Question

1. A steam engine boiler is maintained at 250°C and water is converted into steam. This steam is used to do work and heat is ejected to the surrounding air at temperature 300K. Calculate the maximum efficiency it can have?

Answer 1

Answer:

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Answer 2

Answer:

The maximum efficiency it can have is 42.6%

Explanation:

Carnot Engine:

• It's a theoretical engine that works on the principle of Carnot's Cycle.
• The efficiency of this engine depends only on the temperature of the source($T_{i}$) and sink($T_{f}$).
• Its efficiency($\eta$) is given by

                                     $\eta=(1-\frac{T_{f} }{T_{i} })$

Step 1:

Given that, the Steam engine boiler is maintained at $250^{\circ}C$, which is the temperature of the source. It is also given that heat is ejected at a temperature of $300K$, which is the temperature of the sink.

                                     $T_{i}=250^{\circ}C\\ T_{f}=300K$

Step 2:

The source temperature is in degree Celsius which has to be converted to Kelvin scale. The conversion is given by

                                     $T_{K}=273+T_{c}$

where $T_{K}$ is the temperature in the Kelvin scale and $T_{c}$ is the temperature in the Celsius scale.

Therefore, converting $250^{\circ}C$ to Kelvin scale we get,

                                    $T_{i}=273+250=523K$

Step 3:

Substituting the temperature values in the formula for Efficiency we get,

                                    $\eta=(1-\frac{300}{423})\\\eta=(1-0.57)\\\eta=0.426$

Step 4:

It is sensible to write efficiency in per cent. Therefore,

                                  $\eta=0.426(100)=42.6$

Therefore, the maximum efficiency of the steam engine boiler operating between $250^{\circ}C$ and 300K is 42.6%