A dam is situated at a height of `550m` above sea level and supplies water to a power house which is at a height of `50 m` above sea level. `2000 kg` of water passes through the turbines per second. What would be the maximum electrical power output (in `MW`) os the power house if the whole system were `60%` efficent ?

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Identify the given values
- Height of the dam (H1) = 550 m
- Height of the powerhouse (H2) = 50 m
- Mass of water flowing per second (M) = 2000 kg/s
- Efficiency of the system = 60% = 0.6
- Gravitational acceleration (g) = 10 m/s² (approximate value for simplicity)

### Step 2: Calculate the height difference (ΔH)
\[ \Delta H = H1 - H2 = 550 \, \text{m} - 50 \, \text{m} = 500 \, \text{m} \]

### Step 3: Calculate the maximum power output (P_max)
The formula for maximum power output is given by:
\[ P_{\text{max}} = \frac{M \cdot g \cdot \Delta H}{t} \]
Where:
- \( M \) = mass of water (2000 kg)
- \( g \) = gravitational acceleration (10 m/s²)
- \( \Delta H \) = height difference (500 m)
- \( t \) = time (1 second)

Substituting the values:
\[ P_{\text{max}} = \frac{2000 \, \text{kg} \cdot 10 \, \text{m/s}^2 \cdot 500 \, \text{m}}{1 \, \text{s}} \]
\[ P_{\text{max}} = 2000 \cdot 10 \cdot 500 = 10,000,000 \, \text{W} = 10^7 \, \text{W} \]

### Step 4: Calculate the actual power output considering efficiency
The actual power output (P_actual) can be calculated using the efficiency:
\[ P_{\text{actual}} = \text{Efficiency} \cdot P_{\text{max}} \]
Substituting the values:
\[ P_{\text{actual}} = 0.6 \cdot 10^7 \, \text{W} = 6 \cdot 10^6 \, \text{W} \]

### Step 5: Convert power to Megawatts (MW)
To convert watts to megawatts:
\[ P_{\text{actual}} = \frac{6 \cdot 10^6 \, \text{W}}{10^6} = 6 \, \text{MW} \]

### Final Answer
The maximum electrical power output of the powerhouse, considering the efficiency, is **6 MW**.

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