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The entropy of steam increases in actual steam turbines as a result of irreversibilities. In an effort to control entropy increase, it is proposed to cool the steam in the turbine by running cooling water around the turbine casing. It is argued that this will reduce the entropy and the enthalpy of the steam at the turbine exit and thus increase the work output. How would you evaluate this proposal?

Short Answer

Expert verified
Answer: When evaluating the proposal to control entropy increase in a steam turbine by cooling the steam with water, the key considerations include the first law of thermodynamics, the impact of cooling on steam entropy and enthalpy, and the potential change in work output. The effectiveness of the proposal depends on the efficiency of the cooling process and its ability to balance the energy loss from cooling with the potential benefits to the overall work output of the system. Additionally, it is crucial to investigate specific conditions and consider the system as a whole.

Step by step solution

01

Consider the first law of thermodynamics

First, we need to recall the first law of thermodynamics, which states that energy cannot be created or destroyed, but can only change forms. In the case of the steam turbine, the energy coming in is in the form of heat, which is transferred to the steam, and it is then converted into mechanical work at the turbine exit. By cooling the steam, the proposal aims to reduce the amount of entropy and enthalpy, but this action must comply with the first law of thermodynamics.
02

Analyze the impact of cooling on steam entropy

Entropy is a measure of the disorder or randomness of a system and is related to the amount of heat that a substance can absorb or release. By cooling the steam, we are effectively removing heat from it, which would result in a decrease in entropy. However, we must consider that the cooling process is likely to involve some irreversible processes, such as friction and heat transfer with the surroundings, which would result in some entropy generation. Nevertheless, the overall impact of cooling on steam entropy must be evaluated.
03

Analyze the impact of cooling on steam enthalpy

Enthalpy is a measure of the total energy content of a substance and is a crucial property in evaluating the work output of a system. As the steam is cooled, its enthalpy would reduce due to the removal of heat. The decrease in enthalpy at the turbine exit is expected to affect the work output since the work output of a turbine is typically determined by the difference in enthalpy between the inlet and outlet points.
04

Evaluate the change in work output

The work output of a steam turbine can be expressed by the following equation: W_{out} = m(\Delta h), where W_{out} is the work output, m is the mass flow rate, and \Delta h is the change in enthalpy between the inlet and outlet points. By cooling the steam, the enthalpy of the steam at the turbine exit would reduce. However, if the cooling process is inefficient and results in significant heat loss to the surroundings, then it may not be beneficial to the overall work output of the system.
05

Conclusion

In conclusion, the proposal to control entropy increase in a steam turbine by cooling the steam with water may lead to a decrease in entropy and enthalpy. However, the impact on work output depends on the efficiency of the cooling process and its ability to balance the energy loss from cooling with the potential benefits. To properly evaluate the proposal, one would need to investigate specific conditions and consider the system as a whole, taking into account the first law of thermodynamics and potential changes in work output.

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Most popular questions from this chapter

Consider an ideal steam regenerative Rankine cycle with two feedwater heaters, one closed and one open. Steam enters the turbine at \(10 \mathrm{MPa}\) and \(600^{\circ} \mathrm{C}\) and exhausts to the condenser at \(10 \mathrm{kPa}\). Steam is extracted from the turbine at 1.2 MPa for the closed feedwater heater and at 0.6 MPa for the open one. The feedwater is heated to the condensation temperature of the extracted steam in the closed feedwater heater. The extracted steam leaves the closed feedwater heater as a saturated liquid, which is subsequently throttled to the open feedwater heater. Show the cycle on a \(T-s\) diagram with respect to saturation lines, and determine \((a)\) the mass flow rate of steam through the boiler for a net power output of \(400 \mathrm{MW}\) and \((b)\) the thermal efficiency of the cycle.

The gas-turbine portion of a combined gas-steam power plant has a pressure ratio of \(16 .\) Air enters the compressor at \(300 \mathrm{K}\) at a rate of \(14 \mathrm{kg} / \mathrm{s}\) and is heated to \(1500 \mathrm{K}\) in the combustion chamber. The combustion gases leaving the gas turbine are used to heat the steam to \(400^{\circ} \mathrm{C}\) at \(10 \mathrm{MPa}\) in a heat exchanger. The combustion gases leave the heat exchanger at 420 K. The steam leaving the turbine is condensed at 15 kPa. Assuming all the compression and expansion processes to be isentropic, determine \((a)\) the mass flow rate of the steam, \((b)\) the net power output, and \((c)\) the thermal efficiency of the combined cycle. For air, assume constant specific heats at room temperature.

Why is the Carnot cycle not a realistic model for steam power plants?

An ideal Rankine steam cycle modified with two closed feedwater heaters is shown below. The power cycle receives \(75 \mathrm{kg} / \mathrm{s}\) of steam at the high pressure inlet to the turbine. The feedwater heater exit states for the boiler feedwater and the condensed steam are the normally assumed ideal states. The fraction of mass entering the high pressure turbine at state 5 that is extracted for the feedwater heater operating at \(1400 \mathrm{kPa}\) is \(y=0.1446 .\) Use the data provided in the tables given below to (a) Sketch the \(T\) -s diagram for the ideal cycle. (b) Determine the fraction of mass, \(z\), that is extracted for the closed feedwater heater operating at the \(245 \mathrm{kPa}\) extraction pressure. (c) Determine the required cooling water flow rate, in \(\mathrm{kg} / \mathrm{s}\), to keep the cooling water temperature rise in the condenser to \(10^{\circ} \mathrm{C}\). Assume \(c_{p}=4.18 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) for cooling water (d) Determine the net power output and the thermal efficiency of the plant.

Consider an ideal reheat-regenerative Rankine cycle with one open feedwater heater. The boiler pressure is \(10 \mathrm{MPa}\), the condenser pressure is \(15 \mathrm{kPa}\), the reheater pressure is \(1 \mathrm{MPa}\), and the feedwater pressure is \(0.6 \mathrm{MPa}\). Steam enters both the high- and low- pressure turbines at \(500^{\circ} \mathrm{C} .\) Show the cycle on a \(T\) -s diagram with respect to saturation lines, and determine (a) the fraction of steam extracted for regeneration and \((b)\) the thermal efficiency of the cycle.

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