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Chapter 5: Problem 2

A Carnot heat engine produces \(10 \mathrm{hp}\) by transferring energy between two reservoirs at \(40^{\circ} \mathrm{F}\) and \(212^{\circ} \mathrm{F}\). Calculate the rate of heat transfer from the high-temperature reservoir. SOLUTION The engine efficiency is $$ The efficiency is also given by \(\eta=\dot{W} / \dot{Q}_{H}\). Thus, we see that $$ \dot{Q}_{H}=\frac{\dot{W}}{\eta}=\frac{(10 \mathrm{hp})(2545 \mathrm{Btu} / \mathrm{hr} / \mathrm{hp})}{0.2560}=99.410 \mathrm{Btu} / \mathrm{hr} $$ \eta=1-\frac{T_{L}}{T_{H}}=1-\frac{500}{672}=0.2560 $$

Short Answer

Expert verified
The rate of heat transfer from the high-temperature reservoir is 99.410 BTU/hr.

Step by step solution

01

Convert Temperatures

Convert the temperatures from Fahrenheit to Rankine. Add 459.67 to each temperature: .
02

Calculate Carnot Efficiency

Use the formula for Carnot efficiency: .
03

Convert Work Rate to BTU/hr

Since 1 horsepower equals 2545 BTU/hr, convert: .
04

Find Rate of Heat Transfer

Using the formula .

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Key Concepts

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

Carnot efficiency
The Carnot heat engine is a theoretical device that defines the maximum possible efficiency any engine can have. It operates between two heat reservoirs at different temperatures. To understand its efficiency, we use the Carnot efficiency formula, which is defined as: \[ \text{Carnot efficiency} = 1 - \frac{T_L}{T_H} \]. Here, \(T_L\) is the absolute temperature of the low-temperature reservoir, and \(T_H\) is the absolute temperature of the high-temperature reservoir. This formula tells us the fraction of heat absorbed from the high-temperature reservoir that is converted into work. The rest is expelled to the low-temperature reservoir. The efficiency increases if the temperature difference between the two reservoirs increases. In our exercise, the temperatures were given in Fahrenheit: \(40^{\tiny{\text{掳F}}}\) and \(212^{\tiny{\text{掳F}}}\). Converting these to Rankine (since \(掳R = 掳F + 459.67\)) gives us \(500 \tiny{\text{掳R}}\) and \(672 \tiny{\text{掳R}}\), respectively. Using the Carnot efficiency formula, we obtain: \[ \text{Carnot efficiency} = 1 - \frac{500}{672} = 0.2560 \].
Heat transfer rate
To determine the rate of heat transfer from the high-temperature reservoir, we need to relate it to the work output of the engine and its efficiency. The relationship is given by \[ \text{畏} = \frac{\text{Work output rate} (W)}{\text{Heat transfer rate} (Q_H)} \].Rearranging this, we get: \[ Q_H = \frac{W}{\text{畏}} \]. In our exercise, the work output rate is specified as \(10\) horse power (hp). First, convert horsepower to British thermal units (BTUs) per hour. Knowing that \(1\) hp equals \(2545\) BTU/hr, we calculate: \[ 10 \text{ hp} = 10 \times 2545 \text{ BTU/hr} = 25450 \text{ BTU/hr} \]. Now using the efficiency (畏) value obtained earlier (\(\text{畏} = 0.2560\)), we find the heat transfer rate \(Q_H\): \[ Q_H = \frac{25450 \text{ BTU/hr}}{0.2560} \text{ BTU/hr} \] Finally, we get \( Q_H \text{ 鈮� 99336 \text{ BTU/hr}} \). This value gives us the rate at which heat must be transferred from the high-temperature reservoir to maintain the engine's power output.
Temperature conversion
Temperature conversion is an essential step in using the Carnot efficiency formula, as it requires temperatures in absolute units. The commonly used Fahrenheit (掳F) scale needs to be converted to Rankine (掳R) for this type of thermodynamic calculation. The conversion is straightforward: \[ T(\text{掳R}) = T(\text{掳F}) + 459.67 \]. For the problem, the low and high temperatures given were \(40^{\tiny{\text{掳F}}}\) and \(212^{\tiny{\text{掳F}}}\), respectively. Applying the conversion, we get: \[ 40^{\tiny{\text{掳F}}} + 459.67 = 500^{\tiny{\text{掳R}}} \] and \[ 212^{\tiny{\text{掳F}}} + 459.67 = 672^{\tiny{\text{掳R}}} \]. Now, with the temperatures in Rankine, we proceed to evaluate the Carnot efficiency and further calculate the work and heat transfer rates. Converting temperatures to the right units ensures accurate application of thermodynamic principles and formulas.

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Most popular questions from this chapter

Two Carnot engines operate in series between two reservoirs maintained at \(350^{\circ} \mathrm{C}\) and \(50^{\circ} \mathrm{C}\), respectively. The energy rejected by the first engine is input into the second engine. If the first engine's efficiency is 20 percent greater than the second engine's efficiency, calculate the intermediate temperature. The efficiencies of the two engines are $$ \eta_{1}=1-\frac{T}{623} \quad \eta_{2}=1-\frac{323}{T} $$ where \(T\) is the unknown intermediate temperature. It is given that \(\eta_{1}=\eta_{2}+0.2 \eta_{2}\). Substituting for \(\eta_{1}\) and \(\eta_{2}\) results in $$ 1-\frac{T}{623}=1.2\left(1-\frac{323}{T}\right) $$ or $$ T^{2}+124.6 T-241500=0 \quad \therefore T=\frac{-124.6+\sqrt{124.6^{2}-4(-241500)}}{2}=433 \mathrm{~K} \quad \text { or } \quad 160^{\circ} \mathrm{C} $$

A power utility company desires to use the hot groundwater from a hot spring to power a heat engine. If the groundwater is at \(95^{\circ} \mathrm{C}\), estimate the maximum power output if a mass flux of \(0.2 \mathrm{~kg} / \mathrm{s}\) is possible. The atmosphere is at \(20^{\circ} \mathrm{C}\). The maximum possible efficiency is $$ \eta=1-\frac{T_{L}}{T_{H}}=1-\frac{293}{368}=0.2038 $$ assuming the water is rejected at atmospheric temperature. The rate of heat transfer from the energy source is $$ \dot{Q}_{H}=\dot{m} C_{p} \Delta T=(0.2)(4.18)(95-20)=62.7 \mathrm{~kW} $$ The maximum power output is then $$ \dot{W}=\eta \dot{Q}_{H}=(0.2038)(62.7)=12.8 \mathrm{~kW} $$

An inventor proposes an engine that operates between the \(27^{\circ} \mathrm{C}\) warm surface layer of the ocean and a \(10^{\circ} \mathrm{C}\) layer a few meters down. The inventor claims that the engine produces \(100 \mathrm{~kW}\) by pumping \(20 \mathrm{~kg} / \mathrm{s}\) of seawater. Is this possible? Assume \(\left(C_{p}\right)_{\text {seawater }} \cong 4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). The maximum temperature drop for the seawater is \(17^{\circ} \mathrm{C}\). The maximum rate of heat transfer from the hightemperature water is then $$ \dot{Q}_{H}=\dot{m} C_{p} \Delta T=(20)(4.18)(17)=1421 \mathrm{~kW} $$ The efficiency of the proposed engine is then: \(\eta=\dot{W} / \dot{Q}_{H}=100 / 1421=0.0704\) or \(7.04 \%\). The efficiency of a Carnot engine operating between the same two temperatures is $$ \eta=1-\frac{T_{L}}{T_{H}}=1-\frac{283}{300}=0.0567 \quad \text { or } \quad 5.67 \% $$ The proposed engine's efficiency exceeds that of a Carnot engine; hence, the inventor's claim is impossible.

A heat engine operates on a Carnot cycle with an efficiency of 75 percent. What COP would a refrigerator operating on the same cycle have? The low temperature is \(0^{\circ} \mathrm{C}\). The efficiency of the heat engine is given by \(\eta=1-T_{L} / T_{H}\). Hence, $$ T_{H}=\frac{T_{L}}{1-\eta}=\frac{273}{1-0.75}=1092 \mathrm{~K} $$ The COP for the refrigerator is then $$ \mathrm{COP}_{R}=\frac{T_{L}}{T_{H}-T_{L}}=\frac{273}{1092-273}=0.3333 $$

A refrigerator is rated at a COP of 4. The refrigerated space that it cools requires a peak cooling rate of \(30000 \mathrm{~kJ} / \mathrm{h}\). What size electrical motor (rated in horsepower) is required for the refrigerator? SOLUTION The definition of the \(\mathrm{COP}\) for a refrigerator is \(\mathrm{COP}_{R}=\dot{Q}_{L} / \dot{W}_{\text {net }}\). The net power required is then $$ \dot{W}_{\text {net }}=\frac{\dot{Q}_{L}}{\operatorname{COP}_{R}}=\frac{30000 / 3600}{4}=2.083 \mathrm{~kW} \quad \text { or } \quad 2.793 \mathrm{hp} $$
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