IDEAL GAS: p = RT (11.1) du = c v dT (11.2) dh= c p dT (11.3)

1. 1st and 2nd LAWS: Q-W = U Tds = du +pdv (11.10a) Tds = h –vdp (11.10b) IDEAL GAS: p = RT (11.1) du = cvdT (11.2) dh= cpdT (11.3) h = u + RT; dh = du RdT; cpdT = cvdT + RdT; cp = cv + R (11.4) Ideal Gas and s=0 IDEAL GAS + 1st + 2nd LAWS ds = du/T + pdv/T = cvdT/T + Rdv/v s2 – s1 = cv ln (T2/T1) + R ln (v2/v1) (11.11a) ds = dh/T - vdp/T = cpdT/T + Rdp/p s2 – s1 = cp ln (T2/T1) - R ln (p2/p1) (11.11b) s2 – s1 = cv ln (T2/T1) + R ln (v2/v1) = cv ln(p2 1/p12) + (cp-cv) ln (v2/v1) s2 – s1 = cv ln(p2/p1) + cv ln(v2/v1) + cp ln (v2/v1) - cv ln (v2/v1) s2 – s1 = cv ln(p2/p1) + cp ln (v2/v1) (11.11c) (11.20a) (11.20b) (11.20c)

3. Ideal Gas 5kg; s; p1=300kPa T1=60oC 5kg; s; p2=150kPa T2= ?

4. Ideal Gas IDEAL GAS + ADIABATIC + REVERSIBLE Tvk-1 = T/(k-1) = c(11.12a) Tp(1-k)/k = c(11.12b) pvk = p/k = c(11.12c) 5kg; s; p1=300kPa T1=60oC isentropic 5kg; s; p2=150kPa T2= ?

5. Tp(1-k)/k = c(11.12b) T1(Ko)p1(1-k)/k = T2(Ko)p2(1-k)/k T1 = 333K; p1 = 300,000 Pa T2 = 273K; p2 = 150,000 Pa ?

7. Know: p1, T1, p2, T2 irreversible What is s2-s1? 200,000 Pa 388K s2 irreversible 100,000 Pa 273K s1

8. Know: p1, T1, p2, T2 & irreversible What is s2-s1? Valid for any process between equilibrium states dQ + dW = dE Tds = du + vdp IDEAL GAS & cv and cp = const s2-s1 = cvln(T2/T1) + Rln(v2/v1) (11.11a) s2-s1 = cpln(T2/T1) - Rln(p2/p1) (11.11b) s2-s1 = cvln(p2/p1) + cpln(v2/v1) (11.11c)

9. s2-s1 = cpln(T2/T1) - Rln(p2/p1) (11.11b) = 103[J/kg-K] ln(388/273) – 287[J/kg-K] ln(200,000/100,000) = 134 J/(kg-K) ?

11. Know T1, p1= p2,T2 IDEAL GAS s2-s1 = ? T1 = 858K p1 = 4.5 MPa s1 T2 = 15C p2 = p1 s2

12. 0 s2-s1 = cpln(T2/T1) - Rln(p2/p1) (11.11b) s2-s1 = 1000 ln([273+15]/858) s2-s1 = -1.09 kJ/(kg-K) 1 T 2 s IDEAL GAS & cv and cp = const s2-s1 = cvln(T2/T1) + Rln(v2/v1) (11.11a) s2-s1 = cpln(T2/T1) - Rln(p2/p1) (11.11b) s2-s1 = cvln(p2/p1) + cpln(v2/v1) (11.11c)

15. s2 – s1 = cp ln (T2/T1) - R ln (p2/p1) (11.11b)

16. What equation has q in it?

17. q = dh = cpdT q = cp(T2-T1) q = -572kJ/kg

19. Find po

20. Find po From Table A-3, pg 719 z =12.5 km; p/pSL = 0.1776 / SL = 0.2361 SL = 1.225 kg/m3 pSL = 101.3 kPa k = 1.4 po = 28.85 kPa

21. V from B.E. = ? V for compressible (=Mc) = ?

22. Find plane speed assuming incompressible* B.E. po = p + ½  V (p, T and are for z=12.5 km) po = 28.85kPa; p = 17.99kPa;  = 0.2892kg/m3 V = 274.1 m/s Find plane speed assuming compressible flow. V =Mc = M(kRT)1/2 V = 250.8 m/s ~ 9% error

24. Find To and po

25. Find To and po p0 = 184 psia To = 996oF

26. dm/dt = VA = ?

27. Find mass flow rate Know p, T, M • dm/dt = VA = 174 lbm/sec • A = 2 ft2 • V = Mc = M(kRT)1/2 • R = Ru/Mm for air = 1717 ft2(s2-R) = 8314 m2(s2-K) • For R = 1717 ft2/(s2-R) , T must be in Rankine (460 +60 = 520R) • V = 3.0(1.4*1717*520*)1/2 = 3354 ft/sec  = p/(RT) • = 5[lbf/in2][32.2lbm/lbf][144in2/ft2]/(1717[ft2/(s2-R)]520R)  = 0.0260 lbm/ft3

29. M1 T1 p1 M2>M1 T2>T1 p1>p2 flow Q added

30. s2 – s1 = cv ln (T2/T1) + R ln (v2/v1) (11.11a) s2 – s1 = cp ln (T2/T1) - R ln (p2/p1) (11.11b) s2 – s1 = cv ln(p2/p1) + cp ln (v2/v1) (11.11c)

31. K N O W s2 – s1 = cp ln (T2/T1) - R ln (p2/p1) (11.11b) KNOW  = p/(RT)

32. At location 1 po1 = 1.0MPa[1 + 0.2*(0.2)2]3.5 = 1.028 MPa At location 2 po2 = 862.7kPa[1 + 0.2*(0.4)2]3.5 = 0.9632kPa At location 1 To1 = 580K[1 + 0.2*(0.2)2] = 584.6K At location 2 To2 = 1727K[1 + 0.2*(0.4)2] = 1782K

33. s2 – s1 = cp ln (T2/T1) - R ln (p2/p1) (11.11b) 1004 J/kg-K 1727/580 0.8627/1.0 287 J/kg-k s2 – s1 = 1138 J/kg-K

34. ? s2 – s1 = 1138 J/kg-K po1 = 1.028 MPa po2 = 0.9632kPa To1 = 584.6K To2 = 1782K

37. Can consider ideal gas T1 = 1573oK; p1 = 2.0 MPa T2 = 773oK; p2 = 101 kPa u = ?; h = ?; s = ?

38. Valid for any process between equilibrium states dQ + dW = dE Tds = du + vdp IDEAL GAS & cv and cp = const s2-s1 = cvln(T2/T1) + Rln(v2/v1) (11.11a) s2-s1 = cpln(T2/T1) - Rln(p2/p1) (11.11b) =143 J/(kg-K) s2-s1 = cvln(p2/p1) + cpln(v2/v1) (11.11c) u = cVT(11.2) h = cpT(11.3)

39. increasing pressure

41. Steady, adiabatic flow of air, dm/dt = 0.5 kg/sec, through a turbine. At inlet, V1 = 0, T1 = 1300C, p1 = 2.0 mPa (abs) At outlet, V2 = 200 m/s, T2 = 500C, p2 = 101 kPa Label state points on a Ts diagram:

42. If isentropic: T2 = T1 (p2/p1)(k-1)/k = 670K (397C)  500C So not isentropic!

43. What is power produced by turbine? Steady, adiabatic flow of air, dm/dt = 0.5 kg/sec, through a turbine. At inlet, V1 = 0, T1 = 1300C, p1 = 2.0 mPa (abs); At outlet, V2 = 200 m/s, T2 = 500C, p2 = 101 kPa 0 dW/dt + dQ/dt = (dm/dt) [(h2 + (V2)2/2 + gz2) - (h1 + (V1)2/2 + gz1)] z2 = z1 h2 – h1 = cp (T2 – T1)

45. Can speed of car at 60 mph and 120 mph be considered incompressible? [0 - 1]/0 = ? < 5% then we consider incompressible M = ? < 0.3 the answer is yes!

46. 0 = 1{ 1 + [(k-1)/2]M12}1/(k-1) M1 = V1/c1 c1 = (kRT1)1/2 [0 - 1]/0 = 0.3% M = 0.0782 V1 = 60 mph = 26.8 m/s; R = 287 J/(kg-K); 1 = p1/(RT1) = 1.201 kg.m3;

47. 0 = 1{ 1 + [(k-1)/2]M12}1/(k-1) M1 = V2/c1 c1 = (kRT1)1/2 [0 - 1]/0 = 1.21% M = 0.156 V1 = 120 mph = 53.6 m/s; R = 287 J/(kg-K); 1 = p1/(RT1) = 1.201 kg.m3;

49. Know p0, p and T and are asked to find V of aircraft.

50. Know p0, p and T and are asked to find V of aircraft. M = V/c so V = Mc c = (kRT)1/2 po/p = (1 + [(k-1)/2] M2)k/(k-1)