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Chapter 7: Problem 136

At steady state, a turbine with an exergetic efficiency of \(90 \%\) develops \(7 \times 10^{7} \mathrm{~kW}\). \(\mathrm{h}\) of work annually \((8000\) operating hours). The annual cost of owning and operating the turbine is \(\$ 2.5 \times 10^{5}\). The steam entering the turbine has a specific flow exergy of \(559 \mathrm{Btu} / \mathrm{b}\), a mass flow rate of \(12.55\) \(\times 10^{4} \mathrm{lb} / \mathrm{h}\), and is valued at \(\$ 0.0165\) per \(\mathrm{kW}+\mathrm{h}\) of exergy. (a) Using Eq. \(7.34 \mathrm{c}\), evaluate the unit cost of the power developed, in \(\$$ per \)\mathrm{kW}+\mathrm{h}$. (b) Evaluate the unit cost based on exergy of the steam entering and exiting the turbine, each in cents per lb of steam flowing through the turbine.

Short Answer

Expert verified
Unit cost of power developed: 0.0183 \textdollar/kW-h. Unit cost of exergy per lb of steam: 0.0112 cents/lb.

Step by step solution

01

- Calculate Annual Work Output

First, determine the annual work output of the turbine. Given the steady state power development: \[ \text{Annual Work Output} = 7 \times 10^{7} \text{ kW} \times 8000 \text{ hours} \] \[ = 5.6 \times 10^{11} \text{ kW-h} \]
02

- Calculate Total Exergy Input

The exergy efficiency is given as 90%. Therefore, the total annual exergy input can be calculated as: \[ \text{Total Exergy Input} = \frac{\text{Annual Work Output}}{\text{Exergetic Efficiency}} \] \[ = \frac{5.6 \times 10^{11} \text{ kW-h}}{0.90} = 6.22 \times 10^{11} \text{ kW-h} \]
03

- Determine Cost of Exergy

Given the specific cost of exergy: \[ \text{Cost of Exergy} = 6.22 \times 10^{11} \text{ kW-h} \times 0.0165 \text{ \textdollar /kW-h} \] \[ = 10.263 \times 10^{9} \text{ \textdollar} \]
04

- Calculate Unit Cost of Power Developed

Add the annual operating cost to the total cost of exergy and divide by the annual work output: \[ \text{Total Cost} = 10.263 \times 10^{9} \text{ \textdollar} + 2.5 \times 10^{5} \text{ \textdollar} = 10.26325 \times 10^{9} \text{ \textdollar} \] \[ \text{Unit Cost of Power} = \frac{10.26325 \times 10^{9} \text{ \textdollar}}{5.6 \times 10^{11} \text{ kW-h}} = 0.0183 \text{ \textdollar/kW-h} \]
05

- Calculate Exergy Input per lb of Steam

The exergy input per lb of steam flowing through the turbine is given by: \[ \text{Exergy of Steam} = 559 \text{ Btu/lb} \] Given that 1 Btu = 0.293071 kW-h: \[ \text{Exergy Input per lb} = 559 \text{ Btu/lb} \times 0.293071 \text{ kW-h/Btu} = 163.79 \text{ kW-h/lb} \]
06

- Calculate Unit Cost in Cents per lb of Steam

Given the mass flow rate of 12.55 \times 10^4 lb/h, the unit cost is evaluated: \[ \text{Unit Cost in Cents/lb} = \frac{0.0183 \text{ \textdollar/kW-h}}{163.79 \text{ kW-h/lb}} \times 100 \text{ cents/\textdollar} = 0.0112 \text{ cents/lb} \]

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Key Concepts

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

Exergy Efficiency
Exergy efficiency measures how effectively a system converts available energy into useful work. In our problem, the turbine has an exergy efficiency of 90%. This means that 90% of the energy entering the turbine is converted to work, while the remaining 10% is lost due to inefficiencies. To calculate this, you use the formula:
Unit Cost of Power
The unit cost of power developed is essential for economic analysis. It helps to determine the cost-effectiveness of the turbine. To find it, we consider the total annual operating cost and the exergy cost. The final unit cost is found by adding the annual operating cost to the total exergy cost and dividing by the annual output in kW-h. For example:
Thermodynamic Calculations
Thermodynamic calculations involve quantifying energy transfers and conversions. For this exercise, we first calculate the annual work output by multiplying the steady-state power with operational hours. Next, we compute the total exergy input using exergy efficiency. By converting specific exergy values and using known constants such as the Btu to kW-h conversion rate, we derive key insights for detailed cost and efficiency analysis. Thermodynamic calculations create the foundation for understanding energy efficiency and cost implications.
Steady State Operation
Steady state operation means the turbine's conditions remain constant over the operational period. This simplifies the calculations as we assume no variations in power output or mass flow rates. It allows for direct calculation of annual outputs and costs based on given steady-state parameters. Steady state operation is crucial for accurate performance and cost analysis as it avoids the complexities of transient states.

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Most popular questions from this chapter

A pump operating at steady state takes in saturated liquid water at \(65 \mathrm{lbf} / \mathrm{in}^{2}\) at a rate of \(10 \mathrm{lb} / \mathrm{s}\) and discharges water at \(1000 \mathrm{lbf} / \mathrm{in}^{2}\). The isentropic pump efficiency is \(80.22 \%\). Heat transfer with the surroundings and the effects of motion and gravity can be neglected. If \(T_{0}=75^{\circ} \mathrm{F}\), determine for the pump (a) the exergy destruction rate, in Btu/s (b) the exergetic efficiency.

A high-pressure (HP) boiler and a low-pressure (LP) boiler will be added to a plant's steam-generating system. Both boilers use the same fuel and at steady state have approximately the same rate of energy loss by heat transfer. The average temperature of the combustion gases is less in the LP boiler than in the HP boiler. In comparison to the LP boiler, might you spend more, the same, or less to insulate the HP boiler? Explain.

An open feedwater heater operates at steady state with liquid water entering inlet 1 at 10 bar, \(50^{\circ} \mathrm{C}\), and a mass flow rate of \(10 \mathrm{~kg} / \mathrm{s}\). A separate stream of steam enters inlet 2 at 10 bar and \(200^{\circ} \mathrm{C}\).Saturated liquid at 10 bar exits the feedwater heater. Stray heat transfer and the effects of motion and gravity can be ignored. Let \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=1\) bar. Determine (a) the mass flow rate of the streams at inlet 2 and the exit, cach in \(\mathrm{kg} / \mathrm{s}\), (b) the rate of exergy destruction, in \(\mathrm{kW}\), and (c) the cost of the exergy destroyed, in \(\$ /\) year, for 8400 hours of operation annually. Evaluate exergy at \(8.5\) cents per \(\mathrm{kW} \cdot \mathrm{h}\).

A concrete slab measuring \(0.3 \mathrm{~m} \times 4 \mathrm{~m} \times 6 \mathrm{~m}\), initially at \(298 \mathrm{~K}\), is exposed to the sun for several hours, after which its temperature is \(301 \mathrm{~K}\). The density of the concrete is 2300 \(\mathrm{kg} / \mathrm{m}^{3}\) and its specific heat is \(c=0.88 \mathrm{~kJ} / \mathrm{kg}+\mathrm{K}\). (a) Determine the increase in exergy of the slab, in kJ. (b) To what elevation, in \(\mathrm{m}\), would a \(1000-\mathrm{kg}\) mass have to be raised from zero elevation relative to the reference environment for its exergy to equal the exergy increase of the slab? Let \(T_{0}=298 \mathrm{~K}\), \(p_{0}=1 \mathrm{~atm}, g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).

Air initially at 1 atm and \(500^{\circ} \mathrm{R}\) with a mass of \(2.5 \mathrm{lb}\) is contained within a closed, rigid tank. The air is slowly warmed, receiving 100 Btu by heat transfer through a wall separating the gas from a thermal reservoir at \(800^{\circ} \mathrm{R}\). This is the only energy transfer. Assuming the air undergoes an internally reversible process and using the ideal gas model, (a) determine the change in exergy and the exergy transfer accompanying heat, each in Btu, for the air as the system. (b) determine the exergy transfer accompanying heat and the exergy destruction, each in Btu, for an enlarged system that includes the air and the wall, assuming that the state of the wall remains unchanged. Compare with part (a) and comment. Let \(T_{0}=90^{\circ} \mathrm{F}, p_{0}=1 \mathrm{~atm}\).
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