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Chapter 7: Problem 75

At steady state, an insulated steam turbine develops work at a rate of \(786 \mathrm{~kJ} / \mathrm{kg}\) of steam flowing through the turbine. Steam enters at \(5515 \mathrm{kPa}\) and \(540^{\circ} \mathrm{C}\) and exits at \(100 \mathrm{kPa}\). Evaluate the isentropic turbine efficiency and the exergetic turbine efficiency. Ignore the effects of motion and gravity. Let \(T_{0}=15^{\circ} \mathrm{C}, p_{0}=100 \mathrm{kPa}\).

Short Answer

Expert verified
Isentropic efficiency is 61.86%. Exergetic efficiency is 87.9%.

Step by step solution

01

- Determine the isentropic efficiency

To find the isentropic turbine efficiency, we need the actual work produced by the turbine compared to the work that would be produced if the process were isentropic. The formula is \( \text{Efficiency} = \frac{\text{Actual Work}}{\text{Isentropic Work}} \). The actual work given is 786 kJ/kg.
02

- Find the inlet and exit states

From the steam tables, find the enthalpy and entropy at the inlet state of the steam (5515 kPa and 540掳C). Denoting the enthalpy and entropy at the inlet as \(h_{in}\) and \(s_{in}\) respectively. For steam at 5515 kPa and 540掳C, \(h_{in} = 3583.1 \, \text{kJ/kg}\) and \(s_{in} = 7.023 \, \text{kJ/kg.K}\).
03

- Calculate the isentropic final state

For the isentropic process, entropy remains constant. Thus, using the steam tables find the enthalpy at the exit state where pressure is 100 kPa and entropy is \(s_{in}\) (7.023 kJ/kg路K). For steam at 100 kPa with the same entropy \(s_{out} = 7.023 \, \text{kJ/kg.K}\), the enthalpy \(h_{out,isentropic} = 2312.3 \, \text{kJ/kg}\).
04

- Compute isentropic work

The isentropic work done by the turbine is given by the difference in enthalpy between the inlet and isentropic exit states: \( W_{isentropic} = h_{in} - h_{out,isentropic} = 3583.1 - 2312.3 = 1270.8 \, \text{kJ/kg}\).
05

- Isentropic efficiency calculation

Using \( \text{Efficiency}_{isentropic} = \frac{Actual Work}{Isentropic Work} = \frac{786 \, \text{kJ/kg}}{1270.8 \, \text{kJ/kg}} = 0.6186 = 61.86\text{\%}\).
06

- Determine the exergy destruction

Exergy destruction is given by the formula \( Ex_{destroyed} = T_0 \times \text{Wasted Entropy} \). Calculate the entropy change of the streams. For the given problem, \(T_0\) is 15掳C = 288K. Entropy at the inlet is 7.023 kJ/kg.K and at the exit state (using actual process) use steam tables to find \(s_{out}\) at 100 kPa and corresponding temperature (approximately ~7.4 kJ/kg.K). Then, \(Ex_{destroyed} = 288K \times (7.4 - 7.023) \, \text{kJ/kg.K} = 108.192 \, \text{kJ/kg}\).
07

- Exergetic efficiency calculation

The exergetic efficiency is calculated using \( \text{Efficiency}_{exergy} = \frac{W}{W + Ex_{destroyed}} = \frac{786}{786 + 108.192} = 0.879 \text{,}\ or \text{87.9\%}\).

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Key Concepts

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

steam turbine
A steam turbine is a device that extracts thermal energy from pressurized steam and converts it into mechanical work. In this exercise, we are analyzing an insulated steam turbine at steady state. The steam enters the turbine at very high pressure and temperature, and it's crucial to understand some basic properties to evaluate its performance.
To begin with, steam enters at 5515 kPa and 540掳C and exits at 100 kPa. The mechanical work developed in this process is given as 786 kJ/kg of steam.
Understanding these values helps us calculate key efficiencies and analyze the overall performance of the turbine.
enthalpy
Enthalpy is a measurement of the total energy content in a thermodynamic system. It includes internal energy and energy associated with pressure and volume. For steam turbines, knowing the enthalpy values at different points is essential.
In this problem, the inlet enthalpy ( \(h_{in}\)) is given as 3583.1 kJ/kg. We need to find the enthalpy at the exit for an isentropic process ( \(h_{out,isentropic}\)), which is 2312.3 kJ/kg. By knowing the enthalpy, we can compute the actual work and efficiency of the turbine.
  • Inlet Enthalpy, \(h_{in}\) = 3583.1 kJ/kg
  • Isentropic Exit Enthalpy, \(h_{out,isentropic}\) = 2312.3 kJ/kg
entropy
Entropy is a measure of the disorder or randomness in a system. For an isentropic process, the entropy remains constant, which is a crucial aspect for calculating turbine efficiency.
In this exercise, the entropy at the inlet ( \(s_{in}\)) is 7.023 kJ/kg.K. Since the process is isentropic, the entropy at the exit for the ideal case ( \(s_{out}\)) remains the same. When dealing with real processes, we observe a small increase in entropy due to inefficiencies.
  • Inlet Entropy, \(s_{in}\)= 7.023 kJ/kg.K
  • Exit Entropy (Actual), \(s_{out}\) 鈮� 7.4 kJ/kg.K
exergy destruction
Exergy destruction is an indication of the waste energy in a thermodynamic process. It quantifies the useful work potential lost due to irreversibilities.
In this problem, we first determine the change in entropy. Using the given temperature, we calculate the exergy destruction as follows:
\(Ex_{destroyed}\) = \(T_{0}\) 脳 (Entropy increase). Given \(T_{0}\) is 288K, this translates to:
\(Ex_{destroyed}\) = 288K 脳 (7.4 - 7.023) kJ/kg.K = 108.192 kJ/kg. This value indicates the loss of useful work potential due to the process inefficiencies.
exergetic efficiency
Exergetic efficiency is a measure of the efficiency of a process in utilizing the available energy. It incorporates both the work output and the exergy destruction.
In this case, the exergetic efficiency is calculated using:
\(Efficiency_{exergy}\) = \(\frac{W}{W + Ex_{destroyed}}\). Plugging in the numbers:
\(Efficiency_{exergy}\) = \(\frac{786}{786 + 108.192}\) 鈮� 87.9%. This means that 87.9% of the available energy is converted into useful work, with the remaining 12.1% lost due to inefficiencies.
  • Exergetic Efficiency = 87.9%
  • Exergy Destruction = 108.192 kJ/kg

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Most popular questions from this chapter

A compressor operating at steady state takes in \(1 \mathrm{~kg} / \mathrm{s}\) of air at 1 bar and \(25^{\circ} \mathrm{C}\) and compresses it to 8 bar and \(160^{\circ} \mathrm{C}\). The power input to the compressor is \(230 \mathrm{~kW}\), and heat transfer occurs from the compressor to the surroundings at an average surface temperature of \(50^{\circ} \mathrm{C}\). (a) Perform a full exergy accounting of the power input to the compressor. (b) Devise and evaluate an exergetic efficiency for the compressor. (c) Evaluating exergy at 8 cents per \(\mathrm{kW} \cdot \mathrm{h}\), determine the hourly costs of the power input, exergy loss associated with heat transfer, and exergy destruction. Neglect the effects of motion and gravity. Let \(T_{0}=25^{\circ} \mathrm{C}\), \(p_{\text {? }}=1\) bar.

Air at 1 bar, \(17^{\circ} \mathrm{C}\), and a mass flow rate of \(0.3 \mathrm{~kg} / \mathrm{s}\) enters an insulated compressor operating at steady state and exits at 3 bar, \(147^{\circ} \mathrm{C}\). Determine the power required by the compressor and the rate of exergy destruction, each in \(\mathrm{kW}\). Ignore the effects of motion and gravity. Let \(T_{0}=17^{\circ} \mathrm{C}\), \(p_{0}=1\) bar.

Refrigerant 134 a enters a water-jacketed compressor operating at steady state at \(10^{\circ} \mathrm{C}, 3.2\) bar and exits at \(70^{\circ} \mathrm{C}, 10\) bar. Cooling water enters as a separate stream at \(20^{\circ} \mathrm{C}\) and exits at \(32^{\circ} \mathrm{C}\) with no significant change in pressure. The refrigerant mass flow rate is \(1.63 \mathrm{~kg} / \mathrm{s}\), and the power input to the compressor is \(55.2 \mathrm{~kJ}\) per \(\mathrm{kg}\) of refrigerant flowing. Assuming no heat transfer between the outer surface of the water jacket and the surroundings, and neglecting the effects of motion and gravity (a) determine the mass flow rate of cooling water, in \(\mathrm{kg} / \mathrm{s}\), and (b) perform a full exergy accounting, in \(\mathrm{kW}\), based on the compressor power input and comment. Let \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=1\) bar.

Determine the exergy, in \(\mathrm{kJ}\), at \(0.7\) bar, \(90^{\circ} \mathrm{C}\) for \(1 \mathrm{~kg}\) of (a) water, (b) Refrigerant \(134 \mathrm{a}\), (c) air as an ideal gas with \(c_{p}\) constant. In each case, the mass is at rest and zero evaluation relative to an exergy reference environment for which \(T_{0}=20^{\circ}\) C \(p_{0}=1\) bar.

At steady state, a turbine with an exergetic efficiency of \(85 \%\) develops \(18 \times 10^{7} \mathrm{~kW} \cdot \mathrm{h}\) of work annually \((8000\) operating hours). The annual cost of owning and operating the turbine is \(\$ 5.0 \times 10^{5}\). The steam entering the turbine has a specific flow exergy of \(1500 \mathrm{~kJ} / \mathrm{kg}\), a mass flow rate of \(40 \mathrm{~kg} / \mathrm{s}\), and is valued at \(\$ 0.0182\) per \(\mathrm{kW} \cdot \mathrm{h}\) of exergy. (a) Evaluate the unit cost of the power developed, in \(\$$ per \)\mathrm{kW} \cdot \mathrm{h}\(. (b) Evaluate the unit cost based on exergy of the steam entering and exiting the turbine, each in cents per \)\mathrm{kg}$ of steam flowing through the turbine.
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