/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 A power cycle receives energy by... [FREE SOLUTION] | 91Ó°ÊÓ

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A power cycle receives energy by heat transfer from the combustion of fuel at a rate of \(300 \mathrm{MW}\). The thermal efficiency of the cycle is \(33.3 \%\). (a) Determine the net rate power is developed, in MW. (b) For 8000 hours of operation annually, determine the net work output, in \(\mathrm{kW} \cdot \mathrm{h}\) per year. (c) Evaluating the net work output at \(\$ 0.08\) per \(\mathrm{kW} \cdot \mathrm{h}\), determine the value of the net work, in \$/year.

Short Answer

Expert verified
(a) 99.9 MW (b) 799,200,000 kW â‹… h/year (c) 63,936,000 $/year

Step by step solution

01

Calculate Net Power Developed

The net power developed (\text{P}_{\text{net}}) can be calculated using the thermal efficiency (\text{Eff}) and the rate of energy input by heat transfer (\text{Q}_{\text{in}}). The formula is: \[ \text{P}_{\text{net}} = \text{Q}_{\text{in}} \times \text{Eff} \] Given \[ \text{Q}_{\text{in}} = 300 \text{ MW} \text{ and } \text{Eff} = 0.333 \], Substitute the values into the formula: \[ \text{P}_{\text{net}} = 300 \text{ MW} \times 0.333 = 99.9 \text{ MW} \]
02

Find Net Work Output Annually

To find the annual net work output, (\text{W}_{\text{year}}), multiply the net power developed by the number of operational hours in a year: \[ \text{W}_{\text{year}} = \text{P}_{\text{net}} \times \text{hours of operation} \] Given \[ \text{hours of operation} = 8000 \text{ hours/year} \text{ and } \text{P}_{\text{net}} = 99.9 \text{ MW} = 99,900 \text{ kW} \], nSubstitute the values into the formula: \[ \text{W}_{\text{year}} = 99,900 \text{ kW} \times 8000 \text{ hours/year} = 799,200,000 \text{ kW} \times \text{h/year} \]
03

Calculate the Value of the Net Work Output

The value of the net work output (\text{Value}_{\text{work}}) can be found by multiplying the annual net work output by the price per \text{kW} \times \text{h}: \[ \text{Value}_{\text{work}} = \text{W}_{\text{year}} \times \text{cost per kW} \times \text{h} \] Given \[ \text{cost per kW} \times \text{h} = 0.08 \text{ dollars} \], Substitute the values into the formula: \[ \text{Value}_{\text{work}} = 799,200,000 \text{ kW} \times \text{h/year} \times 0.08 \text{ dollars} / \text{kW} \times \text{h} = 63,936,000 \text{ dollars/year} \]

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Key Concepts

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

Net Power Developed
Net power developed is a crucial concept in understanding how efficient a power cycle is. It tells you how much usable power is produced by a system after accounting for energy losses. To calculate it, you need to know the thermal efficiency (\text{Eff}) of the system and the rate of energy input (\text{Q}_{\text{in}}). The formula used is: \(\text{P}_{\text{net}} = \text{Q}_{\text{in}} \times \text{Eff}\)In our exercise, the rate of energy input is \(\text{Q}_{\text{in}} = 300 \text{ MW}\)and the thermal efficiency is \(\text{Eff} = 0.333\). Substituting these values into the formula, we get: \(\text{P}_{\text{net}} = 300 \text{ MW} \times 0.333 = 99.9 \text{ MW}\). That means the net power developed is \(99.9 \text{ MW}\). This value is important as it represents the actual amount of power available for use from the power cycle.
Annual Net Work Output
The annual net work output is another key parameter that helps you understand the total energy produced by the system over a year. It's important in evaluating how much energy can be harvested for various uses. To find this, you multiply the net power developed by the number of hours the system operates in a year. The formula is: \(\text{W}_{\text{year}} = \text{P}_{\text{net}} \times \text{hours of operation}\)In this exercise, the net power developed is \(\text{P}_{\text{net}} = 99.9 \text{ MW} = 99,900 \text{ kW}\) and the system operates for \(8000 \text{ hours/year}\). Plugging these values into the formula, we get: \(\text{W}_{\text{year}} = 99,900 \text{ kW} \times 8000 \text{ hours/year} = 799,200,000 \text{ kW} \times \text{h/year}\). So, the annual net work output is \(799,200,000 \text{ kW} \times \text{h/year}\). This tells us how much work (or energy) the system produces over a whole year, which is useful for planning and economic analysis.
Value of Net Work
The value of net work is crucial for understanding the economic benefits of a power system. It helps you determine how much money can be made from the energy produced. To calculate this, you need to multiply the annual net work output by the cost per unit of energy (\text{kW} \times \text{h}). The formula is: \(\text{Value}_{\text{work}} = \text{W}_{\text{year}} \times \text{cost per kW} \times \text{h}\)From our exercise, the annual net work output is \(799,200,000 \text{ kW} \times \text{h/year}\) and the cost per \text{kW} \times \text{h} is \(\text{0.08 \text{ dollars}}\). Substituting these values, we get: \(\text{Value}_{\text{work}} = 799,200,000 \text{ kW} \times \text{h/year} \times 0.08 \text{ dollars} / \text{kW} \times \text{h} = 63,936,000 \text{ dollars/year}\). So, the value of the net work output is \(63,936,000 \text{ dollars/year}\), which represents the potential income generated from the electricity produced by the power cycle.

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Most popular questions from this chapter

An object whose mass is \(0.5 \mathrm{~kg}\) has a velocity of \(30 \mathrm{~m} / \mathrm{s}\). Determine (a) the final velocity, in \(\mathrm{m} / \mathrm{s}\), if the kinetic energy of the object decreases by \(130 \mathrm{~J}\). (b) the change in elevation, in \(\mathrm{ft}\), associated with a \(130 \mathrm{~J}\) change in potential energy. Let \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).

An electric generator coupled to a windmill produces an average electric power output of \(15 \mathrm{~kW}\). The power is used to charge a storage battery. Heat transfer from the battery to the surroundings occurs at a constant rate of \(1.8 \mathrm{~kW}\). Determine, for \(8 \mathrm{~h}\) of operation (a) the total amount of energy stored in the battery, in \(\mathrm{kJ}\). (b) the value of the stored energy, in \(\$$ if electricity is valued at \)\$ 0.08\( per \)\mathrm{kW} \cdot \mathrm{h}$.

An airplane whose mass is \(5000 \mathrm{~kg}\) is flying with a velocity of \(150 \mathrm{~m} / \mathrm{s}\) at an altitude of \(10,000 \mathrm{~m}\), both measured relative to the surface of the earth. The acceleration of gravity can be taken as constant at \(g=9.78 \mathrm{~m} / \mathrm{s}^{2}\). (a) Calculate the kinetic and potential energies of the airplane, both in \(\mathrm{kJ}\). (b) If the kinetic energy increased by \(10,000 \mathrm{~kJ}\) with no change in elevation, what would be the final velocity, in \(\mathrm{m} / \mathrm{s}\) ?

An object of mass \(1000 \mathrm{~kg}\), initially having a velocity of \(100 \mathrm{~m} / \mathrm{s}\), decelerates to a final velocity of \(20 \mathrm{~m} / \mathrm{s}\). What is the change in kinetic energy of the object, in \(\mathrm{kJ}\) ?

An object whose mass is \(400 \mathrm{~kg}\) is located at an elevation of \(25 \mathrm{~m}\) above the surface of the earth. For \(g=9.78 \mathrm{~m} / \mathrm{s}^{2}\), determine the gravitational potential energy of the object, in \(\mathrm{kJ}\), relative to the surface of the earth.

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