Question

Water is falling on the blades of a turbine from a height of $40\mathrm{m}$. If the speed of the river before the fall is $10\mathrm{m}/\mathrm{s}$ and the area of the blade of the turbine is $2\mathrm{sq}·\mathrm{m}$. What is the power generated every second?

• A. $27\mathrm{MW}$
• B. $2.7\mathrm{MW}$
• C. $5.4\mathrm{MW}$
• D. $5.4\times {10}^4\mathrm{MW}$

Answer 1

The correct option is A

$27\mathrm{MW}$

Step 1: Given parameters

The height of water falling on the blades of turbine is $40\mathrm{m}$.

Speed of river before fall $\left(u\right)$ is $10\mathrm{m}/\mathrm{s}$.

The area of the blade $\left(A\right)$ is $2\mathrm{sq}·\mathrm{m}$.

Step 2: Calculate the velocity $\left(v\right)$ when water reaches the blade of the turbine

Use the third equation of motion to calculate the fall velocity,

$$\begin{gathered} v^2-u^2=2as \\ {v}^2-{10}^2=2\left(10\right)\left(40\right) \\ {v}^2=800+100 \\ {v}^2=900 \\ v=\sqrt{{30}^2} \\ v=30\mathrm{m}/\mathrm{s} \end{gathered}$$

Step 3: Calculate the rate of flow

The rate of flow of water is given by,

$\begin{array}{rcl}Q& =& AV\\ & =& 2\times 30\\ & =& 60{\mathrm{m}}^3/\mathrm{s}\end{array}$

Step 4: Calculate the power generated every second

The kinetic energy generated per second is given by,

$\begin{array}{rcl}K.E.& =& \frac{1}{2}Q{\rho }_w{v}^2\end{array}$

Where, ${\rho }_w$ is the density of water.

$\begin{array}{rcl}K.E.& =& \frac{1}{2}Q{\rho }_w{v}^2\\ & =& \frac{1}{2}\times 60\times 1000\times {\left(30\right)}^2\\ & =& 27000000\mathrm{J}\end{array}$

The kinetic energy generated on the blade is equal to the power generated per second.

$P=27000000\mathrm{J}/\mathrm{s}$

Convert power generated into megawatt $\left(\mathrm{MW}\right)$ as follows:

$\begin{array}{rcl}1000000\mathrm{J}/\mathrm{s}& =& 1\mathrm{MW}\\ 27000000\mathrm{J}/\mathrm{s}& =& 27\mathrm{MW}\end{array}$

Hence, the power generated every second is $27\mathrm{MW}$.

Hence, option (A) is correct.