Water falls from a height of `60 m` at the rate `15 kg//s` to operate a turbine. The losses due to frictional forces are `10%` of energy . How much power is generated to by the turbine? (g=10 m//s^(2))`.

Options

A. 12.3 kW
B. 7 kW
C. 8.1 kW
D. 10.2 kW

Explanation

To solve the problem of how much power is generated by the turbine when water falls from a height of 60 m at a rate of 15 kg/s, we can follow these steps:

### Step 1: Calculate the gravitational potential energy per second
The power generated by the falling water can be calculated using the formula for gravitational potential energy, which is given by:

\[ P = \frac{mgh}{t} \]

Where:
- \( P \) is the power (in watts),
- \( m \) is the mass of water falling per second (in kg/s),
- \( g \) is the acceleration due to gravity (in m/s²),
- \( h \) is the height (in meters),
- \( t \) is the time (in seconds).

Given:
- \( m = 15 \, \text{kg/s} \)
- \( g = 10 \, \text{m/s}^2 \)
- \( h = 60 \, \text{m} \)

Substituting the values into the formula:

\[ P = mgh = 15 \, \text{kg/s} \times 10 \, \text{m/s}^2 \times 60 \, \text{m} \]

### Step 2: Calculate the power before losses
Now, we calculate the power:

\[ P = 15 \times 10 \times 60 = 9000 \, \text{watts} \]

### Step 3: Account for energy losses
Since there are losses due to frictional forces, which are 10% of the energy, we need to calculate the effective power generated by the turbine. The efficiency of the turbine is 90% (100% - 10%).

Thus, the actual power output \( P_{\text{output}} \) is given by:

\[ P_{\text{output}} = P \times \text{Efficiency} \]

Substituting the values:

\[ P_{\text{output}} = 9000 \, \text{watts} \times \frac{90}{100} = 9000 \times 0.9 = 8100 \, \text{watts} \]

### Step 4: Convert watts to kilowatts
Finally, to express the power in kilowatts:

\[ P_{\text{output}} = \frac{8100}{1000} = 8.1 \, \text{kW} \]

### Final Answer
The power generated by the turbine is **8.1 kW**.

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