Collection of Solved Problems in Physics

Work Performed by a Steam Engine

Task number: 1803

• Hint 1

  If we want the efficiency to be maximal, the engine must work according to Carnot cycle.

• Hint 2

  Find the definition of an efficiency of a heat engine in general, and the definition of Carnot engine efficiency.

• Efficiency of a heat engine

  The efficiency η of a heat engine is defined as

  \[\eta=\frac{W}{Q_1},\]

  where W is the performed work and Q₁ is the heat supplied per one cycle. Work W performed by the engine can also be expressed as the difference between the heat supplied by the hot reservoir Q₁ and the heat Q₂ given off to the cold reservoir. Then the efficiency η is given by the following relationship

  \[\eta=\frac{Q_1-Q_2}{Q_1}.\]

• Carnot cycle efficiency

  The efficiency of Carnot engine is described as follows

  \[\eta=\frac{T_1-T_2}{T_1},\]

  where T₁ is the temperature of the hot reservoir and T₂ is the temperature the cold reservoir.

• Analysis

  First we write down the efficiency of a heat engine as the performed work divided by the supplied heat.

  The efficiency of the engine is maximal provided that the engine operates according to the Carnot cycle. In such a case it would perform maximum work with the same heat supplied. The efficiency of Carnot cycle depends only on the temperatures between which it operates.

  By comparing both terms we obtain the expression for the efficiency, from which we evaluate the performed work.

• Given values

  t1 = 120 °C => T1 = 393 K	temperature of steam (hot reservoir)
  t2 = 40 °C => T2 = 313 K	temperature of cold reservoir
  Q = 4.2 kJ = 4.2·103 J	heat consumption
  W = ?	maximum work performed by engine

• Solution

  The efficiency η of a heat engine is defined as the ratio of performed work W and supplied heat Q. Therefore it applies

  \[\eta=\frac{W}{Q}.\]

  Maximum efficiency is achieved when the engine operates according to Carnot cycle; the efficiency η depends only on the temperature T₁ of the hot reservoir and the temperature T₂ of the cold reservoir. Therefore the following applies

  \[\eta=\frac{T_1-T_2}{T_1}.\]

  In our case, the temperature of the hot reservoir corresponds to the temperature of the steam in the cylinder of the steam engine.

  By comparing the two equations for efficiency we obtain an equation

  \[\frac{W}{Q}=\frac{T_1-T_2}{T_1}.\]

  We evaluate the work performed by the engine from this equation:

  \[W=\frac{Q\left(T_1-T_2\right)}{T_1}.\]

• Numerical solution

  \[W=\frac{Q\left(T_1-T_2\right)}{T_1}= \frac{4.2\cdot{ 10^3}\cdot \left(393-3135\right)}{393}\,\mathrm{J}\,\dot{=}\,850\,\mathrm{J}\]

• Answer

  Maximum work performed by the engine is approximately 850 J.