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Valter правка от 19 авг. 2026 г., 15:10

правка #20158 позже →

en/5.9.13.md +44 −0

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		+	### Statement
		+
		+	$5.9.13.$ [Insert the problem statement]
		+
		+	### 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
		+
		+	<b>1. Actual efficiency of the steam engine</b>
		+
		+	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:
		+	$$\eta = \frac{14700 \cdot 3600}{3.3 \cdot 10^7 \cdot 8.1} \approx 0.198 = 19.8\%$$
		+
		+	<b>2. Efficiency of an ideal heat engine (Carnot cycle)</b>
		+
		+	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).
		+
		+	$$\eta_{\text{ideal}} = \frac{T_1 - T_2}{T_1}$$
		+	$$\eta_{\text{ideal}} = \frac{473 - 331}{473} = \frac{142}{473} \approx 0.300 = 30\%$$
		+
		+	#### Answer
		+	The efficiency of the steam engine is $19.8\%$, and the efficiency of the ideal heat engine is $30\%$.
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$5.9.13.$ [Insert the problem statement]

			### 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

			<b>1. Actual efficiency of the steam engine</b>

			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:
			$$\eta = \frac{14700 \cdot 3600}{3.3 \cdot 10^7 \cdot 8.1} \approx 0.198 = 19.8\%$$

			<b>2. Efficiency of an ideal heat engine (Carnot cycle)</b>

			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).

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

			#### Answer
			The efficiency of the steam engine is $19.8\%$, and the efficiency of the ideal heat engine is $30\%$.

			#### Answer

			[Insert a concise answer or boxed result]