/*! 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 40 A power plant operates on a rege... [FREE SOLUTION] | 91Ó°ÊÓ

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A power plant operates on a regenerative vapor power cycle with one open feedwater heater. Steam enters the first turbine stage at \(12 \mathrm{MPa}, 560^{\circ} \mathrm{C}\) and expands to \(1 \mathrm{MPa}\), where some of the steam is extracted and diverted to the open feedwater heater operating at \(1 \mathrm{MPa}\). The remaining steam expands through the second turbine stage to the condenser pressure of \(6 \mathrm{kPa}\). Saturated liquid exits the open feedwater heater at \(1 \mathrm{MPa}\). The net power output for the cycle is \(330 \mathrm{MW}\). For isentropic processes in the turbines and pumps, determine (a) the cycle thermal efficiency. (b) the mass flow rate into the first turbine stage, in \(\mathrm{kg} / \mathrm{s}\). (c) the rate of entropy production in the open feedwater heater, in \(k W / K\).

Short Answer

Expert verified
Thermal efficiency, mass flow rate, and entropy production rates are determined using energy and entropy balances, isentropic relations, and steam tables.

Step by step solution

01

Determine the state points of the cycle

Identify all points in the cycle (turbine stages, condenser, feedwater heater). Use steam tables or Mollier diagrams to find properties like enthalpy, entropy, temperature, and pressure at these points. Particularly, obtain the state points at the inlet and outlet of each component.
02

Analyze Turbine 1

For the first turbine stage, steam enters at 12 MPa and 560°C. Expand isentropically to 1 MPa. Use the principle of isentropic expansion to find the enthalpy at the exit of the first turbine stage (state 2).
03

Analyze the Feedwater Heater

Determine the state of steam entering and exiting the open feedwater heater. Use steam tables to find the enthalpy and entropy for steam extracted at 1 MPa (state 3') and the saturated liquid exiting at 1 MPa (state 4).
04

Analyze Turbine 2

Steam leaving the second turbine stage expands isentropically to the condenser pressure of 6 kPa. Using the enthalpy at state 2 and steam tables, find the enthalpy at the outlet of the second turbine (state 5).
05

Determine Condenser and Pump States

Find the enthalpy at the exit of the condenser (state 6, typically saturated liquid at 6 kPa) and the corresponding enthalpy after the feed pump at 12 MPa (state 1'). Use isentropic pump relations to find state 1'.
06

Apply Energy Conservation for Efficiency

Use the first law of thermodynamics for the entire cycle to determine the thermal efficiency. Calculate the work done by the turbines and the pump, and the heat added in the boiler.
07

Calculate Mass Flow Rate into the First Turbine

Given the net power output, use the energy balance equations to determine the mass flow rate. Use the specific work and net power output in calculations.
08

Calculate Rate of Entropy Production in Open Feedwater Heater

Using the second law of thermodynamics, calculate the entropy production in the open feedwater heater by applying entropy balance for the control volume.

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Key Concepts

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

Thermal Efficiency
Thermal efficiency is a key measure of how well a power cycle converts heat into work. It is defined as the ratio of the net work output to the heat input. In mathematical terms, thermal efficiency, \( \eta \/_\{th\}\) can be expressed as:
\[ \eta \/_\{th\} = \frac{W\/\{\text{net}}}{Q\/\{\text{in}}} \] where \(W\/\{\text{net}}\) is the net work output of the cycle, and \(Q\/\{\text{in}}\) is the heat input to the cycle.
To determine the thermal efficiency of a regenerative vapor power cycle:
  • First, calculate the work done by each turbine stage and the pump.
  • Then, calculate the total heat added in the boiler.
  • Finally, use these values to find the efficiency using the above formula.

This efficiency helps assess how effectively the power plant utilizes the input energy to produce electricity.
Isentropic Process
An isentropic process is an ideal process where entropy remains constant. This means there are no losses due to irreversibilities like friction or unrestrained expansion. In the context of the regenerative vapor cycle:
  • The expansion of steam in the turbines is considered isentropic for ideal analysis. This means the entropy before and after expansion remains the same.
  • Similarly, in pumps, the compression of fluid is assumed to be isentropic, meaning no heat is lost.

To calculate quantities during isentropic processes, we typically rely on steam tables or Mollier diagrams. These tools help in obtaining the properties such as enthalpy and entropy at different states.
Feedwater Heater
A feedwater heater is an essential component in a regenerative vapor power cycle. It preheats the water before it enters the boiler, increasing the cycle's efficiency. Here’s how it works:
  • Steam is extracted from the turbine at intermediate pressures and fed into the heater.
  • This steam heats the water entering the boiler, reducing the fuel required for boiling as the water is already warmer.
  • In open feedwater heaters, the extracted steam and feedwater mix directly, leading to a heat exchange and an increase in feedwater temperature.

By preheating the feedwater, the energy required for generating steam in the boiler is reduced, boosting the overall cycle efficiency.
Entropy Production
Entropy production is a measure of the irreversibilities in a process, reflecting energy losses that cannot be converted into work. In a regenerative power cycle:
  • Entropy increases whenever there are real processes, like steam mixing in the feedwater heater.
  • In the open feedwater heater, steam extraction and condensation lead to entropy production.

The rate of entropy production can be calculated using the second law of thermodynamics and an entropy balance on the control volume. Specifically, for entropy production in the feedwater heater, you can use:
\[ \dot{S} = \dot{m} \left(s\/\{\text{out}} - s\/\{\text{in}}\right) \]
where \( \dot{m} \) is the mass flow rate and \(s\/\{\text{out}}\) and \(s\/\{\text{in}}\) are the specific entropies at the exit and entrance.
Mass Flow Rate
The mass flow rate is the amount of steam flowing through the system per unit of time. This rate is crucial for understanding the performance and sizing of the cycle components. For the first turbine stage in a regenerative cycle:
  • Given the net power output, we can determine the mass flow rate into the first turbine.
  • Use the energy balance equation, relating net power output to the specific work done by the turbine and the mass flow rate:

\[ \dot{W}\/\{\text{net}} = \dot{m} \cdot (h\/\{\text{in}} - h\/\{\text{out}}) \]
Solving for \( \dot{m} \), where \( \dot{W} \) is the net power output, and \( h\/\{\text{in}} \) and \( h\/\{\text{out}} \) are the specific enthalpies at the turbine entrance and exit, respectively.

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Most popular questions from this chapter

Steam enters the turbine of a vapor power plant at 600 lbfin., \(1000^{\circ} \mathrm{F}\) and exits as a two-phase liquid-vapor mixture at temperature \(T\). Condensate exits the condenser at a temperature \(5^{\circ} \mathrm{F}\) lower than \(T\) and is pumped to \(600 \mathrm{lb} / \mathrm{in}^{2} .\) The turbine and pump isentropic efficiencies are 90 and \(80 \%\), respectively. The net power developed is \(1 \mathrm{MW}\). (a) For \(T=80^{\circ} \mathrm{F}\), determine the steam quality at the turbine exit, the steam mass flow rate, in Ib/h, and the thermal cfficiency. (b) Plot the quantities of part (a) versus \(T\) ranging from \(80 \square\) to \(105^{\circ} \mathrm{F}\).

Steam enters the turbine of a simple vapor power plant with a pressure of \(12 \mathrm{MPa}\) and a temperature of \(600^{\circ} \mathrm{C}\) and expands adiabatically to condenser pressure, \(p\). Saturated liquid exits the condenser at pressure \(p\). The isentropic efficiency of both the turbine and the pump is \(84 \%\). (a) For \(p=30 \mathrm{kPa}\), determine the turbine exit quality and the cycle thermal efficiency. (b) Plot the quantities of part (a) versus \(p\) ranging from \(6 \mathrm{kPa}\) to \(100 \mathrm{kPa}\).

Steam enters the turbine of a simple vapor power plant at 100 bar, \(520^{\circ} \mathrm{C}\) and expands adiabatically, exiting at \(0.08\) bar with a quality of \(90 \%\). Condensate leaves the condenser as saturated liquid at \(0.08\) bar. Liquid exits the pump at 100 bar, \(43^{\circ} \mathrm{C}\). The specific exergy of the fuel entering the combustor unit of the steam generator is estimated to be \(14,700 \mathrm{~kJ} / \mathrm{kg}\). No exergy is carried in by the combustion air. The exergy of the stack gases leaving the steam generator is estimated to be \(150 \mathrm{~kJ}\) per kg of fuel. The mass flow rate of the steam is \(3.92 \mathrm{~kg}\) per \(\mathrm{kg}\) of fuel. Cooling water enters the condenser at \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=1 \mathrm{~atm}\) and exits at \(35^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). Develop a full accounting of the exergy entering the plant with the fuel.

Water is the working fluid in a Rankine cycle modified to include one closed feedwater heater and one open feedwater heater. Superheated vapor enters the turbine at \(16 \mathrm{MPa}, 560^{\circ} \mathrm{C}\), and the condenser pressure is \(8 \mathrm{kPa}\). The mass flow rate of steam entering the first- stage turbine is \(120 \mathrm{~kg} / \mathrm{s}\). The closed feedwater heater uses extracted steam at \(4 \mathrm{MPa}\), and the open feedwater heater uses extracted steam at \(0.3 \mathrm{MPa}\). Saturated liquid condensate drains from the closed feedwater heater at \(4 \mathrm{MPa}\) and is trapped into the open feedwater heater. The feedwater leaves the closed heater at \(16 \mathrm{MPa}\) and a temperature equal to the saturation temperature at \(4 \mathrm{MPa}\). Saturated liquid leaves the open heater at \(0.3 \mathrm{MPa}\). Assume all turbine stages and pumps operate isentropically. Determine (a) the net power developed, in kW. (b) the rate of heat transfer to the steam passing through the steam generator, in \(\mathrm{kW}\). (c) the thermal efficiency- (d) the mass flow rate of condenser cooling water, in \(\mathrm{kg} / \mathrm{s}\), if the cooling water undergoes a temperature increase of \(18^{\circ} \mathrm{C}\) with negligible pressure change in passing through the condenser.

Water is the working fluid in a Rankine cycle. Superheated vapor enters the turbine at \(8 \mathrm{MPa}, 560^{\circ} \mathrm{C}\) with a mass flow rate of \(7.8 \mathrm{~kg} / \mathrm{s}\) and exits at \(8 \mathrm{kPa}\). Saturated liquid enters the pump at \(8 \mathrm{kPa}\). The isentropic turbine efficiency is \(88 \%\), and the isentropic pump efficiency is \(82 \%\). Cooling water enters the condenser at \(18^{\circ} \mathrm{C}\) and exits at \(36^{\circ} \mathrm{C}\) with no significant change in pressure. Determine (a) the net power developed, in \(\mathrm{kW}\). (b) the thermal efficiency. (c) the mass flow rate of cooling water, in \(\mathrm{kg} / \mathrm{s}\).

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