Question

In the runner of a reaction-type hydraulic turbine, the followings are given: r

J
​
=25 cm,α
l
​
=30
∘
, α
2
​
=90
∘
, cross-sectional area perpendicular to the absolute velocity c
l
​
is As=0. 125 m
2
, loss of head hL=15 m, leakage efficiency η
x
​
=0. 95, the number of revolutions of the runner is n=300rpm, the flow rate is Q=3 m
3
/s and the tangential velocity coefficient at the outlet is k
n2
​
=0. 3. Determine a) Net head (H
0
​
), b) Hydraulic efficiency (η
ℏ
​
), c) Relative velocity at the runner input (w
l
​
) and tangential velocity at the outlet (u
2
​
), d) For 100 m head (H
∘
​

∘
), find the number of revolutions (n
′
) under the best efficiency conditions

Answer 1

a) To find the net head (H0), we use the following formula:

H0 = H - hL

where H is the total head and hL is the loss of head. We are given that hL = 15 m. To find H, we use the following formula:

H = (V^2)/(2g) + z

where V is the absolute velocity at the runner input, g is the acceleration due to gravity, and z is the vertical distance between the centerline of the runner and the free surface of the water. Since the runner is a reaction-type turbine, we can assume that the velocity triangles are axial and that the absolute velocity at the runner input is equal to the relative velocity. We can also assume that the flow is incompressible and that the velocity of the water is negligible at the inlet and outlet of the turbine.

From the given information, we know that the cross-sectional area perpendicular to the absolute velocity at the runner input is As = 0.125 m^2, the flow rate is Q = 3 m^3/s, and the tangential velocity coefficient at the outlet is k_n2 = 0.3. We can use these values to find the absolute velocity at the runner input:

V = Q/As = 3/0.125 = 24 m/s

We can then use the formula for total head to find H:

H = (V^2)/(2g) + z = (24^2)/(2*9.81) + 25/2 = 156.5 m

Finally, we can use the formula for net head to find H0:

H0 = H - hL = 156.5 - 15 = 141.5 m

Therefore, the net head is 141.5 m.

b) To find the hydraulic efficiency (η_ℏ), we use the following formula:

η_ℏ = (H0*η_x)/(Q*g)

where η_x is the leakage efficiency. We are given that η_x = 0.95. Substituting the given values, we get:

η_ℏ = (141.5*0.95)/(3*9.81) = 0.459

Therefore, the hydraulic efficiency is 0.459 or 45.9%.

c) To find the relative velocity at the runner input (w_l), we use the following formula:

w_l = V/cos(α_2)

where α_2 is the angle between the absolute velocity and the tangent to the runner at the outlet. We are given that α_2 = 90°, so cos(α_2) = 0. Substituting the given values, we get:

w_l = V/cos(α_2) = undefined

The relative velocity at the runner input is undefined because the denominator in the formula is zero.

To find the tangential velocity at the outlet (u_2), we use the following formula:

u_2 = k_n2*V

where k_n2 is the tangential velocity coefficient at the outlet. We are given that k_n2 = 0.3. Substituting the given values, we get:

u_2 = k_n2*V = 0.3*24 = 7.2 m/s

Therefore, the tangential velocity at the outlet is 7.2 m/s.

d) To find the number of revolutions (n') under the best efficiency conditions for a head of 100 m, we can use the following formula:

η_ℏ = (H0'*η_x)/(Q'*g)

where H0' is the net head, Q' is the flow rate, and g is the acceleration due to gravity. We want to find n' such that η_ℏ is maximized for a head of 100 m. Since the hydraulic efficiency is a function of the number of revolutions, we need to find the value of n' that maximizes η_ℏ.

To do this, we can plot η_ℏ as a function of n' and find the maximum value. However, this is a time-consuming process. Alternatively, we can use the following approximation:

n' = n*(H0'/H0)^0.5

where n is the given number of revolutions and H0 is the given net head. This approximation is based on the fact that the hydraulic efficiency is proportional to the square root of the net head and inversely proportional to the square root of the number of revolutions.

Substituting the given values, we get:

n' = 300*(100/141.5)^0.5 = 258.5 rpm

Therefore, for a head of 100 m, the number of revolutions under the best efficiency conditions is approximately 258.5 rpm.

Answer 2

a) To determine the net head, we can use the following formula:

H0 = H + hL

where H is the total head and hL is the head loss. We are given that hL = 15 m, so we need to find H.

To find H, we can use the following formula:

H = (w2/2g) + (p2 - p1)/ρg + z2 - z1

where w is the flow rate, g is the acceleration due to gravity, p is the pressure, ρ is the density of the fluid, z is the height, and the subscripts 1 and 2 refer to two different points in the system.

We can assume that the turbine is operating at steady state, which means that the pressure and height at the inlet and outlet of the turbine are the same. Therefore, we can simplify the formula to:

H = w2/2g

Substituting the given values, we get:

H = (3 m3/s)2 / (2 x 9.81 m/s2) = 45.98 m

Therefore, the net head is:

H0 = 45.98 m + 15 m = 60.98 m

b) To determine the hydraulic efficiency, we can use the following formula:

ηℏ = (H0 × Q) / (g × As × H∘)

where H∘ is the available head, which is given as 100 m.

Substituting the given values, we get:

ηℏ = (60.98 m × 3 m3/s) / (9.81 m/s2 × 0.125 m2 × 100 m) = 0.147 or 14.7%

c) To determine the relative velocity at the runner input (wl) and the tangential velocity at the outlet (u2), we can use the following formulas:

wl = Q / As

u2 = k n2 √(2gH0)

Substituting the given values, we get:

wl = 3 m3/s / 0.125 m2 = 24 m/s

u2 = 0.3 x 300 rpm x (2π/60) x √(2 x 9.81 m/s2 x 60.98 m) = 36.68 m/s

d) To find the number of revolutions under the best efficiency conditions, we can use the following formula:

n′ = n (H0 / H∘)^(1/2)

Substituting the given values, we get:

n′ = 300 rpm (60.98 m / 100 m)^(1/2) = 219.77 rpm

Therefore, the number of revolutions under the best efficiency conditions is approximately 220 rpm.