River Dam generates electricity with water taken from a depth of 110 m and with an average flow rate of 740 m3/s.
a) Calculate the power in this flow. Note that the velocity and height of the water are not changed appreciably by the generators, and the pressure at the exit point is at atmospheric level. Take the density of water to be 1000 kg/m3.
answer in MW
b) What is the speed of the water at the exit point?
answer in m/s
c) What is the radius of the exit hole in the dam?
answer in m
Density of water = 1000 kg/m^3
pressure difference = density *g * h =10^3*9.81*110 = 1.08*10^6 Newtons/m^2
power = pressure*Q = 1.08*10^6 * 740
= 8*10^8 Joules meters/second or Watts
or 800 megawatts
b) (1/2) rho v^2 = rho g h
v = sqrt (2 g h)
c) pi r^2 v = 740
a) To calculate the power generated by the river dam, we can use the formula:
Power = Flow rate x Height x Gravity
where:
- Flow rate is the average flow rate of the water in m^3/s
- Height is the depth of the water in meters
- Gravity is the acceleration due to gravity, which is approximately 9.8 m/s^2
In this case, the flow rate is 740 m^3/s and the height is 110 m. The values are given in the question.
Power = 740 m^3/s x 110 m x 9.8 m/s^2
Power = 867,720 W
To convert the power into megawatts (MW), divide by 1,000,000:
Power = 867,720 W / 1,000,000
Power = 0.86772 MW
Therefore, the power generated by the river dam is approximately 0.86772 MW.
b) To find the speed of the water at the exit point, we can use Bernoulli's equation, which states that the sum of the pressure, density, and kinetic energy per unit volume of a fluid along a streamline is constant.
At the exit point, the pressure is at atmospheric level, so we can neglect the pressure term in Bernoulli's equation.
The equation can be written as:
Pressure + (Density x Gravity x Height) + (Density x (Velocity^2 / 2)) = Constant
Since the velocity of the water at the exit point is what we want to find, we can rearrange the equation and solve for the velocity:
(Density x Gravity x Height) + (Density x (Velocity^2 / 2)) = Constant
Density x Gravity x Height = Density x (Velocity^2 / 2)
Gravity x Height = Velocity^2 / 2
2 x Gravity x Height = Velocity^2
Velocity = sqrt(2 x Gravity x Height)
Velocity = sqrt(2 x 9.8 m/s^2 x 110 m)
Velocity = 47.01 m/s
Therefore, the speed of the water at the exit point is approximately 47.01 m/s.
c) The radius of the exit hole can be calculated using the equation for the flow rate of liquid through an orifice:
Flow rate = Coefficient x Area x Velocity
where:
- Flow rate is the average flow rate of the water in m^3/s (given as 740 m^3/s)
- Coefficient is a dimensionless constant
- Area is the cross-sectional area of the exit hole
- Velocity is the speed of the water at the exit point (which we found to be approximately 47.01 m/s)
Since the coefficient and area are unknown, we'll use the equation to calculate the area:
Area = Flow rate / (Coefficient x Velocity)
The coefficient depends on the shape of the exit hole. For a circular hole, the coefficient is typically around 0.61.
Area = 740 m^3/s / (0.61 x 47.01 m/s)
Area = 25.65 m^2
To find the radius, we can use the formula for the area of a circle:
Area = π x Radius^2
25.65 m^2 = π x Radius^2
Radius^2 = 25.65 m^2 / π
Radius^2 = 8.17 m^2
Radius = sqrt(8.17 m^2)
Radius = 2.86 m
Therefore, the radius of the exit hole in the dam is approximately 2.86 meters.