Question 7.DE.7: The following data refer to a single-stage axial flow gas tu......

(1) Ψ = \frac{C_{pg} (T_{01} – T_{03})} {U^2} = \frac{(1147)(145)} {345^2} = 1.4
Using velocity diagram
U/Ca = \tanβ_3 – \tanα_3
or \tanβ_3 = \frac{1}{Φ}  + \tan α_3
= \frac{1} {0.75} + tan 12°
β_3 = 57.1°
From Equations (7.14) and (7.15), we have

Ψ = 2Φ(\tanβ_2 + \tanβ_3)        (7.14)

Λ = \frac{Φ}{2}(\tanβ_3 – \tanβ_2)       (7.15)
Ψ = Φ(\tanβ_2 + \tanβ_3)

and
Λ = \frac{Φ}{2}(\tanβ_3 – \tanβ_2)
From which
\tanβ_3 = \frac{1}{2Φ}(Ψ + 2Λ)
Therefore
tan 57.1° = \frac{1}{2 \times 0.75} (1.4 + 2Λ)
Hence
Λ = 0.4595
(2) \tanβ_2 = \frac{1}{Φ}(Ψ – 2Λ)
=\frac{1} {2 \times 0.75} (1.4 – [2][0.459])
β_2 = 17.8°
\tanα_2 = \tanβ_2 + \frac{1}{Φ}
= tan 17.8° + \frac{1} {0.75} = 0.321 + 1.33 = 1.654
α_2 = 58.8°
(3)  Ca_{1} = UΦ
= (345)(0.75) = 258.75 m/s
C_2 = \frac{Ca_1} {\cosα_2} = \frac{258.75}{\cos 58.8°} = 499.49 m/s
T_{02} – T_2 = \frac{C_2^2 }{2C_p} = \frac{499.49^2}{(2)(1147)} = 108.76 K
T_2 – T_{2s} = \frac{(T_N)(499.49^2)} {(2)(1147)} = \frac{(0.05)(499.49^2)} {(2)(1147)} = 5.438 K
T_{2s} = T_2 – 5.438
T_2 = 1100 – 108.76 = 991.24 K
T_{2s} = 991.24 – 5.438 = 985.8 K
\frac{p_{01}}{ {p_2}} = \left(\frac{T_{01}} {T_{2s}}\right)^{γ/(γ – 1)}

p_2 = 4 × \left(\frac{985.8} {1100}\right)^4 = 2.58
ρ_2 = \frac{p_2} {RT_2} = \frac{(2.58)(100)} {(0.287)(991.24)} = 0.911 kg /m³
(4) Nozzle throat area = \frac{m}{ρ_1C_1} = \frac{24} {(0.907)(499.49)} = 0.053 m²
A_1 = \frac{m} {ρ_1Ca_1} = \frac{24} {(0.907)(258.75)} = 0.102 m²