on downstream pressure, pb.

  35. Note: diverging section decelerates subsonic flow, but accelerates supersonic flow. What is does for sonic flow depends on downstream pressure, pb. There are two Mach numbers, one < 1, one >1 for a given C-D nozzle which still supports isentropic flow.

  36. Flow can not expand isentropically to pb so expand through a shock. Flows are referred to as being overexpanded because pressure p in nozzle <pb. When M>1 and isentropic then flow is said to be at design conditions. Lowering pb further will have no effect upstream, where flow remains isentropic. Flow will go through 3-D irreversible expansion. Flow is called underexpanded, since additional expansion takes place outside the nozzle.

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