Soal 9.4
Udara mengalir secara isentropis melalui sebuah talang. Pada penampang 1 luas, kecepatan, tekanan, dan suhunya adalah A1 = 1 ft2 , V1 = 607 ft/s, p1 = 12.017 lbf/ft2, dan T1 = 857 R. tentukan (a) T0, (b) Ma1, (c) Po, dan (d) A*. kalau A2 = 0,75 ft2 tentukan Ma2 dan P2 kalau V2 adalah (e) subsunik atau (f) supersonic.
Penyelesaian dengan V1 dan T1 diketahui, maka persamaan energy (9.23) memberikan
T0= T1+(V_1^2)/〖2c〗_p =857 + 〖((607))/(2(6010))〗^2= 857+30.653=887,653◦ R jawab(a)
Kepesatan bunyi a1 ≈ 49(857)1/2 = 1434,45 ft/s; maka
Ma1 = V1/a1 =607/1434.45 = 0,42 jawab(b)
dengan diketahuinya Ma1, maka
P_0/P_1 =(1 + 0,2 Ma12)3,5= 1,129
Sehingga P0 = 1,129P1 = 1,129(12.017) = 13.567,2 lbf/ft2 jawab(c)
Dengan cara yang sama, dari persamaan(9.45)
A_1/A^* = 〖[1+0,2(0,42)^2]〗^(3 )/(1,728(0,42)) =1,529
Sehingga A*= A1/1,529 = 〖1 ft〗^2/1,529 = 0,654 ft2 jawab(d)
Leher ini harus benar benar ada dalam talang itu untuk mengembangkan Ma1 subsonik menjadi aliran supersonic di baian hilir.
Dengan A2 = 0,75 ft2 , kita dapat menghitung A2/A* = 0,75/0,654 = 1,147. Untuk aliran subsonic, kita gunakan persamaan (9.48b) untuk memperkirakan
Ma2 ≈ 1-0,88(ln 1,147)0,45 = 0,64 jawab(e)
Sehingga P2=P0/[1+0,2(0,64)^2 ]^3,5 =(13.567,2 )/1,317= 10.302 lbf/ft2 jawab(e)
Untuk keadaan (e) tidak ada leher sonic misalnya luas talang belum bekurang secukupnya untuk menciptakan aliran supersonic. Sepertinya itu dapat terjadi kea rah hilir, seperti diperlihatkan dalam gambar C9.4. sebaliknya , kalau aliran pada penampang 2 itu supersonic, kita menggunakan persamaan(9.48) untuk memperoleh taksiran
Ma2≈ 1 + 1,2 ( 1,147 – 1 )1/2 = 1,46 jawab(f)
Sehingga P2 = (13.567,2 )/[1+0,2(1,46)^2 ]^3,5 =(13.567,2 )/3,46=3921 lbf/ft2 jawab(f)
Jawaban- jawaban diatas mempunyai ketepatan lebih bagus dari 1%. Perhatikan bahwa arus tekanan aliran supersonic itu jauh lebih rendah daripada keadaan subsonic pada penampang 2, untuk luas penampang yang sama, dan leher sonic (A*= 0,654 ft2) sudah harus terjadi diantara penampang 1 dan 2.
Soal 9.6
Air flow from a reservior where p = 317 kPa and T = 517 K trough a throat to section 1 in fig. E9.6, where there is a normal shock wave. Compute (a) P1, (b) P2 (c) P02,(d) A2*, (e) P03, (f) A3*, (g)P3, (h) T03, and (i) T3.
Solution
The reservoir conditions are the stagnation properties, which, for assumed one dimensional adiabatic frictionless flow, hold trough the throat up to section 1
P01 = 317 kPa T01 = 517 K
Ashock wave cannot exist unless Ma1 is supersonic; therefore the flow must have accelerated through a throat which is sonic
At = A1*= 1 m2
We can now find the mach number Ma1 from the known isentropic area ratio
A_t/(A_1*)=(2 m^2)/(1 m^2 )=2
From Eq (9.48c)
Ma1 ≈ 1 + 1,2(2.0 – 1 )1/2 = 2,20
further iteration with E1. (9.45) would give Ma1 = 2.1972, showing that Eq. (9.48c) gives satisfactory accuracy. The pressure P1 follows from the isentropic relation (9.28)
P_01/P_1 = [1 + 0.2(2.20)2]3.5 = 10.7
Or P1 = (317 kPa)/10.7 = 29.6 kPa answer(a)
The pressure P2 is now obtained from Ma1 and the normal shock relation (9.55)
P_2/P_1 =1/2.4[2.8(2.20)2 – 0,4] = 5,48
Or P2 = 5.48(29.6) = 162 kPa answer(b)In similiar manner, for Ma1 = p02/p01 = 0.628 from Eq. (9.58) and A2*/A1* = 1.592 from Eq. (9.59)
P02 = 0.628(317 kPa) = 199 kPa answer(c)
A2* = 1.592(1 m2) = 1.592 m2 answer(d)
The flow from section 2 to 3 is isentropic (but at higher entropy than the flow upstream of the shock). Thus
P03 = P02 = 199 kPa answer(e)
A3* = A2* = 1.592 m2 answer(f)
knowing A3*, we can now compute p3 by finding Ma3 and without borhering to find Ma2 (which) happens to equal 0.547). the area ratio at section 3 is
A_3/(A_3*)= (3 m^2)/(1.592 m^2 )=1.884
Then, since Ma3 is known to be subsonic because it is downstream of a normal shock, we use Eq. (9.48a) to estimate
Ma3 ≈ (1+0.27/(1.884)^2)/(1.728(1.884)) = 0.330
The pressure P3 then follows from the isentropic relation (9.28)
P_03/P_3 = [1 + 0.2(0.330)2]3.5 = 1.078
Or P3 = (199 Kpa)/1.078 = 185 kPa answer(g)
Meanwhile, the flow is diabatic throughout the duct; thus
T01 = T02 = T03 = 517 K answer(h)
Therefore, finally, from the adiabatic relation (9.26)
T_03/T_3 = 1 + 0.2(0.330)2 = 1.022
Or T3 = (517 K)/1.022 = 505 K answer(i)
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