91Ó°ÊÓ

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.

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.