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en/5.9.13.md
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| + | ### Statement | ||
| + | |||
| + | $5.9.13.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | ### Statement | ||
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| + | $5.9.13.$ 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] | |||