Question

In an axial flow compressor air enters the compressor at stagnation conditions of 1 bar and 290 K. Air enters with an absolute velocity of 145 m/s axially into the first stage of the compressor and axial velocity remains constant through the stage. The rotational speed is 5500 rpm and stagnation temperature rise is 22 K. The radius of rotor-blade has a hub to tip ratio of 0.5. The stage work done factor is 0.92, and the isentropic efficiency of the stage is 0.90. Assume for air Cp=1005 kJ/(kg·K) and γ= 1.4

Determine the followings. List your assumptions.

i. The tip radius and corresponding rotor angles at the tip, if the inlet Mach number for the relative velocity at the tip is limited to 0.96.
ii. The mass flow at compressor inlet.
iii. The stagnation pressure ratio of the stage and power required by the first stage.
iv. The rotor angles at the root section.

Answer 1

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°

Step-by-step 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°