An engineer claims to have made an engine delivering 10 kW power with fuel consumption of `1gs^(-1)` . The calorific value of fuel is 2 k cal / g . His claim

Options

A. valid
B. invalid
C. dependent on engine design
D. dependent on load

Explanation

To determine whether the engineer's claim about the engine is valid, we need to compare the power output of the engine to the power input derived from the fuel consumption. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Power output of the engine (P_output) = 10 kW
- Fuel consumption rate = 1 g/s
- Calorific value of fuel = 2 kcal/g

### Step 2: Calculate the input power from the fuel
To find the input power, we will first convert the calorific value from kilocalories to calories and then calculate the power input.

1. Convert calorific value:
\[ \text{Calorific value} = 2 \text{ kcal/g} = 2 \times 1000 \text{ cal/g} = 2000 \text{ cal/g} \]

2. Calculate the power input (P_input):
\[ P_{\text{input}} = \text{Calorific value} \times \text{Fuel consumption rate} \]
\[ P_{\text{input}} = 2000 \text{ cal/g} \times 1 \text{ g/s} = 2000 \text{ cal/s} \]

### Step 3: Convert calories per second to watts
1 calorie is equivalent to 4.184 joules. Therefore, we convert calories per second to watts:
\[ P_{\text{input}} = 2000 \text{ cal/s} \times 4.184 \text{ J/cal} = 8368 \text{ J/s} = 8.368 \text{ kW} \]

### Step 4: Compare output power and input power
- Output power (P_output) = 10 kW
- Input power (P_input) = 8.368 kW

### Step 5: Conclusion
Since the output power (10 kW) is greater than the input power (8.368 kW), this is not physically possible according to the law of conservation of energy. Therefore, the engineer's claim is invalid.

### Final Answer
The engineer's claim is **invalid**.

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