Solution

Statement

$5.9.13.$ A steam engine with a power of $14.7 \text{ kW}$ consumes $8.1 \text{ kg}$ of coal per hour of operation, with a specific heat of combustion of $3.3 \cdot 10^7 \text{ J/kg}$. The temperature of the boiler is $200^\circ\text{C}$, and the cooler is $58^\circ\text{C}$. Find the efficiency of this engine and compare it with the efficiency of an ideal heat engine.

Solution

1. Actual efficiency of the steam engine

The efficiency ($\eta$) is defined as the ratio of the useful work $W$ performed by the engine to the amount of heat supplied $Q$:
$$\eta = \frac{W}{Q}$$

The useful work over time $t$ can be expressed via the constant power $P$ of the engine:
$$W = P t$$
The supplied heat $Q$ is the heat released during the combustion of mass $m$ of coal:
$$Q = q m$$
where $q$ is the specific heat of combustion.

Substituting the expressions for work and heat into the base formula:
$$\eta = \frac{P t}{q m}$$

Converting values to SI units: $P = 14.7 \cdot 10^3 \text{ W}$, $t = 3600 \text{ s}$, $q = 3.3 \cdot 10^7 \text{ J/kg}$, $m = 8.1 \text{ kg}$, and calculating:
$$\eta = \frac{14700 \cdot 3600}{3.3 \cdot 10^7 \cdot 8.1} \approx 0.198 = 19.8%$$

2. Efficiency of an ideal heat engine (Carnot cycle)

The maximum possible efficiency for given temperatures is calculated using the Carnot formula. First, let's convert the temperatures to the absolute Kelvin scale:
$T_1 = 200^\circ\text{C} + 273 = 473 \text{ K}$ (heater/boiler temperature).
$T_2 = 58^\circ\text{C} + 273 = 331 \text{ K}$ (cooler temperature).

$$\eta_{\text{ideal}} = \frac{T_1 - T_2}{T_1}$$
$$\eta_{\text{ideal}} = \frac{473 - 331}{473} = \frac{142}{473} \approx 0.300 = 30%$$