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. `8.1 kW`
B. `10.2 kW`
C. `12.3 kW`
D. `7.0 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 potential energy (PE) of the falling water can be calculated using the formula:
\[ \text{PE} = mgh \]
where:
- \( m \) is the mass flow rate (in kg/s),
- \( g \) is the acceleration due to gravity (in m/s²),
- \( h \) is the height (in meters).

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

Substituting the values:
\[ \text{PE} = 15 \, \text{kg/s} \times 10 \, \text{m/s}^2 \times 60 \, \text{m} \]

### Step 2: Calculate the power input
The power input (P_input) is the work done per unit time, which is equivalent to the potential energy calculated above:
\[ P_{\text{input}} = \text{PE} = 15 \times 10 \times 60 \]

Calculating this:
\[ P_{\text{input}} = 9000 \, \text{W} \, \text{or} \, 9 \, \text{kW} \]

### Step 3: Account for energy losses
The problem states that there are losses due to frictional forces amounting to 10% of the energy. Therefore, the efficiency (η) of the turbine is:
\[ \eta = 1 - 0.1 = 0.9 \]

### Step 4: Calculate the power generated
The power generated (P_generated) can be calculated using the efficiency:
\[ P_{\text{generated}} = P_{\text{input}} \times \eta \]
Substituting the values:
\[ P_{\text{generated}} = 9000 \, \text{W} \times 0.9 \]

Calculating this:
\[ P_{\text{generated}} = 8100 \, \text{W} \, \text{or} \, 8.1 \, \text{kW} \]

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

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