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Chapter 7: Problem 122

Saturated water vapor at \(400 \mathrm{lbf} / \mathrm{in}^{2}\) enters an insulated turbine operating at steady state. At the turbine exit the pressure is \(0.6\) lbf/in. \({ }^{2}\) The work developed is 306 Btu per pound of steam passing through the turbine. Kinetic and potential energy effects can be neglected. Let \(T_{0}=60^{\circ} \mathrm{F}\), \(p_{0}=1 \mathrm{~atm}\). Determine (a) the exergy destruction rate, in Btu per pound of steam expanding through the turbine. (b) the isentropic turbine efficiency- (c) the exergetic turbine efficiency.

Short Answer

Expert verified
The exergy destruction rate, isentropic turbine efficiency, and exergetic turbine efficiency are calculated step-by-step by determining enthalpy and entropy changes and applying relevant formulas.

Step by step solution

01

- Understand Given Data

Collect and note all given data: - Initial state (inlet): Saturated water vapor - Inlet pressure: \(400 \mathrm{lbf} / \mathrm{in}^{2}\)- Outlet pressure: \(0.6\ \mathrm{lbf} / \mathrm{in}^{2}\)- Work developed: \(306\ \mathrm{Btu/lb}\)- Surroundings temperature: \(T_0 = 60^{\circ} \mathrm{F}\)- Surroundings pressure: \(p_0 = 1\ \mathrm{atm}\)
02

- Determine Enthalpies at Inlet and Outlet

Use steam tables or Mollier diagram to find enthalpy values:At the inlet (saturated steam), find enthalpy \(h_1\) at \(400\ \mathrm{lbf/in}^2\).At the outlet, with pressure \(0.6\ \mathrm{lbf/in}^2\) and considering the output work, find \(h_2\) using: \[ h_2 = h_1 - W_t \]
03

- Determine Specific Entropies at Inlet and Outlet

Use steam tables to find entropy at the inlet \(s_1\). To find the outlet entropy \(s_2\) for an isentropic process, use: \(s_2\) at the given outlet pressure.
04

- Calculate Exergy Destruction

Determine the exergy destruction rate using the formula: \[ \dot{E}_D = T_0 \cdot s_{gen} \] where entropy generation \(s_{gen}\) is: \[ s_{gen} = s_2 - s_1 \]
05

- Compute Isentropic Turbine Efficiency

Using: \[ \eta_{t,is} = \frac{h_1 - h_2}{h_1 - h_2s} \] where \(h_{2s}\) is the exit enthalpy for isentropic expansion at exit pressure.
06

- Compute Exergetic Turbine Efficiency

Using the formula: \[ \eta_{ex} = \frac{w_{t}}{(h_1 - h_2) - T_0 \cdot(s_2 - s_1)} \]

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Key Concepts

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

Exergy Destruction
Exergy destruction in a turbine happens because of inefficiencies like friction and heat losses.
It represents the loss of the potential to do work.
In our problem, given the initial and final conditions along with the developed work and entropy values, exergy destruction is calculated using:
\[ \text{Exergy Destruction Rate} = T_0 \times s_{\text{gen}} \]Here, entropy generation (\(s_{\text{gen}}\)) is the difference between the specific entropy at the outlet (\(s_2\)) and the specific entropy at the inlet (\(s_1\)).
You can find these entropy values in the steam tables.
This helps quantify the energy lost because the process is not reversible.
Isentropic Efficiency
Isentropic Efficiency is a measure of the actual performance of a turbine relative to an ideal isentropic turbine.
It compares the actual work output to the maximum possible work output in an ideal, frictionless, and adiabatic process.
The formula is:\[ \eta_{t,is} = \frac{h_1 - h_2}{h_1 - h_{2s}} \]Where:
  • \(h_1\) is the enthalpy at the inlet,
  • \(h_2\) is the actual enthalpy at the turbine exit,
  • \(h_{2s}\) is the enthalpy at the turbine exit for an ideal isentropic process.
The closer the isentropic efficiency is to 1, the more efficient the turbine is.
You'll typically find the enthalpy values using steam tables.
Exergetic Efficiency
Exergetic Efficiency focuses on how well a system uses available energy to do work.
It's calculated as the ratio of the useful work output to the maximum possible available energy difference that could have been converted into work.
For a turbine, it's given by:\[ \eta_{ex} = \frac{w_{t}}{(h_1 - h_2) - T_0 \times (s_2 - s_1)} \]Where:
  • \(w_t\) is the actual work output,
  • \(h_1 - h_2\) is the change in specific enthalpy,
  • \(T_0 \times (s_2 - s_1)\) is the energy loss due to entropy generation.
This efficiency shows the real-world effectiveness of energy use.
Higher values indicate less energy loss in the process.
Saturated Steam
Saturated steam is steam at the temperature where it is about to condense.
It means that the steam is at a specific temperature and pressure where it exists in equilibrium between liquid and vapor phases.
In our exercise, the initial steam entering the turbine is saturated.
This is significant because the properties (enthalpy and entropy) of saturated steam are readily available in steam tables.
Knowing whether the steam is saturated helps in accurately retrieving these values, which are necessary for calculations.
Steam Tables
Steam tables are essential for thermodynamic analysis of systems involving water and steam.
They contain data on properties like enthalpy (\(h\)), entropy (\(s\)), pressure, and temperature for various states of water and steam.
In solving our exercise, steam tables help find the necessary properties at different pressure levels.
For example, we look up the enthalpy and entropy for saturated steam at 400 lbf/in2 and find the state at 0.6 lbf/in2 after expansion.
Using these tables ensures accurate and reliable data for the calculations.
Entropy Generation
Entropy generation occurs when there are irreversibilities in a process.
It leads to increased disorder in the system, reducing the ability to do work.
In the context of turbines, entropy generation is the deviation from the ideal isentropic process.
The formula to find entropy generation (\(s_{\text{gen}}\)) is:
\[ s_{\text{gen}} = s_2 - s_1 \]Where:
  • \(s_1\) is the entropy at the inlet,
  • \(s_2\) is the entropy at the outlet.
Higher entropy generation points to higher losses in the turbine.
Minimizing entropy generation is key to improving the efficiency of the system.

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Most popular questions from this chapter

Air enters a turbine operating at steady state at a pressure of \(75 \mathrm{lbf} / \mathrm{in} .^{2}\), a temperature of \(800^{\circ} \mathrm{R}\), and a velocity of \(400 \mathrm{ft} / \mathrm{s}\) At the exit, the conditions are \(15 \mathrm{lbf}^{2} .^{2}, 600^{\circ} \mathrm{R}\), and \(100 \mathrm{ft} / \mathrm{s}\). There is no significant change in elevation. Heat transfer from the turbine to its surroundings at a rate of 10 Btu per lb of air flowing takes place at an average surface temperature of \(700^{\circ} \mathrm{R}\). (a) Determine, in Btu per lb of air passing through the turbine, the work developed and the exergy destruction rate. (b) Expand the boundary of the control volume to include both the turbine and a portion of its immediate surroundings so that heat transfer occurs at a temperature \(T_{0}\). Determine, in Btu per lb of air passing through the turbine, the work developed and the exergy destruction rate. (c) Explain why the exergy destruction rates in parts (a) and (b) are different. Let \(T_{0}=40^{\circ} \mathrm{F}, p_{0}=15 \mathrm{lbf} / \mathrm{in}^{2}\)

Two solid blocks, each having mass \(m\) and specific heat \(c\), and initially at temperatures \(T_{1}\) and \(T_{2}\), respectively, are brought into contact, insulated on their outer surfaces, and allowed to come into thermal equilibrium. (a) Derive an expression for the exergy destruction in terms of \(m, c, T_{1}, T_{2}\), and the temperature of the environment, \(T_{0}\) (b) Demonstrate that the exergy destruction cannot be negative. (c) What is the source of exergy destruction in this case?

Oxygen \(\left(\mathrm{O}_{2}\right)\) enters a well-insulated nozzle operating at steady state at \(80 \mathrm{lbf}\) in. \({ }^{2}, 1100^{\circ} \mathrm{R}, 90 \mathrm{ft} / \mathrm{s}\) At the nozle exit, the pressure is 1 lbf/in. \({ }^{2}\) The isentropic nozle efficiency is \(85 \%\). For the nozle, determine the exit velocity, in \(\mathrm{m} / \mathrm{s}\), and the exergy destruction rate, in Btu per lb of oxygen flowing. Let \(T_{0}=70^{\circ} \mathrm{F}, p_{0}=14.7 \mathrm{lbf} / \mathrm{in}^{2}\)

Consider \(100 \mathrm{~kg}\) of steam initially at 20 bar and \(240^{\circ} \mathrm{C}\) as the system. Determine the change in exergy, in \(\mathrm{kJ}\), for each of the following processes: (a) The system is heated at constant pressure until its volume doubles. (b) The system expands isothermally until its volume doubles. Let \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=1\) bar and ignore the effects of motion and gravity.

\(7.92\) Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) gas enters a turbine operating at steady state at 50 bar, \(500 \mathrm{~K}\) with a velocity of \(50 \mathrm{~m} / \mathrm{s}\). The inlet area is \(0.02 \mathrm{~m}^{2}\). At the exit, the pressure is 20 bar, the temperature is \(440 \mathrm{~K}\), and the velocity is \(10 \mathrm{~m} / \mathrm{s}\) The power developed by the turbine is \(3 \mathrm{MW}\), and heat transfer occurs across a portion of the surface where the average temperature is \(462 \mathrm{~K}\). Assume ideal gas behavior for the carbon dioxide and neglect the effect of gravity. Let \(T_{0}=298 \mathrm{~K}, p_{0}=1\) bar. (a) Determine the rate of heat transfer, in \(\mathrm{kW}\). (b) Perform a full exergy accounting, in \(\mathrm{kW}\), based on the net rate exergy is carried into the turbine by the carbon dioxide.
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