Chapter09

CHAPTER 9

Compressible Flow 9.1

c p = 0.24

Btu

778

o

lbm- R

ft-lb lbm ft-lb 32.2 = 6012 Btu slug slug-o R

cv = c p − R = 6012 − 1716 = 4296 = 0.171

9.2

c p = cv + R.

ft - lb ft - lb 1 Btu 1 slug = 4296 o slug - R slug - o R 778 ft - lb 32.2 lbm

Btu lbm- o R c p = kcv .

∴cp =

 1 + R or c p  1 −  = R.  k k

cp

∴ c p = Rk / ( k − 1). 9.3

If ∆s = 0, Eq. 9.1.9 can be written as cp

p  T  or ln 2  = ln 2   T1   p1  It follows that, using c p = cv + R and c p / cv = k , T p c p ln 2 = R ln 2 T1 p1

T2  p2  =  T1  p1  Using Eq. 9.1.7,

R /c p

p  = 2  p1  1−

1−

R

1 k

.

1

p  T2 p 2 ρ 1  p 2  k ρ = =  or 1 =  2  T1 ρ 2 p 1  p 1  ρ2  p1  Finally, this can be written as

− 1/ k

.

k

p2  ρ2  =  . p1  ρ1 

9.4

Substitute Eq. 4.5.18 into Eq. 4.5.17 and neglect potential energy change: & −W & S V22 − V12 p 2 p 1 Q = + − +~ u2 − ~ u1 . & m 2 ρ 2 ρ1

200

Enthalpy is defined in Thermodynamics as h = ~ u + pv = u~ + p / ρ. Therefore, & −W & S V22 − V12 Q = + h2 − h 1 . & m 2 Assume the fluid is an ideal gas with constant specific heat so that ∆h = c p ∆T. Then & −W & Q V 2 − V12 S = 2 + c p (T2 − T1 ). & m 2 Next, let c p = cv + R and k = c p / cv so that c p / R = k ( k − 1). Then, with the ideal gas

law T = p / Rρ , the first law takes the form & −W & p  Q V22 − V12 k  p2 S = + − 1 .  & m 2 k − 1  ρ2 ρ 1  9.5

Differentiate pρ − k = c using d( xy) = ydx + xdy: ρ − k dp − pkρ − k − 1 dρ = 0.

Rewrite: dp p =k . dρ ρ 9.6

The speed of sound is given by c = dp / dρ . For an isothermal process TR = p / ρ = K, where K is a constant. This can be differentiated: dp = Kdρ = RTdρ. Hence, the speed of sound is c = RT .

9.7

& S = 0 is: Eq. 9.1.4 with Q& = W

V2 + c p T = cons't. 2 V2 (V + ∆V ) 2 V 2 + 2V∆V + ( ∆V ) 2 + c pT = + c p (T + ∆T ) = + c p T + c p ∆T . 2 2 2

2V ∆V ( ∆V )2 + + c p ∆T . ∴ −V ∆V = c p ∆T = ∆h. 2 2 We neglected (∆V)2 . The velocity of a small wave is V = c. ∴0 =

9.8

For water dp ρ = 2110 × 10 6 Pa dρ Since ρ = 1000 kg / m 3 , we see that 201

∴ ∆h = − c∆V .

c = dp / dρ = 2110 × 10 6 / 1000 = 1453 m / s

9.9

∆p dp 2110 × 10 6 For water c = ≅ = = 1453 m / s. ∆ρ dρ 1000 L = velocity × time = 1453 × 0.6 = 872 m.

9.10

Since c = 1450 m/s for the small wave, the time increment is d 10 ∆t = = = 0.0069 seconds c 1450

9.11

a) M =

V 200 = = 0.588. c 1. 4 × 287 × 288 b) M = 600/ 1.4 × 1716× 466 = 0.567. c) M = 200 / 1.4 × 287 × 223 = 0.668. d) M = 600/ 1.4 ×1716 × 392 = 0.618. e) M = 200 / 1.4 × 287 × 238 = 0.647.

9.12

c = kRT = 1. 4 × 287 × 263 = 256 m / s.

∴d = ct = 256 × 1.21 = 309 m.

9.13

a) Assume T = 20°C: c = kRT = 1. 4 × 287 × 293 = 343 m /s. d = c ∆t = 343 × 2 = 686 m b) Assume T = 70°F: c = kRT = 1. 4 × 1716 × 530 = 1130 fps. d = c∆t = 1130 × 2 = 2260 ft. For every second that passes, the lightning flashed about 1000 ft away. Count 5 seconds and it is approximately one mile away.

9.14

c = 1.4 × 287 × 263 = 256 m / s.

sin α =

1 c = . M V

1000 sin α = 0.256. ∴ tan α = 0.2648 = . ∴ L = 3776 m L 3776 ∆t = = 3.776 s. 1000

202

V 1000 m

L

9.15

Use Eq. 9.2.13: c a) = sin α or V = V

1.4 × 287 × 288 = 908 m / s sin 22 o

c = sin α or V = V

1.4 × 1716 × 519 = 2980 fps sin 22 o

b)

9.16

∆p ∆p 0. 3 =− =− = −0.113 fps. ρc ρ kRT .00237 1.4 × 1716 × 519 V2 (V = ∆V) 2 Energy Eq: + c pT = + c p ( T + ∆T ). ∴ 0 = V∆V + c p ∆T . 2 2 ∴ ∆T = −c∆V / c p = − 1. 4 × 1716 × 519 ( −.113) /( 0.24 × 778 × 32.2) = 0.021o F. Btu ft - lb lbm ft - lb Note: c p =.24 × 778 × 32.2 =.24 × 778 × 32.2 . o lbm - F Btu slug slug - o F Eq. 9.2.4: ∆V = −

Then 9.17

ft 2 / sec 2 ft 2 − lb - sec 2 − o F o = = F. ft - lb / (slug - o F) sec 2 − ft - lb - ft

(units can be a pain!)

a) ρAV = ρAV + ρAdV + AVdρ + AdρdV + ρVdA + ρdAdV + VdρdA + dρdAdV Keep only the first order terms (the higher order terms—those with more than one differential quantity—will be negligible): 0 = ρAdV + AVdρ + ρVdA Divide by ρAV: dV dρ dA + + =0 V ρ A b) Expand the r.h.s. of Eq. 9.3.5 (keep only first order terms): V2 k p V 2 + 2VdV k p + dp + = + . 2 k−1ρ 2 k − 1 ρ + dρ Hence, 2VdV k  p + dp p  0= + −   2 k − 1  ρ + dρ ρ 

= VdV +

k  ρp + ρdp − pρ − pdρ    k −1 ρ 2 + ρdρ 

= VdV +

k  ρdp − pdρ    k −1 ρ2 

where we neglected ρdρ compared to ρ 2 . For an isentropic process Eq. 9.2.8 gives ρdp = kpdρ , so the above becomes

203

k kpdρ − pdρ k−1 ρ2 p k ( k − 1) pdρ = VdV + = VdV + k 2 dρ 2 k−1 ρ ρ But dρ / ρ = −dV / V − dA / A so that the above equation is p dV dA 0 = VdV + k  − −  ρ V A 0 = VdV +

which can be written as  dV dA  V 2 ρ = − 1 . A  kp  V Since c 2 = kp / ρ , and M = V/c, this is put in the form dA  V 2  dV =  2 − 1 A c  V

dA dV = M2 − 1 A V

(

or

)

c) Substituting in V = M c, c 2 = kRT , and R / c p = ( k − 1) / k , we find

T0 V2 M2 c 2 M 2 kRT = +1= +1= +1 T 2c p T 2c p T 2c p T M 2 k( k − 1) k−1 2 + 1= 1+ M . 2k 2 k /( 1 − k ) k − 1 2  p0  1 + M    k k 2 d) m & =p AM = AM −1 / 2 TR R k − 1 2  T0  1 + M    2 =

k+1

= p0

k k − 1 2  2( 1− k )  M A 1 + M    RT0 2

At the critical area A * , M * = 1. Hence, k+1

k  k + 1  2( 1− k ) & = p0 m A*  .  RT0  2  e) Since m & is constant throughout the nozzle, we can equate Eq. 9.3.17 to Eq. 9.3.18: k +1

p0

k+1

k k − 1 2  2( 1− k ) k   k + 1 2( 1− k ) M A 1 + M  = p0 A*      2  RT0 2 RT0

or k+1

A 1  2 + ( k − 1) M 2  2( k − 1) =  A* M  k +1 

204

9.18

a) p s = patm + 10 = 69.9 + 10 = 79.9 kPa abs. p1 = 69.9 kPa abs.

V 1

p p  V2 p From 1 → s : 1 + 1 = s . ρ s = ρ 1  s  2 ρ1 ρs  p1 

1/k

 79.9  =. 906   69.9 

s

V s=0

1/1. 4

= 0.997 kg / m 3 .

V12 69 900 79 900 ∴ + = . ∴V1 = 77.3 m / s. 2 .906 .997 b) p s = 26.4 + 10 = 36.4 kPa abs. p 1 = 26.4 kPa abs. p p  V2 p From 1 → s : 1 + 1 = s . ρ s = ρ 1  s  2 ρ1 ρ s  p1 

V12 26 400 36 400 + = . 2 .412 .518 9.19

V2 p p a) 1 + 1 = s . 2 ρ1 ρ s

1 /k

 36. 4  = 0. 412   26.4 

1 /1 . 4

= 0. 518 kg / m 3 .

∴V1 = 111 m /s. 1/ k

p  ρ s = ρ1  s   p1 

1/1.4

 105  = 1.22    101 

= 1.254 kg/m3 .

V12 1.4101 000 105 0001.4 + = . ∴V1 = 81.3 m/s. 2 .4 1.22 1.254 .4 V 2 4000 81.3-81 b) 1 = . ∴V1 = 81.0 m/s. % error = × 100 = 0.42%. 2 1.22 81.3 9.20

Is p r < . 5283p 0 ?

0.5283 × 200 = 105.7 kPa.

a) p r < .5283 p 0 .

∴ choked flow. ∴ M e = 1. ∴V e2 = kRTe . p e = 105.7 kPa. 1.4 × 287 Te 1000 × 298 = + 1000 Te . ∴ Te = 248.1 K, Ve = 315. 8 m / s. 2 105.7 & = 1.484 × π ×.01 2 × 315.8 = 0.1473 kg / s. ρe = = 1.484 kg / m 3 . ∴ m .287 × 248.1 1. 4 Ve2 1.4 130 000 130  ρ e  b) p r > .5283 p 0 . ∴ M e < 1. 1000 × 298 = + . =  2 .4 ρe 200  2.338 

200 = 2. 338. ∴ ρ e = 1.7187 kg / m 3 . .287 × 298 & = 1.7187 × π ×.01 2 × 257.9 = 0.1393 kg / s. ∴m ρ0 =

9.21

Is p r < . 5283p 0 ?

∴V e = 257. 9 m / s.

0.5283 × 30 = 15.85 psia.

a) p r < 15.85. ∴ choked flow and M e = 1 , p e = 15.85 psia. V e2 = kRT. 1.4 × 1716 × Te 0.24 × 530 = + 0.24 Te . ∴ Te = 441.7 o R, Ve = 1030 fps. 2( 778 × 32.2)

205

ρe =

15.85 ×144 = 0.003011 slug/ft 3 . 1716 × 441.7 2

 .5  ∴ m& = .003011 × π   × 1030 = 0.01692 slug/sec.  12  b) pr > 15.85. ∴ M e < 1, and pe = 20 psia. 30 ×144 ρ0 = = .00475 slug/ft 3 . 1716 × 530 1/1.4

 20  ∴ ρ e = .00475    30 

= .003556 slug/ft 3 .

Ve2 1.4 20 × 144 + . 2 .4 .003556 2 .5   & =.003556 × π   × 838.9 = 0.01627 slug / sec. ∴Ve = 838. 9 fps. ∴ m  12  0.24 × 530(778 × 32.2) =

ft-lb ft-lb (Note: c p =0.24 Btu/lbm-oR=0.24×778 =0.24× 778×32.2 .) o lbm- R slug-oR

9.22 a) p r < .5283 p 0 . ∴ M e = 1. ∴ p e =.5283 × 200 = 105.7 kPa. Te =.8333 × 298 = 248.3 K. 105.7 ρe = = 1.483 kg / m 3 . V e = 1. 4 × 287 × 248. 3 = 315. 9 m / s. .287 × 248.3 & ∴ m = 1.483 × π ×.01 2 × 315.9 = 0.1472 kg / s. b) p r > .5283 p 0 . ∴ p e = 130 kPa,

pe = 0.65. ∴ M e =.81 , Te =.884T0 p0

130 = 1.719 kg / m 3 , Ve =. 81 1. 4 × 287 × 263. 4 = 263.5 m / s. .287 × 263. 4 & = 1.719 × π ×.01 2 × 263. 5 = 0.1423 kg / s. ∴m ρe =

9.23

a) p r < .5283 p 0 . ∴ M e = 1. ∴ p e =.5283 × 30 = 15.85 psia. Te =. 8333 × 530 = 441.6 o R.

slug 15.85 × 144 =.003012 . Ve = 1.4 × 1716 × 441.6 = 1030 fps. 1716 × 441.6 ft 3 2 .5 & =.003012 × π   × 1030 = 0.01692 slug / sec. m  12  p 20 b) p r > .5283 p 0 . ∴ p e = 20 psia. e = =.6667 . ∴ M e =.785. Te = 0.890T0 . p 0 30 20 × 144 ∴ ρ0 = =.00356. Ve =.785 1. 4 × 1716 × 472 = 836 fps. 1716 × 472 2 .5   & =.00356π   × 836 = 0.01664 slug /sec. ∴m  12  ∴ρe =

206

9.24 p e =.5283 × 400 = 211.3 kPa abs . Te =.8333 × 303 = 252.5 K. 211.3 & = Ve = 1.4 × 287 × 252.5 = 318.5 m / s. ∴ m π ×.05 2 × 318. 5 = 7 .29 kg /s. .287 × 252. 5 9.25 p e =.5283 p 0 = 101 kPa. ∴ p 0 = 191.2 kPa abs. Te =. 8333 × 283 = 235.8 K. 101 & = Ve = 1.4 × 287 × 235. 8 = 307. 8 m / s. ∴ m π ×.03 2 × 307 .8 = 1.30 kg /s. .287 × 235.8 p 0 = 2 × 191.2 = 382.4 kPa abs. p e =.5283 p 0 = 202.0 kPa abs. Te = 235. 8 K. 202 & = Ve = 307.8 m / s since M e = 1. ∴ m π ×.03 2 × 307.8 = 2.60 kg / s. .287 × 235.8 9.26 p e =.5283 p 0 = 14.7 psia .

∴ p 0 = 27. 83 psia . Te =.8333 × 500 = 416. 6o R.

Ve = 1.4 × 1716 × 416.6 =1000 fps. ∴ ρ e = 0.3203 kg/m3 and pe = 199.4 kPa abs. p0 = 2×27.83. pe = 0.5283 p0 = 29.4 psia, & = 0.202 slug / sec. ∴m

9.27

Te = 416.6o R,

Ve = 1000 fps.

Treat the pipeline as a reservoir. Then, p e =. 5283 p 0 = 264.5 kPa abs.

M e = 1 and Ve = 1. 4 × 287(.8333 × 283 ) = 307 .8 m /s. 264.5 & = m × 30 × 10 −4 × 307.8 = 3.61 kg / s. .287 × (.8333 × 283) m & ∆t 3.61 × 6 × 60 ∆ V− = = = 333 m 3 . ρ 264.5 / (.287× .8333 × 283) 1 . 667

1. 667 × 2077 T e

 225  . 667 9.28 5193 × 300 = + 5193 T e . ∴ T e = 225 K. ∴ p e = 200   300  2 =97.45 kPa. Next, Tt = 225 K, p t = 97. 45 kPa; ∴Vt = 1.667 × 2077 × 225 = 882. 6 m / s. 97.45 ρt = = 0.2085 kg/m 3. 0.2085 × π × .032 × 882.6 = ρeπ × 0.0752Ve 2.077 × 225 V 2 1.667 p e 5193 × 300 = e + . 2 . 667 ρ e

 ρe  p e = 200   200 / 2.077 × 300 

1. 667

= 1330ρ e 1. 667 kPa.

Ve2 + 3324 × 10 3 × 9.54Ve− . 667 . 2 6 or 3.116 × 10 = Ve2 + 63 420 × 10 3 Ve−. 667 . Trial - and - error: Ve = 91.8 m / s. =

∴ ρ e = 0.3203 kg/m3 and pe = 199.4 kPa abs.

207

9.29

300 + 100  340  ρ1 = = = 4.757 kg / m 3 . ρ 2 = 4 .757    400  RT1 .287 × 293 p1

1 /1 . 4

= 4 .236 kg / m 3 .

V1 × 4 .757 × 10 2 = V2 × 4.236 × 5 2 . ∴V2 = 4. 492 V1 . V12 V12 1.4 400 000 4.492 2 V12 1.4 340 000 k p 1 V22 k p2 + = + . + = + . 2 k − 1 ρ1 2 k − 1 ρ2 2 . 4 4.757 2 .4 4.236 m ∴V1 = 37.35 . s & = ρ 1 A1V1 = 4.757 × π ×.05 2 × 37.35 = 1. 395 kg / s. ∴m 9.30

ρ1 =

p1 slug ( 45 + 14 .7 )144 = = 0.009634 . RT1 1716 × 520 ft 3

50.7  ρ 2 =.009634   59.7 

1 / 1. 4

=.008573 slug / ft 3 .

V1 ×.009634 × 4 2 = V2 ×.008573 × 2 2 . ∴V2 = 4 .495 V1 . V12 1. 4 59.7 × 144 4.495 2 V12 1. 4 50.7 × 144 + = + . ∴V1 = 121.9 fps. 2 .4 .009634 2 .4 .008573 & =.009634 π × ( 2 / 12 ) 2 × 121.9 = 0.1025 slug / sec . ∴m 9.31

V22 Energy 0 → 2: 1000 × 303 = + 1000 T2 . V2 = 3 kRT2 2 ∴ p2l = 1.627 × 20 = 32.5 kPa.

0

1

2

1.4

5.39  107.9  .4 ∴ p2 = 200  = 0.1740 kg/m 3 .  = 5.390 kPa. ρ 2 = .287 × 107.9  303  2 V V12 Energy 0 → 1: 1000 × 303 = 1 + 1000 . ∴V1 = 318. 4 m / s, T1 = 252.3 K. 2 1.4 × 287  252.3  p1 = 200   303 

1. 4 .4

= 105.4 kPa. ρ 1 =

Continuity: 1.455π ×.05 2 × 318.4 =.174π 9.32

Vt2 = kRT t . 1000 × 293 =

 244  ∴ pt = 500   293 

105.4 = 1.455 kg / m 3 . .287 × 252.3

d 22 × 3 1.4 × 287 × 107.9 . ∴ d 2 = 0.2065 m. 4

1.4 × 287 Tt + 1000 Tt . ∴ Tt = 244 .0 K. V t = 313.1 m / s. 2

1. 4 .4

= 263.5 kPa abs.

∴ ρt =

208

263.5 = 3.763 kg / m 3 . .287 × 244

Ve2 1.4 pe pe 263 500 2 2 + . 3.763 × π × .025 × 313.1 = ρe π × .075 Ve . 1.4 = 2 .4 ρ e ρe 3.7631.4

1000 × 293 =

Ve2 + 1.014 × 106 Ve−.4 . Trial-and-error: Ve = 22.2 m/s, 659 m/s. 2 ∴ ρ e = 5.897 , 0.1987 kg / m 3 . ∴ p e = 494 .2 kPa, 4.29 kPa abs. ∴ 293 000=

9.33

Ae

9.34

M t = 1. ∴ pt =.5283 × 120 = 63.4 psia, Tt =.8333 × 520 = 433.3 o R.

*

A

pe = 0.997 from Table D.1. ∴ pe = 500 ×.997 = 498.5 kPa. p0 p and e = 0.00855 from Table D.1. ∴ pe = 4.28 kPa abs. p0

= 9. ∴

∴ ρ t =.01228

slug . ft 3

& = 1 =.01228 m

π d t2 4

1. 4 × 1716 × 433.3 .

∴ d t = 0. 319 ft.

p 15 = =.125. ∴ M e = 2.014 , Te =.552 × 520 = 287 o R, Ve = 2.014 1.4 × 1716 × 287 p 0 120 = 684 fps. 2 2 π de A π ×.319 = 1.708. ∴ = 1.708 . ∴d e = 0.417 ft. * A 4 4 9.35

M e = 4. For

A A*

= 10.72, pe = .006586 × 2000 = 13.17 kPa, Te = .2381 × 293 = 69.76 K.

A = 10.72 , M e =.0584. * A

∴ p e =.9976 p 0 =.9976 × 2000 = 1995.2 kPa abs.

150

9.36

Let M t = 1. Neglect viscous effects. M1 =

= 0.430. 1.4 × 287 × 303 A A1 π × .052 π dt2 ∴ = 1.5007. ∴ At = = = . ∴ dt = 0.0816 m or 8.16 cm. 1.5007 1.5007 4 A*

9.37

p e = .5283 × 400 = 211.3 kPa abs. Tes =.8333 × 303 = 252.5. 303 − Te .96 = . ∴Te = 254.5 K. ∴Ve = 1.4 × 287 × 254.5 = 319.8 m/s. 303 − 252.5 211.3 ∴ m& = π × .052 × 319.8 = 7.27 kg/s. .287 × 254.5

209

9.38

Isentropic flow. Since k = 1.4 for nitrogen, the isentropic table may be used. A M = 3: = 4.235. * A

t

i M>1

Ve ~= 0 M<1

e

Mt = 1

100 = .9027 kg/m 3 . .297 × 373 & m 10 .00938 ∴ Ai = = = 0.00938 m 2 . ∴ A t = = 0.00221 m 2 . ρ iV i .9027 × 1181 4 .235 At M = 3 , T =. 3571 T0 , p =.02722 p 0 . 373 100 ∴ T0 = Te = = 1044 K or 772 o C . p 0 = = p e = 3670 kPa. .3571 .02722 Vi = 3 1.4 × 297 × 373 = 1181 m/s.

9.39

ρi =

Isentropic flow. Since k = 1.4 for nitrogen, the isentropic table may be used. A M = 3: = 4.235. * A

t

i M>1

Ve ~= 0 M<1

e

Mt = 1

slug 15 × 144 =.001843 . 1776 × 660 ft 3 .2 .0283 Ai = =.0283 ft 2 . ∴ At = = 0.00667 ft 2 . .001843 × 3840 4.235 At M = 3 , T =. 3571 T0 , p =.02722 p 0 . 660 15 ∴ T0 = Te = = 1848 o R or 1388 o F. p0 = p e = = 551 psia. .3571 .02722 Vi = 3 1.4 × 1776 × 660 = 3840 fps.

9.40

9.41

9.42

ρi =

101 =.4198 kg / m 3 . .189 × 1273 80 000 × 9.81 & = ρ AV 2 . Momentum: F = mV = .4198π × .252 V 2 . 6 ∴V = 1260 m/s.

Assume p e = 101 kPa. Then ρ e =

101 =.403 kg / m 3 . (Assume gases are air. ) .287 × 873 100 × 9.81 = .403 × 200 × 10−4 V 2 . ∴V = 349 m/s. & = ρAV 2 . ρ = F = mV

M t = 1.

Ae *

= 4 ; ∴ M e = 2.94 , p e =.02980 p 0 .

A Te = .3665 T0 = .3665 × 300 = 109.95 K,

pe = 100 = .0298 p0 . ∴ p0 = 3356 kPa abs.

210

FB p0A0

Ve

∴Ve = 2.94 1. 4 × 287 × 109.95 = 618 m / s. −100 ∴ FB = π ×.05 2 × 618 2 + 3 356 000π ×.2 2 = 412 000 N. .287 × 109.95

9.43

Assume an isentropic flow; Eq. 9.3.13 provides 1

103 . p  k − 1 2  k −1 = 1+ M  .   p 2 Using k = 1.4 this gives M 2 = 0.0424 or M = 0.206. For standard conditions V = M c = 0.206 14 . × 287 × 288 = 70 m / s. 9.44

a) 0.9850 × 1000 = ρ2V2 . 80 000 − p2 = 0.985 ×1000(V2 − 1000)

 V22 − 1000 2 1.4  p 2 80   + − 287 × 283 = 0.  ρ 1 = =.9850 kg / m 3 .    2 .4  ρ 2 .287 × 283 

V22 10002 1.4 V2 − + ( −985V2 + 1 065 000) − 284 300 = 0 2 2 .4 985 ∴ 3V22 − 3784V2 + 784 300 = 0. ∴V2 = 261 m / s. ρ 2 = 3.774 kg / m 3 . Substitute in and find p 2 = 808 kPa. 1000 808 M1 = = 2.966. T2 = = 746 K or 473 o C . .287 × 3.774 1.4 × 287 × 283 261 M2 = = 0.477 . 1. 4 × 287 × 746 b) M 1 = 1000 / 1.4 × 287 × 283 = 2.97. ∴ M 2 = 0.477. p 2 = 10.12 p 1 = 809.6 kPa. 809.6 T2 = 2.644 × 283 = 748 K or 475 o C . ∴ ρ 2 = = 3.771 kg / m 3 . .287 × 748 9.45

slug 12 × 144 =.002014 . .002014 × 3000 = ρ 2V2 . 1716 × 500 ft 3 Momentum: 12 ×144 − p2 = .002014 × 3000(V2 − 3000).

a) ρ 1 =

V22 − 3000 2 1.4  p2  + − 1716 ×500  = 0.  2 .4  ρ2   V  2 2 6 V2 − 3000 + 7 2  ( 19 ,854 − 6.042V 2 ) − 6.006 × 10 = 0.  6.042  2

6

∴ 6V2 − 23 ,000V2 + 15 × 10 = 0. ∴V2 = 833 fps. ρ 2 = 0.00725

slug ft 3

p2 = 102.9 psia.

M1 =

3000 102.9 × 144 = 2.74. T2 = = 1191 o R or 731 o F. 1716×.00725 1.4 × 1716 × 500

211

.

833 = 0. 492. 1.4 × 1716 × 1191 b) M1 = 3000/ 1.4 ×1716 × 500 = 2.74. ∴ M 2 = 0.493. p2 = 8.592 ×12 = 103.1 psia. 103.1 × 144 T2 = 2.386 × 500 = 1193 o R or 733 o F. ∴ ρ 2 = = 0.00725 slug / ft 3 . 1716 × 1193 M2 =

9.46

ρ 2 p2 T1 2 kM 21 − k + 1 ( k + 1) 2 M 21 ( k + 1)M 21 = = = . ρ 1 p 1 T2 k +1 2 + ( k − 1) M 21 1 + k − 1 M 2  4 kM 2 − 2k + 2 1 1  2  k + 1 p2 k − 1 M 21 = + . (This is Eq. 9.4.12). Substitute into above: 2k p 1 2k

[

 p2  ( k + 1) ( k + 1) + ( k − 1)  p1 ρ2   = = ρ1  p2  4 k + ( k − 1)( k + 1) + ( k − 1) p1   k − 1 + ( k + 1) p 2 / p 1 = . k + 1 + ( k − 1) p 2 / p 1 p For a strong schock in which 2 >> 1, p1

9.47

]

 p2  ( k + 1) ( k + 1) + k − 1 p1   . p 2 2 ( k + 1) + ( k − 1)( k + 1) p1

ρ2 k + 1 = . ρ1 k − 1

Assume standard conditions: T1 = 15 o C, ρ 1 = 101 kPa. ∴V 1 = 2 1.4 × 287 × 288 = 680 m / s. M1 = 2. ∴ M 2 = .5774. T2 = 1.688 × 288 = 486 K. p2 = 4.5 ×101 = 454 kPa.

stationary shock V2

∴V2 = .5774 1.4 × 287 × 486 = 255 m/s. ∴Vinduced = V1 − V2 = 680 − 255 = 425 m/s.

The high pressure and high induced velocity cause extreme damage. 9.48

If M 2 =.5 , then M 1 = 2.645.

∴V1 = 2.645 1.4 × 287 × 293 = 908 m /s. 1600 p 2 = 8.00 × 200 = 1600 kPa abs. ρ 2 = = 8.33 kg / m 3 . .287 × ( 2.285 × 293)

9.49

If M 2 =.5 , then M 1 = 2.645.

∴V1 = 2.645 1.4 × 1716 × 520 = 1118 fps .

p2 = 8.00 × 30 = 240 psia. ρ 2 =

240 × 144 = 0.01695 slug / ft 3 . 1716 × ( 2.285 × 520)

212

V1

9.50

p 1 =.2615 × 101 = 26.4 kPa. T1 = 223.3 K. M 1 = 1000 / 1. 4 × 287 × 223.3 = 3.34. ∴ M 2 =. 4578. p 2 = 12.85 × 26.4 = 339 kPa. T2 = 3.101 × 223.3 = 692.5 K. For isentropic flow from ‚ → €: For M = .458, p = .866 p 0 and T =.960 T0 .

9.51

9.52

∴ p 0 = 339/.866 = 391 kPa abs. T0 = 692.5/.960 = 721 K or 448 o C .

After the shock M 2 =.4752 , p 2 = 10.33 × 800 = 8264 kPa abs. For isentropic flow from ‚ → €: For M = .475, p = .857 p 0 . ∴ p 0 = 8264 /.857 = 9640 kPa abs. A = 4. ∴ M e =.147. p e =.985 p 0 ∴ p 0 = 101/.985 = 102.5 kPa abs. * A M t = 1. p t =.5283 × 102.5 = 54 .15 kPa. Tt =.8333 × 298 = 248. 3 K. 54.15 ∴ ρt = =.7599 kg / m 3 . Vt = 1.4 × 287 × 248.3 = 315.9 m / s. .287 × 248.3 & ∴ m =.7599 × π ×.025 2 × 315.9 = 0.471 kg / s. If throat area is reduced, M t

& =.7599 × π × .02 2 × 315.9 = 0.302 kg / s. remains at 1, ρ t =.7599 kg / m 3 and m 9.53

A = 4. ∴ M 1 = 2.94 , and p 2 / p 1 = 9.918. * A ∴ p 1 = 101 / 9.918 = 10.18 kPa. At M 1 = 2.94 , p / p 0 =.0298. ∴ p 0 = 10.18 /.0298 = 342 kPa abs. M t = 1, p t =.5283 × 342 = 181 kPa abs. Tt =.8333 × 293 = 244 .1 K. p e = 101 kPa = p 2 .

∴Vt = 1.4 × 287 × 244.1 = 313 m / s. M 1 = 2.94 , p 1 = 10.18 kPa abs. T1 =.3665 × 293 = 107.4 K. ∴V1 = 2.94 1.4 × 287 × 107 .4 = 611 m / s. M 2 =. 4788, p e = 101 kPa . Te = T2 = 2.609 × 107 .4 = 280.2 K. ∴V2 =.4788 1.4 × 287 × 280.2 = 161 m / s. 9.54

A = 4. ∴ M 1 = 2.94 , and p 2 / p 1 = 9.918. * A ∴ p 1 = 14 .7 / 9.918 = 1.482 psia. At M 1 = 2.94 , p / p 0 =.0298. ∴ p 0 = 1.482/.0298 = 49.7 psia . p e = 14.7 psia = p2 .

M t = 1, p t =.5283 × 49.7 = 26. 3 psia . Tt =.8333 × 520 = 433. 3 o R. ∴Vt = 1.4 × 1716 × 433.3 = 1020 fps.

M 1 = 2.94 , p 1 = 1.482 psia. T1 =.3665 × 520 = 190.6 o R. ∴V1 = 2.94 1.4 × 1716 × 190.6 = 1989 fps.

213

M 2 =. 4788, p e = 14.7 psia . Te = T2 = 2.609 × 190.6 = 497.3 o R. ∴V2 =.4788 1.4 × 1716 × 497.3 = 523 fps.

9.55

M t = 1. p t =.5283 × 500 = 264 kPa. Tt =.8333 × 298 = 248.3 K. A1 8 2 = = 2.56. ∴ M 1 = 2. 47 , p 1 =.0613 × 500 = 30.65. A* 5 2 T1 = .451 × 298 = 134.4 K. ∴V1 = 2.47 1.4 × 287 × 134 .4 = 574 m / s. M 2 =.516 , p 2 = 6.95 × 30.65 = 213 kPa. T2 = 2.108 × 134.4 = 283.3 K. A After the shock it’s isentropic flow. At M =.516 , * = 1.314 . A 2 π ×.04 p 02 =.511 × 500 = 255.5 kPa. A * = =.003825 m 2 . 1.314 Ae π ×.05 2 = = 2.05. ∴ p e =.940 × 255.5 = 240 kPa abs. = p r . M e =.298. A* .003825 . 2857 213   Te = 283.3 = 273.8 K. ∴Ve =.298 1.4 × 287 × 273.8 = 99 m / s.   240  . 3 /1 . 3

9.56

9.57

655  pt =.546 p0 =.546 × 1200 = 655 kPa. Tt = 673 = 585 K.   1200  655 ∴ ρt = = 2. 42 kg / m 3 . Vt = 1. 3 × 462 × 585 = 593 m /s. ( M t = 1. ) .462 × 585 π × d t2 & = ρ t At Vt . ∴ 4 = 2.42 × m × 593. ∴ dt = 0.060 m or 6 cm. 4 . 3 / 1. 3 101  101  Te = 673 = 380.2 K ∴ ρ e = =.575 kg / m 3 .   1200  .462 × 380.2 2 Ve + 1872 × 380.2 = 1872 × 673. (Energy from € → e .) (c p = 1872 J / kg ⋅ K) 2 πd 2e ∴Ve = 1050 m / s. ∴ 4 =.575 × 1050. ∴ d e = 0.092 m or 9.2 cm. 4

M e = 1. p e =.546 p 0 =.546 × 1000 = 546 kPa. .3

 546  1. 3 Te = 623  = 542 K.  1000 

kg 546 = 2.18 . .462 × 542 m3 πd 2 15 = 2.18 e × 571. ∴ d e = 0.124 m or 12.4 cm. 4

∴ρ e =

Ve = 1.3 × 462 × 542 = 571 m / s.

214

.3

9.58

 81.9  1. 3 o M e = 1. p e =.546 × 150 = 81.9 psia. Te = 1160  = 1009 R.  150 

slug 81.9 × 144 =.00423 . Ve = 1.3 × 2760 × 1009 = 1903 fps. 2762 × 1009 ft 3 πd e2 .25 =.00423 × 1903. ∴ d e = 0.199 ft. or 2.39". 4

∴ρe =

. 3 / 1. 3

9.59

655  M t = 1. p t =.546 × 1200 = 655 kPa. Tt = 673 = 585 K.   1200  655 ∴Vt = 1.3 × 462 × 585 = 593 m / s. ρ t = = 2.42 kg / m 3 . .462 × 585 & = 2.42 × π ×.0075 2 × 593 = 0.254 kg / s per nozzle ∴m 120  Te = 673   1200 

9.60

. 3 /1 . 3

= 396 K.

800 = 2.29. 1. 4 × 287 × 303 From Fig. 9.15, β = 46 o , 79 o . M1 =

V2 V1

a) β = 46o. ∴ M1n = 2.29sin46o = 1.65.

∴ M 2n = .654 = M 2 sin(46o − 20o ). ∴ M 2 = 1. 49.

p 2 = 3.01 × 40 = 120.4 kPa abs. T2 = 1.423 × 303 = 431 K. V2 = 1.4 × 287 × 431 × 1.49 = 620 m / s. b) β = 79 . ∴ M 1n = 2.29 sin 79 o = 2.25. ∴ M 2n =.541 = M 2 sin( 79 o − 20 o ). ∴ M 2 = 0.631. o

p 2 = 5.74 × 40 = 230 kPa abs. T2 = 1.90 × 303 = 576 K. V2 = 1.4 × 287 × 576 × .631 = 303 m / s. a detached shock

c) V1

= 35o

215

= 20o

9.61

β 1 = 40 o . ∴θ = 10 o . M 1n = 2 sin 40 o = 1.29. ∴ M 2n =.791 = M 2 sin(40 o − 10 o ). ∴ M 2 = 1.58. If θ 2 = 10 o then, with M = 1.58 , β 2 = 51o . 1.58 sin 51o = M 2n . ∴ M 2n = 1.23. ∴ M 3n =.824 = M 3 sin( 51o − 10 o ). ∴ M 3 = 1.26. β = β 2 − 10 = 51 − 10 = 41 o .

9.62

M 1n = 3.5 sin 35 o = 2.01. ∴ M 2n =.576. T2 = 1.696 × 303 = 514 K. .576 M2 = = 2.26. θ 1 = 20 o = θ 2 . ∴ β 2 = 47 o . o o sin( 35 − 20 ) M 2n = 2.26 sin 47 o = 1.65. ∴ M 3n =.654 = M 3 sin( 47 o − 20 o ). ∴ M 3 = 1.44. T3 = 1.423 × 514 = 731 K. V3 = M 3 kRT3 = 1.44 1.4 × 287 × 731 = 780 m /s.

9.63

M 1n = 3.5 sin 35 o = 2.01. ∴ M 2n =.576. T2 = 1.696 × 490 = 831o R. .576 M2 = = 2.26. θ 1 = 20 o = θ 2 . ∴ β 2 = 47 o . o o sin( 35 − 20 ) M 2n = 2.26 sin 47 o = 1.65. ∴ M 3n =.654 = M 3 sin( 47 o − 20 o ). ∴ M 3 = 1. 44. T3 = 1.423 × 831 = 1180 o R. V3 = M 3 kRT3 = 1.44 1.4 × 1716 × 1180 = 2420 fps .

9.64

M 1 = 3 , θ = 10 o . ∴ β 1 = 28 o . M 1n = 3 sin 28 o = 1.41. ∴ M 2n =.736. ∴ p 2 = 2.153 × 40 = 86.1 kPa. .736 M2 = = 2. 38. ∴ p 3 = 6.442 × 86.1 = 555 kPa . sin( 28 o − 10 o ) ( p 3 ) normal = 10.33 × 40 = 413 kPa.

9.65

At M 1 = 3 , θ 1 = 49.8 o , µ 1 = 19.47 o . (See Fig. 9.18.) θ 1 + θ 2 = 49.8 + 25 = 74 .8 o . ∴ M 2 = 4.78. p p 1 From isentropic flow table: p2 = p 1 0 2 = 20 × ×.002452 = 1.80 kPa. p1 p0 .02722 T T 1 T2 = T1 0 2 = 253 × × .1795 = 127K or −146o C. µ 2 = 12.08o. T1 T0 .3571 V2 = 4 .78 1.4 × 287 × 127 = 1080 m / s. α = 90 + 25 − 70.53 − 12.08 = 32.4 o .

9.66

θ 1 = 26.4 o . For M = 4 , θ = 65.8 o . (See Fig. 9.18.) ∴θ = 65.8 − 26.4 = 39.4 o . T T 1 T2 = T1 0 2 = 273 ×.2381 = 117 K. ∴V2 = 4 1.4 × 287 × 117 = 867 m / s. T1 T0 .5556 T2 = −156 o C.

216

9.67

θ = 26.4 o . For M = 4 , θ = 65.8 o . ∴θ = 65.8 − 26.4 = 39.4 o . T T 1 T2 = T1 0 2 = 490 × .2381 = 210o R or −250o F. T1 T0 .5556 V2 = 4 1.4 × 1716 × 210 = 2840 fps.

9.68

1 ×.04165 .0585 = 14.24 kPa. o = 2.5 sin 27 = 1.13. ∴ M 2n =.889. ∴ p 2l = 1.32 × 20 = 26.4 kPa.

a) θ 1 = 39.1 o . θ 2 = 39.1 + 5 = 44.1 o . ∴ M u = 2.72. p 2 u = 20 For θ = 5 o and M = 2.5 , β = 27 o . M 1n Ml = M2 =

.889 = 2. 37. sin( 27 o − 5 o )

b) M = 2.72 , θ = 5 o . ∴ β = 25 o . M 1n = 2.72 sin 25 o = 1.15 , M 2n =.875. .875 M 2u = = 2.56. sin( 25 o − 5 o ) For M = 2. 37 , θ = 36.0 o . For θ = 36 + 5 = 41 o , M 2 l = 2.58. c) Force on plate = ( 26.4 − 14.24 ) × 1000 × A = F . F cos 5 o = 1 2 ρ 1V1 A 2 F sin 5 o d) C D = = 1 2 ρ 1V1 A 2 CL =

9.69

12.2×.996 × 1000A

= 0.139. 1 2 × 1.4 × 2.5 × 20 000A 2 12.2 × 1000 A×.0872 = 0.0122. 1 2 × 1.4 × 2.5 × 20 000A 2

F Airfoil surface Drag

β = 19 o . M 1n = 4 sin 19 o = 1.30. ∴ p 2 = 1.805 × 20 = 36.1 kPa. M 2n =.786. .786 M2 = = 3.25. θ 1 = 54.36. θ 2 = 59.36. ∴ M 3 = 3.55. sin( 19 o − 5 o ) shock p p 1 p3 = p 2 0 3 = 36.1 × ×.0122 = 23.4 kPa. p3 M p 2 M2 p2 p0 .0188 M 3 1

 36.1 A − 23.4 A sin 5 o    6.35×.0872 2 2 CD = = = 0.0025. 1 1 2 2 ρV1 A × 1.4 × 4 × 20 2 2

217

Lift

9.70

If θ = 5o with M 1 = 4, then Fig. 9.15 → β = 18o .

M1

o

M1n = 4sin18 = 1.24. ∴ M 2n = .818. ∴ p2l = 1.627 × 20 = 32.5 kPa. .818 ∴ M2 l = = 3.64. sin( 18 o − 5 o )

shock

M2u M2 l

p0 p2 .002177 = 20 p p0 .006586 = 6.61 kPa. o o Lift 32.5 A cos5 − 20 × A / 2 − 6.61 × A / 2 × cos10 CL = = = 0.0854. 1 1 2 2 ρV A × 14 . × 4 × 20 A 2 1 2 Drag 32.5 A sin 5 o − 6.61 × A / 2 × sin 10 o CD = = = 0.010. 1 1 2 2 ρV1 A × 1.4 × 4 × 20 A 2 2

At M 1 = 4 , θ 1 = 65.8 o . At 75.8 o M 2u = 4.88. p 2 u = p 1

218

shock