Review of Steam Turbine Control Valve for Nuclear Plant

Chapter 4: The Kickoff Police force of Thermodynamics for Control Volumes

b) Steam Power Plants

A basic steam power plant consists of four interconnected components, typically equally shown in the figure below. These include a steam turbine to produce mechanical shaft power, a condenser which uses external cooling water to condense the steam to liquid water, a feedwater pump to pump the liquid to a high pressure, and a boiler which is externally heated to boil the water to superheated steam. Unless otherwise specified we assume that the turbine and the pump (as well as all the interconnecting tubing) are adiabatic, and that the condenser exchanges all of its heat with the cooling water.

A Uncomplicated Steam Power Plant Instance - In this case we wish to determine the performance of this basic steam power establish under the conditions shown in the diagram, including the power of the turbine and feedwater pump, heat transfer rates of the boiler and condenser, and thermal efficiency of the organisation.

In this example we wish to evaluate the post-obit:

• Turbine output ability and the power required to drive the feedwater pump

• Oestrus ability supplied to the boiler and that rejected in the condenser to the cooling water

• The thermal efficiency of the ability constitute (η _thursday ), divers every bit the net work washed by the system divided by the oestrus supplied to the boiler.

• The minimum mass flow rate of the cooling water in the condenser required for a specific temperature rise

Do not be intimidated by the complexity of this system. Nosotros volition discover that we can solve each component of this system separately and independently of all the other components, always using the same approach and the same basic equations. We first use the information given in the above schematic to plot the iv processes (1)-(2)-(three)-(iv)-(ane) on the P-h diagram . Notice that the fluid entering and exiting the boiler is at the high pressure level 10 MPa, and similarly that entering and exiting the condenser is at the low pressure 20 kPa. State (i) is given by the intersection of x MPa and 500°C, and state (ii) is given as 20 kPa at 90% quality, Land (3) is given past the intersection of 20 kPa and 40°C, and the feedwater pumping process (3)-(4) follows the constant temperature line, since T4 = T3 = 40°C, .

Notice from the P-h diagram plot how nosotros can become an instant visual appreciation of the system performance, in detail the thermal efficiency of the system past comparing the enthalpy difference of the turbine (1)-(2) to that of the boiler (iv)-(1). Nosotros also notice that the ability required by the feedwater pump (three)-(4) is negligible compared to any other component in the system.

( Notation: We discover it foreign that the only thermodynamics text that we know of that even considered the use of the P-h diagram for steam power plants is Engineering Thermodynamics - Jones and Dugan (1995). It is widely used for refrigeration systems, nonetheless not for steam ability plants.)

Nosotros at present consider each component as a separate control book and apply the free energy equation, starting with the steam turbine. The steam turbine uses the high-pressure - high-temperature steam at the inlet port (1) to produce shaft power by expanding the steam through the turbine blades, and the resulting low-force per unit area - low-temperature steam is rejected to the condenser at port (ii). Detect that we take assumed that the kinetic and potential free energy change of the fluid is negligible, and that the turbine is adiabatic. In fact any rut loss to the surroundings or kinetic energy increase would be at the expense of output power, thus practical systems are designed to minimise these loss effects. The required values of enthalpy for the inlet and outlet ports are determined from the steam tables .

Thus we encounter that nether the conditions shown the steam turbine will produce 8MW of power.

The very low-force per unit area steam at port (two) is at present directed to a condenser in which heat is extracted by cooling water from a nearby river (or a cooling belfry) and the steam is condensed into the subcooled liquid region. The assay of the condenser may require determining the mass period rate of the cooling water needed to limit the temperature ascension to a certain amount - in this instance to x°C. This is shown on the post-obit diagram of the condenser:

Notice that our steam tables do not include the subcooled (or compressed) liquid region that we find at the outlet of the condenser at port (3). In this region we notice from the P-h diagram that over an extremely loftier force per unit area range the enthalpy of the liquid is equal to the saturated liquid enthalpy at the same temperature, thus to a close approximation h _three = h _f@40°C , independent of the pressure.

Thus nosotros see that nether the conditions shown, 17.6 MW of heat is extracted from the steam in the condenser.

I have oftentimes been queried by students every bit to why nosotros have to turn down such a big amount of heat in the condenser causing such a big decrease in thermal efficiency of the power establish. Without going into the philosophical aspects of the 2d Law (which we comprehend after in Chapter 5 , my best reply was provided to me by Randy Sheidler, a senior engineer at the Gavin Power Plant . He stated that the Quaternary Law of Thermodynamics states: " You lot can't pump steam! " , so until we condense all the steam into liquid water past extracting 17.vi MW of heat, we cannot pump information technology to the high pressure to complete the cycle. (Randy could not requite me a reference to the source of this amazing observation).

In society to determine the enthalpy change Δh of the cooling h2o (or in the feedwater pump which follows), we consider the water to be an Incompressible Liquid , and evaluate Δh as follows:

From the steam tables we notice that the specific rut capacity for liquid water C _H2o = 4.xviii kJ/kg°C. Using this analysis nosotros found on the condenser diagram to a higher place that the required mass period charge per unit of the cooling water is 421 kg/s. If this flow rate cannot be supported past a nearby river so a cooling tower must be included in the power constitute blueprint.

Nosotros now consider the feedwater pump every bit follows:

Thus as we suspected from the above P-h diagram plot, the pump power required is extremely low compared to whatever other component in the arrangement, being only 1% that of the turbine output ability produced.

The final component that nosotros consider is the boiler, as follows:

Thus we run across that under the weather condition shown the heat ability required by the boiler is 25.7 MW. This is normally supplied by combustion (or nuclear power). We now accept all the data needed to determine the thermal efficiency of the steam power plant as follows:

Note that the feedwater pump work can normally exist neglected.

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Solved Problem iv.1 - A Supercritical Steam Power Institute with Reheat for Athens, Ohio

Solved Problem 4.ii - An Open Feedwater Heater added to the Supercritical Reheat Steam Power Institute for Athens, Ohio

Solved Problem 4.2 (Alternating) - An Open Feedwater Heater and throttling command valve added to the Supercritical Reheat Steam Power Constitute

Problem four.3 - A Geothermal Hybrid Steam Ability Found

Problem 4.four - Solar Pond Hybrid Steam Power Constitute

Problem 4.five - A Cogeneration Steam Power Institute

Problem 4.6 - An Open Feedwater Heater added to the Cogeneration Steam Power Constitute

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On to Part c) - Refrigerators and Heat Pumps

On to Part d) – Carbon Dioxide Refrigerant (R744)

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Source: https://www.ohio.edu/mechanical/thermo/Intro/Chapt.1_6/Chapter4b.html