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Chapter 20: Problem 19

You design an engine that takes in 1.50 \(\times\) 10\(^4\) J of heat at 650 K in each cycle and rejects heat at a temperature of 290 K. The engine completes 240 cycles in 1 minute. What is the theoretical maximum power output of your engine, in horsepower?

Short Answer

Expert verified
The theoretical maximum power output is approximately 44.56 horsepower.

Step by step solution

01

Identify Given Information

We are given the heat input to the engine, \(Q_H = 1.50 \times 10^4\) J, and the temperatures of the heat reservoirs: \(T_H = 650\) K and \(T_C = 290\) K. The engine completes 240 cycles per minute.
02

Calculate Carnot Efficiency

The Carnot efficiency \(\eta\) for a heat engine is given by \(\eta = 1 - \frac{T_C}{T_H}\). Plug in the values: \(\eta = 1 - \frac{290}{650}\).
03

Determine Carnot Efficiency

Calculating the expression gives us \(\eta = 1 - \frac{290}{650} = 0.5538\), or 55.38%.
04

Calculate Work Done Per Cycle

The work done per cycle \(W\) can be found using \(W = \eta \times Q_H\). So, \(W = 0.5538 \times 1.50 \times 10^4\) J.
05

Compute Work Done Per Cycle

Calculating the product gives \(W = 8307\) J per cycle.
06

Calculate Total Work Done Per Minute

The engine completes 240 cycles per minute, so the total work done per minute is \(240 \times 8307\) J.
07

Determine Total Work Done Per Minute

Multiplying gives a total work of \(1993680\) J per minute.
08

Convert Work Done Per Minute to Power

Power is work done per unit time. Convert 1 minute to 60 seconds to find \(P = \frac{1993680}{60}\) W.
09

Calculate Power Output

Solving the division gives \(P = 33228\) W.
10

Convert Power to Horsepower

1 horsepower is equivalent to 746 watts. Convert the power to horsepower: \( \text{horsepower} = \frac{33228}{746} \).
11

Final Calculation to Horsepower

This calculation yields approximately \(44.56\) horsepower.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Carnot efficiency
The Carnot efficiency serves as a theoretical limit on the efficiency any heat engine can achieve, based on the temperatures of its heat reservoirs. This efficiency is determined by the formula \( \eta = 1 - \frac{T_C}{T_H} \), where \( T_C \) is the temperature of the cold reservoir and \( T_H \) is the temperature of the hot reservoir.
Carnot efficiency reflects the maximum possible efficiency, suggesting that a real engine can never surpass this value. This is due to the inherent limitations imposed by the second law of thermodynamics.
  • Real engines operate below Carnot efficiency.
  • The temperatures must be in Kelvin for calculations.
  • Carnot efficiency is not achievable in practice.
Understanding these concepts helps us gauge how close our engine is to the theoretical maximum, using the given temperatures.
heat reservoirs
Heat reservoirs are systems that retain or dispense vast quantities of heat energy with negligible change in temperature. They provide or absorb heat during the engine cycles. For our engine example:
  • The hot reservoir is at 650 K, dispensing heat to the engine.
  • The cold reservoir is at 290 K, absorbing unused heat.
The substantial temperature difference between reservoirs is what drives the engine, allowing it to convert some heat energy into mechanical work.
Real-life heat reservoirs could be bodies of water, large boilers, or earth itself, maintaining near-constant temperature due to their large capacities.
work done per cycle
Work done per cycle is a crucial concept. It represents the amount of energy converted into useful work in a single cycle of the engine. We calculate it using the Carnot efficiency and the heat input: \( W = \eta \times Q_H \). In our example, we employ the values:
  • Carnot efficiency \( \eta = 0.5538 \).
  • Heat input \( Q_H = 1.50 \times 10^4 \) J.
Plugging these into the formula gives us the work done per cycle, \( W = 8307 \) J.
This measure helps understand the capabilities of the engine in terms of energy conversion per operational cycle.
power output conversion
Power output refers to the rate at which an engine performs work. To find the power output, we first calculate the total work done over a minute, considering the number of cycles. In our engine:
  • Total work per minute is \( 1993680 \) J.
  • Power is then calculated as work over time: \( P = \frac{1993680}{60} \).
This yields a power output of \( 33228 \) watts (W).
To make this value usable in different contexts, we convert it into horsepower, a common unit: 1 horsepower equals 746 watts, thus converting: \( \frac{33228}{746} \approx 44.56 \) horsepower. This conversion gives a comprehensive idea of the engine's capacity in more relatable terms.

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Most popular questions from this chapter

Compare the entropy change of the warmer water to that of the colder water during one cycle of the heat engine, assuming an ideal Carnot cycle. (a) The entropy does not change during one cycle in either case. (b) The entropy of both increases, but the entropy of the colder water increases by more because its initial temperature is lower. (c) The entropy of the warmer water decreases by more than the entropy of the colder water increases, because some of the heat removed from the warmer water goes to the work done by the engine. (d) The entropy of the warmer water decreases by the same amount that the entropy of the colder water increases.

A Carnot engine is operated between two heat reservoirs at temperatures of 520 \(K\) and 300 \(K\). (a) If the engine receives 6.45 \(kJ\) of heat energy from the reservoir at 520 \(K\) in each cycle, how many joules per cycle does it discard to the reservoir at 300 \(K\)? (b) How much mechanical work is performed by the engine during each cycle? (c) What is the thermal efficiency of the engine?

A Carnot engine whose high-temperature reservoir is at 620 K takes in 550 J of heat at this temperature in each cycle and gives up 335 J to the low- temperature reservoir. (a) How much mechanical work does the engine perform during each cycle? What is (b) the temperature of the low-temperature reservoir; (c) the thermal efficiency of the cycle?

An object of mass \(m_1\), specific heat \(c_1\), and temperature \(T_1\) is placed in contact with a second object of mass \(m_2\), specific heat \(c_2\), and temperature \(T_2\) > \(T_1\). As a result, the temperature of the first object increases to \(T\) and the temperature of the second object decreases to \(T'\). (a) Show that the entropy increase of the system is $$\Delta S = m_1c_1 ln {T \over T_1} + m_2c_2 ln {T' \over T_2}$$ and show that energy conservation requires that $$m_1c_1 (T - T_1) = m_2c_2 (T_2 - T')$$ (b) Show that the entropy change \(\Delta S\), considered as a function of \(T\), is a \(maximum\) if \(T = T'\), which is just the condition of thermodynamic equilibrium. (c) Discuss the result of part (b) in terms of the idea of entropy as a measure of randomness.

A diesel engine performs 2200 J of mechanical work and discards 4300 J of heat each cycle. (a) How much heat must be supplied to the engine in each cycle? (b) What is the thermal efficiency of the engine?
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