1. Ch. 11: Introduction to Compressible Flow When a fixed mass of air is heated from 20 o C to 100 o C, what is change in…. p 1, h 1, s 1,  1, u 1, Vol 1 20 o C p 2, h 2, s 2,  2, u 2, Vol 2 100 o C …. Constant s? constant p? constant volume?… STATE 1 STATE 2

2. Ch. 11: Introduction to Compressible Flow When a fixed mass of air is heated from 20 o C to 100 o C – What is the change in enthalpy? Change in entropy (constant volume)? Change in entropy (constant pressure)? If isentropic change in pressure? If isentropic change in density?

3. p =  RT [R=R univ /m mole ] (11.1) du = c v dT (11.2) u 2 - u 1 = c v (T 2 – T 1 ) (11.7a) dh = c p dT (11.3) h 2 - h 1 = c p (T 2 – T 1 ) (11.7b) IDEAL, CALORICALLY PERFECT GAS

4. h = u + pv IDEAL GAS h = u + RT dh = du + RdT IDEAL GAS du = c v dT & dh = c p dT c p dT = c v dT + R dT c p – c v = R Eq. (11.4)

5. c p - c v = R (11.4) k  c p /c v ([k=  ]) (11.5) c p = kR/(k-1) (11.6a) c v = R/(k-1) (11.6b) IDEAL GAS

6. Ideal calorically perfect gas – constant c p, c v p =  RT;c p = dh/dT; c v = du/dT s 2 – s 1 = c v ln(T 2 /T 1 ) - Rln(  2 /  1 ) s 2 – s 1 = c p ln(T 2 /T 1 ) - Rln(p 2 /p 1 ) always true dq + dw = du ds =  q/T| rev Tds = du - pdv = dh – vdp

7. s 2 – s 1 = c v ln(T 2 /T 1 ) - Rln(  2 /  1 ) s 2 – s 1 = c p ln(T 2 /T 1 ) - Rln(p 2 /p 1 ) Ideal / Calorically Perfect Gas Handy if need to find change in entropy

8. Ideal / Calorically Perfect Gas C v = du/dT; C p = dh/dT; p =  RT = (1/ v)RT Tds = du + pdv = dh –vdp ds = du/T +  RTdv/T ds = c v dT/T + (R/v)dv s 2 – s 1 = c v ln(T 2 /T 1 ) + Rln(v 2 /v 1 ) s 2 – s 1 = c v ln(T 2 /T 1 ) - Rln(  2 /  1 )

9. Ideal / Calorically Perfect Gas C v = du/dT; C p = dh/dT; p =  RT = (1/ v)RT Tds = du + pdv = dh –vdp ds = du/T +  RTdv/T ds = c v dT/T + (R/v)dv Note: don’t be alarmed that c v and dv in same equation! c v = du/dT is ALWAYS TRUE for ideal gas

10. Tds = du + pdv = dh –vdp ds = dh/T – vdp/T ds = C p dT/T - (RT/[pT])dp s 2 – s 1 = C p ln(T 2 /T 1 ) - Rln(p 2 /p 1 ) Ideal / Calorically Perfect Gas C v = du/dT; C p = dh/dT; p =  RT = (1/ v)RT

11. Tds = du + pdv = dh –vdp ds = dh/T – vdp/T ds = C p dT/T - (RT/[pT])dp Ideal / Calorically Perfect Gas C v = du/dT; C p = dh/dT; p =  RT = (1/ v)RT Note: don’t be alarmed that c p and dp are in same equation! c p = dh/dT is ALWAYS TRUE for ideal gas

12. Isentropic Ideal / Calorically Perfect Gas Handy if isentropic  2 /  1 = (T 2 /T 1 ) 1/(k-1) p 2 /p 1 = (T 2 /T 1 ) k/(k-1) (  2 /  1 ) k = p 2 /p 1 ; p 2 /  2 k = const c = kRT

13. s 2 – s 1 = C v ln(T 2 /T 1 ) - Rln(  2 /  1 ) If isentropic s 2 – s 1 = 0 ln(T 2 /T 1 ) C v = ln(  2 /  1 ) R c p – c v = R; R/c v = k – 1  2 /  1 = (T 2 /T 1 ) c v /R = (T 2 /T 1 ) 1/(k-1) assumptions ISENROPIC & IDEAL GAS & constant c p, c v

14. s 2 – s 1 = c p ln(T 2 /T 1 ) - Rln(p 2 /p 1 ) If isentropic s 2 – s 1 = 0 ln(T 2 /T 1 ) c p = ln(p 2 /p 1 ) R c p – c v = R; R/c p = 1- 1/k p 2 /p 1 = (T 2 /T 1 ) c p/ R = (T 2 /T 1 ) k/(k-1) assumptions ISENROPIC & IDEAL GAS & constant c p, c v

15.  2 /  1 = (T 2 /T 1 ) 1/(k-1) p 2 /p 1 = (T 2 /T 1 ) k/(k-1) assumptions ISENROPIC & IDEAL GAS & constant c p, c v (  2 /  1 ) k = p 2 /p 1 p 2 /  2 k = p 1 /  1 k = constant

16. Ch. 11: Introduction to Compressible Flow When a fixed mass of air is heated from 20 o C to 100 o C – What is the change in enthalpy? h 2 – h 1 = C p (T 2 - T 1 )

17. Ch. 11: Introduction to Compressible Flow When a fixed mass of air is heated from 20 o C to 100 o C – Change in entropy (constant volume)? s 2 – s 1 = C v ln(T 2 /T 1 )

18. Ch. 11: Introduction to Compressible Flow When a fixed mass of air is heated from 20 o C to 100 o C – Change in entropy (constant pressure)? s 2 – s 1 = C p ln(T 2 /T 1 )

19. Ch. 11: Introduction to Compressible Flow When a fixed mass of air is heated from 20 o C to 100 o C – If isentropic change in density?  2 /  1 = (T 2 /T 1 ) 1/(k-1)

20. Ch. 11: Introduction to Compressible Flow When a fixed mass of air is heated from 20 o C to 100 o C – If isentropic change in pressure? p 2 /p 1 = (T 2 /T 1 ) k/(k-1)

21. Stagnation Reference (V=0) (refers to “total” pressure (p o ), temperature (T o ) or density (  o ) if flow brought isentropically to rest)

22. 11-3 REFERENCE STATE: LOCAL ISENTROPIC STAGNATION PROPERTIES Since p, T, , u, h, s, V are all changing along the flow, the concept of stagnation conditions is extremely useful in that it defines a convenient reference state for a flowing fluid. To obtain a useful final state, restrictions must be put on the deceleration process. For an isentropic (adiabatic and no friction) deceleration there are unique stagnation T o, p o,  o, u o, s o, h o (V o =0) properties.

23. 1-D, energy equation for adiabatic and no shaft or viscous work Eq. (8.28); h lT = [u 2 -u 1 ] -  Q/m (p 2 /  2 ) + u 2 + ½ V 2 2 + gz 2 = (p 1 /  1 ) + u 1 + ½ V 1 2 + gz 1 Isentropic process 0 Definition: h = u + pv = u + p/  ; assume z 2 = z 1 h 2 + ½ V 2 2 = h 1 + ½ V 1 2 = h o + 0 h o – h 1 = ½ V 1 2

24. 1-D, energy equation for adiabatic and no shaft or viscous work (8.28, h lT = [u 2 -u 1 ] -  Q/m) h o - h 1 = ½ V 1 2 h o – h 1 = c p (T o – T 1 ) ½ V 1 2 = c p (T o – T 1 ) ½ V 1 2 + c p T 1 = c p T o T o = {½ V 1 2 + c p T 1 }/c p T 0 = ½ V 1 2 /c p + T 1 = ½ V 2 /c p + T

25. T 0 = ½ V 1 2 /c p + T = T (1 + V 2 /[2c p T]) c p = kR/(k-1) T 0 = T (1 + V 2 /[2kRT/{(k-1)}) T 0 = T (1 + (k-1)V 2 /[2kRT]) c 2 = kRT T 0 = T (1 + (k-1)V 2 /[2c 2 ]) M = V 2 / c 2 T 0 = T (1 + [(k-1)/2] M 2 )

26. T o /T = 1 + {(k-1)/2} M 2 Steady, no body forces, one-dimensional, frictionless, ideal, calorically perfect, adiabatic, isentropic

27.  /  o = (T/T o ) 1/(k-1) T o /T = 1 + {(k-1)/2} M 2  /  o = (1 + {(k-1)/2} M 2 ) 1/(k-1) Steady, no body forces, one-dimensional, frictionless, ideal, calorically perfect, adiabatic, isentropic

28. p/p 0 = (T/T o ) k/(k-1) T o /T = 1 + {(k-1)/2} M 2 p/p 0 = (1 + {(k-1)/2} M 2 ) k/(k-1) Steady, no body forces, one-dimensional, frictionless, ideal, calorically perfect, adiabatic, isentropic

29. p =  RT; c p = dh/dT; c v = du/dT s 2 – s 1 = c v ln(T 2 /T 1 ) - Rln(  2 /  1 ) s 2 – s 1 = c p ln(T 2 /T 1 ) - Rln(p 2 /p 1 )  2 /  1 = (T 2 /T 1 ) 1/(k-1) ; p 2 /p 1 = (T 2 /T 1 ) k/(k-1) ; p 2 /  2 k = const; c = kRT p 0 /p = (1 + {(k-1)/2} M 2 ) k/(k-1) ;  o /  = (1 + {(k-1)/2} M 2 ) 1/(k-1) T o /T = 1 + {(k-1)/2} M 2 Ideal & constant c p & c v Ideal & constant c p & c v & isentropic Ideal & constant c p & c v & isentropic + …

30. Stagnation condition not useful for velocity Use critical condition – when M = 1, V* = c* (critical speed is the speed obtained when flow is isentropically accelerated or decelerated until M = 1) At critical conditions, the isentropic stagnation quantities become: p 0 /p* = (1+{(k-1)/2} 1 2 ) k/(k-1) = {(k+1)/2} k/(k-1)  o /  = (1+{(k-1)/2} 1 2 ) 1/(k-1) = {(k+1)/2} 1/(k-1) T o /T = 1 + {(k-1)/2} 1 2 = (k+1)/2 p 0 /p = (1 + {(k-1)/2} M 2 ) k/(k-1) ;  o /  = (1 + {(k-1)/2} M 2 ) 1/(k-1) T o /T = 1 + {(k-1)/2} M 2