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Chapter 6: Problem 17

A heat engine has a heat input of \(3 \times 10^{4} \mathrm{Btu} / \mathrm{h}\) and a thermal efficiency of 40 percent. Calculate the power it will produce, in hp.

Short Answer

Expert verified
Question: A heat engine has a heat input of \(3 \times 10^{4} \mathrm{Btu} / \mathrm{h}\) and a thermal efficiency of 0.4. Calculate its mechanical power output in horsepower (hp). Answer: The heat engine's mechanical power output is approximately 4.71 hp.

Step by step solution

01

Calculate the total work produced from the heat input.

To find the work produced from the heat input, multiply the heat input by the thermal efficiency: Work = Heat Input × Thermal Efficiency Work = \(3 \times 10^{4} \mathrm{Btu} / \mathrm{h}\) × 0.4
02

Calculate the work produced.

Perform the multiplication: Work = \(1.2 \times 10^{4} \mathrm{Btu} / \mathrm{h}\)
03

Convert Btu/h to hp.

To convert from Btu/h to hp, use the conversion factor: 1 hp = 2,544.43 Btu/h So, divide the work by the conversion factor: Power (hp) = Work (Btu/h) / Conversion Factor (Btu/h per hp) Power (hp) = \((1.2 \times 10^{4} \mathrm{Btu} / \mathrm{h})\) / 2,544.43 hp
04

Calculate the power in hp.

Perform the division: Power (hp) ≈ 4.71 hp The heat engine will produce approximately 4.71 horsepower.

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Most popular questions from this chapter

The cargo space of a refrigerated truck whose inner dimensions are \(12 \mathrm{m} \times 2.3 \mathrm{m} \times 3.5 \mathrm{m}\) is to be precooled from \(25^{\circ} \mathrm{C}\) to an average temperature of \(5^{\circ} \mathrm{C}\). The construction of the truck is such that a transmission heat gain occurs at a rate of \(120 \mathrm{W} /^{\circ} \mathrm{C}\). If the ambient temperature is \(25^{\circ} \mathrm{C}\) determine how long it will take for a system with a refrigeration capacity of \(11 \mathrm{kW}\) to precool this truck.

An air-conditioning system operating on the reversed Carnot cycle is required to remove heat from the house at a rate of \(32 \mathrm{kJ} / \mathrm{s}\) to maintain its temperature constant at \(20^{\circ} \mathrm{C}\) If the temperature of the outdoors is \(35^{\circ} \mathrm{C},\) the power required to operate this air-conditioning system is \((a) 0.58 \mathrm{kW}\) (b) \(3.20 \mathrm{kW}\) \((c) 1.56 \mathrm{kW}\) \((d) 2.26 \mathrm{kW}\) \((e) 1.64 \mathrm{kW}\)

A window air conditioner that consumes \(1 \mathrm{kW}\) of electricity when running and has a coefficient of performance of 3 is placed in the middle of a room, and is plugged in. The rate of cooling or heating this air conditioner will provide to the air in the room when running is \((a) 3 \mathrm{kJ} / \mathrm{s},\) cooling \((b) 1 \mathrm{kJ} / \mathrm{s},\) cooling \((c) 0.33 \mathrm{kJ} / \mathrm{s},\) heating \((d) 1 \mathrm{kJ} / \mathrm{s},\) heating \((e) 3 \mathrm{kJ} / \mathrm{s},\) heating

It is well established that the thermal efficiency of a heat engine increases as the temperature \(T_{L}\) at which heat is rejected from the heat engine decreases. In an effort to increase the efficiency of a power plant, somebody suggests refrigerating the cooling water before it enters the condenser, where heat rejection takes place. Would you be in favor of this idea? Why?

The "Energy Guide" label of a refrigerator states that the refrigerator will consume \(\$ 170\) worth of electricity per year under normal use if the cost of electricity is \(\$ 0.125 / \mathrm{kWh}\). If the electricity consumed by the lightbulb is negligible and the refrigerator consumes \(400 \mathrm{W}\) when running, determine the fraction of the time the refrigerator will run.
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