Vehicles arrive at a recreational park booth at a uniform deterministic rate of 5 veh/min. If uniform deterministic processing o
1 answer:
You might be interested in
Answer:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Explanation:
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°
Answer:
The input power is -26.58 kW.
Explanation:
Explanation is in the following attachment
