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Answer:
i) r_t = 0.5101 m
ii) m' = 106.73 kg/s
iii) R_s = 1.26
P = 2359.8 kW
iv) β2 = 55.63°
Explanation:
We are given;
Stagnation pressure; T_01 = 290 K
Inlet velocity; C1 = 145 m/s
Cp for air = 1005 kJ/(kg·K)
Mach number; M = 0.96
Ratio of specific heats; γ = 1.4
Stagnation pressure; P_01 = 1 bar
rotational speed; N = 5500 rpm
Work done factor; τ = 0.92
Isentropic effjciency; η = 0.9
Stagnation temperature rise; ΔT_s = 22 K
i) Formula for Stagnation temperature is given as;
T_01 = T1 + C1/(2Cp)
Thus,making T1 the subject, we havw;
T1 = T_01 - C1/(2Cp)
Plugging in the relevant values, we have;
T1 = 290 - (145/(2 × 1005))
T1 = 289.93 K
Formula for the mach number relative to the tip is given by;
M = V1/√(γRT1)
Where V1 is relative velocity at the tip and R is a gas constant with a value of 287 J/Kg.K
Thus;
V1 = M√(γRT1)
V1 = 0.96√(1.4 × 287 × 289.93)
V1 = 0.96 × 341.312
V1 = 327.66 m/s
Now, tip speed is gotten from the velocity triangle in the image attached by the formula;
U_t = √(V1² - C1²)
U_t = √(327.66² - 145²)
U_t = √86336.0756
U_t = 293.83 m/s
Now relationship between tip speed and tip radius is given by;
U_t = (2πN/60)r_t
Where r_t is tip radius.
Thus;
r_t = (60 × U_t)/(2πN)
r_t = (60 × 293.83)/(2π × 5500)
r_t = 0.5101 m
ii) Now mean radius from derivations is; r_m = 1.5h
While relationship between mean radius and tip radius is;
r_m = r_t - h/2
Thus;
1.5h = 0.5101 - 0.5h
1.5h + 0.5h = 0.5101
2h = 0.5101
h = 0.5101/2
h = 0.2551
So, r_m = 1.5 × 0.2551
r_m = 0.3827 m
Formula for the area is;
A = 2πr_m × h
A = 2π × 0.3827 × 0.2551
A = 0.6134 m²
Isentropic relationship between pressure and temperature gives;
P1 = P_01(T1/T_01)^(γ/(γ - 1))
P1 = 1(289.93/290)^(1.4/(1.4 - 1))
P1 = 0.9992 bar = 0.9992 × 10^(5) N/m²
Formula for density is;
ρ1 = P1/(RT1)
ρ1 = 0.9992 × 10^(5)/(287 × 289.93)
ρ1 = 1.2 kg/m³
Mass flow rate at compressor inlet is;
m' = ρ1 × A × C1
m' = 1.2 × 0.6134 × 145
m' = 106.73 kg/s
iii) stagnation pressure ratio is given as;
R_s = (1 + ηΔT_s/T_01)^(γ/(γ - 1))
R_s = (1 + (0.9 × 22/290))^(1.4/(1.4 - 1))
R_s = 1.26
Work is;
W = C_p × ΔT_s
W = 1005 × 22
W = 22110 J/Kg
Power is;
P = W × m'
P = 22110 × 106.73
P = 2359800.3 W
P = 2359.8 kW
iv) We want to find the rotor angle.
now;
Tan β1 = U_t/C1
tan β1 = 293.83/145
tan β1 = 2.0264
β1 = tan^(-1) 2.0264
β1 = 63.73°
Formula for Stagnation pressure rise is given by;
ΔT_s = (τ•U_t•C1/C_p) × tan(β1 - β2)
Plugging in the relevant values;
22 = (0.92 × 293.83 × 145/1005) × (tan 63.73 - tan β2)
(tan 63.73 - tan β2) = 0.5641
2.0264 - 0.5641 = tan β2
tan β2 = 1.4623
β2 = tan^(-1) 1.4623
β2 = 55.63°
