Question

A wind turbine is rotating at 15 rpm under steady winds flowing through the turbine at a rate of 42,000 kg/s. The tip velocity of the turbine blade is measured to be 250 km/h. If 180 kW power is produced by the turbine, determine (a) the average velocity of the air and (b) the conversion efficiency of the turbine. Take the density of air to be 1.31kg/m3.

Answer 1

Answer:

a) 5.22 m/s

b) 31.4 %

Explanation:

f = rotating speed = 15 rpm = 15/60 =0.25 rps

m = Mass flow rate of air = 42000 kg/s

v = Tip velocity = 250 km/h = 250/3.6 = 69.44 m/s

W = Work output = 180 kW

A = Swept area of wind turbine

r = Radius of wind turbine

η = Efficiency

$r=\frac{v}{2\pi f}\\\Rightarrow r=\frac{250\div 3.6}{2\pi 0.25}=\frac{250}{1.8\pi}$

$A=\pi r^2\\\Rightarrow A=\pi\left(\frac{250}{1.8\pi}\right)^2\\\Rightarrow A=\left(\frac{250}{1.8}\right)^2\frac{1}{\pi}$

$m=\rho V\\\Rightarrow m=\rho vA\\\Rightarrow v=\frac{m}{\rho A}\\\Rightarrow v=\frac{42000}{1.31 \left(\frac{250}{1.8}\right)^2\frac{1}{\pi}}\\\Rightarrow v=5.22\ m/s$

∴ The average velocity of the air is 5.22 m/s

$E=m\frac{v^2}{2}=42000\frac{5.22^2}{2}\\\Rightarrow E=572538.92$

$\eta=\frac{W}{E}=\frac{180000}{572538.92}\\\Rightarrow \eta =0.314$

∴ Conversion efficiency of the turbine is 0.314 or 31.4 %

Answer 2

Answer:

(a) 5.223 m/s

(b) 0.314

Explanation:

The mass flow rate (m⁺) of the turbine is related to the density of air(ρ), velocity of air (vₐ) and the cross-sectional area (A) of the turbine as follows;

m⁺ = ρ x vₐ x A                -----------------(i)

But;

A = π x r²

Substitute A = π x r² into equation (i) as follows;

m⁺ = ρ x vₐ x π x r²       --------------------(ii)

But;

The radius (r) of the turbine is the ratio of the velocity (vₓ) of the turbine to the angular velocity (ω) of its blades. i.e

r = vₓ / ω

Where;

ω = 2 π f                   [f = frequency i.e number of oscillations/revolutions per second of the blades of the turbine]

=> r = vₓ / (2 π f )

Substitute the value r = vₓ / (2 π f ) into equation (ii) as follows;

m⁺ = ρ x vₐ x π x [vₓ / (2 π f )]²             ------------------(iii)

(a) Now from the question,

m⁺ = mass flow rate = 42,000 kg/s

ρ = density of air = 1.31 kg/m³

vₓ = 250km/h

= 250 x 1000/3600 m/s

= 69.44m/s

∴ vₓ = 69.44m/s

f = number of revolutions per second

=  15rpm

= 15 revolutions per minute

= 15/60 revolutions per second

∴ f = 0.25 revolutions per second

Take π = 3.142 and substitute these values into equation (iii) as follows;

m⁺ = ρ x vₐ x π x [vₓ / (2 π f )]²

42000 = 1.31 x vₐ x 3.142 x [69.44 / (2 x 3.142 x 0.25)]²

42000 = 1.31 x vₐ x 3.142 x [69.44 / 1.571]²

42000 = 1.31 x vₐ x 3.142 x [44.20]²

42000 = 1.31 x vₐ x 3.142 x [1953.64]

42000 = 8041.22 x vₐ

Solve for vₐ;

vₐ = 42000/8041.22

vₐ = 5.223 m/s

Therefore, the average velocity of the air is 5.223m/s

(b) First, let's get the kinetic energy power (Pₓ) of the system as follows;

Pₓ = KE / t

Where;

t = time

Kinetic energy (KE) = (1 / 2) x m x v²

=> Pₓ = (1 / 2) x m x v² / t

=> Pₓ = (1 / 2) x (m / t) x v²   -----------------(iv)

Where;

m/t is the mass flow rate = m⁺ = 42000 kg/s

v = velocity of air = vₐ = 5.223 m/s

Substitute these values into equation (iv) as follows;

=> Pₓ = (1 / 2) x (42000) x (5.223)²

=> Pₓ = (1 / 2) x (42000) x (27.280)

=> Pₓ = 572880 W

=> Pₓ = 572.880 kW

Also, the power output (P₀) produced by the turbine is 180kW according to the question.

Now, let's calculate the conversion efficiency (η) of the turbine as follows;

Efficiency = Power Output / Kinetic Energy Power

η = P₀ / Pₓ

Substitute the values of P₀ and Pₓ into the equation to give;

η = 180 kW / 572.880kW

η = 0.314

Therefore the conversion efficiency is 0.314