1. Isentropic Flow In A Converging Nozzle

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

3. M can equal 1 only where dA = 0 Can one find M > 1 upstream of exit? x = 0 Converging Nozzle M = 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) ~ How the mass flow changes with decreasing back pressure in a converging nozzle? Converging Nozzle p b may or may not equal p e

5. p b = p e if M e <1p b =p o ; V=0 p b =0; V=max Converging nozzle operating at various back pressures. Plotting mass flow as a function of p b /p o as p b decreases from vacuum (p b /p o = 0) to when the nozzle is closed (p b /p o = 1; then p o everywhere and flow = 0). pbpb pepe

6. M e = 1 M e < 1 p o /p = [1 + {(k-1)/2}M 2 ] k/(k-1) p e */p o = {2/(k+1)} k/(k-1) ; k=1.4, p e */p o =0.528 =tVtAt=tVtAt 0 0.528p o p 0 Mass Flow Rate vs Back Pressure

7. p o /p e = [1 + {(k-1)/2}M 2 ] k/(k-1) p o /p e * = [1 + {(k-1)/2}] k/(k-1) p o /p e * = [(k + 1)/2}] k/(k-1) p e */p o = [2/(k+1)] k/(k-1) k=1.4, p e */p o =0.528 p b = p e p* = p e

8. What is the maximum mass flux through nozzle? dm/dt max = dm/dt choked =  * V * A * = f(A e, T o, p o, k, R) A * = A e V * = c * = [kRT*] 1/2 = [{2k/(k+1)}RT o ] 1/2 Eq. 11.22  * = p*/(RT*) p* = p o [2/(k+1)] k/(k-1) Eq. 11.21a T* = T o [2/(k+1)] Eq. 11.21b

9. dm/dt max = dm/dt choked =  * V * A *  * V * A * = {p*/(RT*)}{[{2k/(k+1)}RT o ] 1/2 }A e  * V * A * = {p o [2/(k+1)] k/(k-1) }/(R{T o [2/(k+1)]){[{2k/(k+1)}RT o ] 1/2 }A e  * V * A * = A e p o (RT o ) (-1+1/2) [k 1/2 ] [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 * = A e p o (RT o ) -1/2 k 1/2 [2/(k+1)] [k+1]/(2[k-1])  * V * A * = A e p o (k/[RT o ] -1/2 ) k 1/2 [2/(k+1)] [k+1]/(2[k-1])

10. dm/dt max = dm/dt choked =  * V * A *  * V * A * = {p*/(RT*)}{[{2k/(k+1)}RT o ] 1/2 }A e  * V * A * = {p o [2/(k+1)] k/(k-1) }/(R{T o [2/(k+1)]){[{2k/(k+1)}RT o ] 1/2 }A e  * V * A * = A e p o (RT o ) (-1+1/2) [k 1/2 ] [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 * = A e p o (RT o ) -1/2 k 1/2 [2/(k+1)] [k+1]/(2[k-1])  * V * A * = A e p o (k/[RT o ] -1/2 ) k 1/2 [2/(k+1)] [k+1]/(2[k-1])

11. dm/dt max = dm/dt choked =  * V * A *  * V * A * = {p*/(RT*)}{[{2k/(k+1)}RT o ] 1/2 }A e  * V * A * = {p o [2/(k+1)] k/(k-1) }/(R{T o [2/(k+1)]){[{2k/(k+1)}RT o ] 1/2 }A e  * V * A * = A e p o (RT o ) (-1+1/2) [k 1/2 ] [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 * = A e p o (RT o ) -1/2 k 1/2 [2/(k+1)] [k+1]/(2[k-1])  * V * A * = A e p o (k/[RT o ] -1/2 ) k 1/2 [2/(k+1)] [k+1]/(2[k-1])

12. Note! Maximum mass flow rate will depend on: the exit area A e, properties of the gas, k and R, conditions in the reservoir, p o and T o but not the pressure at the exit, p e. k = 1.4 R = 287 mks = 0.04A e p o /T o 1/2

13. As long as p b /p o > 0.528 then p e = p b and M at throat is < 1. p e If p b /p o is = 0.528 then = p b and M at throat is = 1. If p b /p o is < 0.528 then p e < p b and M at throat is = 1. (Air so k = 1)

14. p b = p o ; M e < 1 p b < p o ; M e < 1 M e = 1 p b < p o ; M e < 1 M e = 1  s > 0  s = 0

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

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

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

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

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

20. Regime I: 1  p b /p o  p*/p o isentropic and p e =p b Regime II: p b /p o < p*/p o Isentropic to throat M e =1 p e = p* > p b nonisentropic expansion occurs after leaving throat

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

24. Given: A e = 6 cm 2 ; p o = 120 kPa; T o = 400K; p b = 90kPa Find: p e ; dm/dt

25. Check for choking: p*/p o = 0.528 p* = 0.528 p o = 0.528 (120 kPa) = 63.4 kPa If back pressure is less than this than flow is choked. p b = 90 kPa > p* so flow is not choked and p e = p b = 90 kPa Given: A e = 6 cm 2 ; p o = 120 kPa; T o = 400K; p b = 90kPa Find: p e ; dm/dt

26. dm/dt =  VA =  e V e A e Given: A e = 6 cm 2 ; p o = 120 kPa; T o = 400K; p b = p e = 90kPa Find: p e ; dm/dt Isentropic relations: p o /p e = [1+(k-1)M e 2 /2] k/(k-1) ; T o /T e = 1 + (k-1)M e 2 /2;  o /  e = [1 + (k-1)M e 2 /2] 1/(k-1) p e =  e RT e M e = V e /c e = V e / (kRT e ) 1/2 A e /A * = (1/M e ){[(1+(k-1)M e 2 ]/(k+1)/2} (k+1)/2(k-1)

27. dm/dt =  VA =  e V e A e Given: A e = 6 cm 2 ; p o = 120 kPa; T o = 400K; p b = p e = 90kPa Find: p e ; dm/dt p o /p e = [1+(k-1)M e 2 /2} k/(k-1) M e 2 = 5[(p o /p e ) 2/7 –1] = 0.428 M e = 0.654 T o /T e = 1 + (k-1)M e 2 /2 T e = T o / [1 + 0.2 M e 2 ] = 400/{1 + 0.2(0.654) 2 } = 368K

28. dm/dt =  VA =  e V e A e Given: A e = 6 cm 2 ; p o = 120 kPa; T o = 400K; p b = p e = 90kPa Find: p e ; dm/dt M e = 0.654 kPa; T e = 368K M e = V e /c e = V e / (kRT e ) 1/2 V e = M e (kRT e ) 1/2 = 0.654[(1.4)(287)(268)] 1/2 V e = 251 m/s  e = p e /(RT e ) = 90,000/{(287)(368)} = 0.851 kg/m 3  e V e A e = 0.128 kg/s

29. Given: A e = 6 cm 2 ; p o = 120 kPa; T o = 400K; p b = 90kPa = 45 kPa Find: p e ; dm/dt

30. Check for choking: p*/p o = 0.528 p* = 0.528 p o = 0.528 (120 kPa) = 63.4 kPa If back pressure is less than this than flow is choked. p b = 45 kPa < p* so flow is choked and p e  p b ; p e = p* = 63.4 kPa Given: A e = 6 cm 2 ; p o = 120 kPa; T o = 400K; p b = 45kPa Find: p e ; dm/dt

31. dm/dt choked = 0.04A e p o /T o 1/2 Given: A e = 6 cm 2 ; p o = 120 kPa; T o = 400K; p b = p e = 90kPa Find: p e ; dm/dt k = 1.4 R = 287 mks = 0.04A e p o /T o 1/2  e V e A e = 0.04(0.0006)120000/400 1/2 = 0.144 kg/s

32. THE END … until next time

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

42. CONVERGINGCONVERGING-DIVERGING ISENTROPICISENTROPIC

43. 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 p b where flow is choked; M t =1 i, ii and iii are all isentropic flows But replace A e by A t.

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

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

46. Lowering p b 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. When M>1 and isentropic then flow is said to be at design conditions. Flow can not expand isentropically to p b so expand through a shock. Flows are referred to as being overexpanded because pressure p in nozzle <p b.