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**Design &Types of Steam Turbines**

Prof. Osama El Masry

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**Design Characteristics for a Steam Turbine**

Custom design Thermal output Fuel flexibility Reliability and life Size range Emissions

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**Performance and Efficiency Enhancements**

Electrical Efficiency Thermodynamic Efficiency Efficiency Enhancements

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Condensing Turbine The condensing turbine processes result in maximum power and electrical generation efficiency from the steam supply and boiler fuel Inlet pressure is relatively high and exhaust pressure is largely reduced The power output of condensing turbines is sensitive to ambient conditions

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Condensing Turbine Increasing condensation temp. from 38 oC to 45°C, gives 6.5% less power output When condensation temp. is reduced to 27°C the power output is increased by 9.5%

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Example 5.1 In a condensing turbine cycle, the turbine generates 3.5 MW of Electric power with inlet steam at 27 bar & 71oC of superheat, the turbine efficiency is 75%, and the condenser pressure is 0.07 bar. The plant has also a separate low-pressure boiler, which generates saturated steam at 2.7 bar from feed water at 130oC and the boiler provides the heating capacity of 70x103 MJ/h. Calculate the steam mass flow rate and heat added in the two boilers if the boiler efficiency is 84% in both boilers.

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Solution For the condensing turbine E=3.5 x 1000=ms1(h1-h2)ηT h1= 3000 kJ/kg h2=2o65 kJ/kg ms1= 5 kg/s Q1= ms1(h1-h`pc)/ηb=5( )/0.84=16.88 MW For the heating boiler H=70 x 106/3600== ms2 (h3-h4) = ms2( x 4.187) ms2=8.94kg/s Q2= ms2(h3-h4)/ηb=70 x 106/3600 x 0.84=23.1 MW mtotal=13.94 kg/s Qtotal=40 MW

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**Back-pressure Turbine**

The term “back-pressure” refers to turbines that exhaust steam at atmospheric pressures and above A back-pressure turbine exhausts its entire flow of steam to the industrial or facility process

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**Back-pressure Turbine**

Combined Heat and Power is the main application For industrial plants: H.P steam flows through the turbine to a low pressure steam tank and then desuperheated (by small jet) to dry and saturated condition and then allowed to flow to the process where it gives off its latent heat

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Example 5.2 For the plant in example 5.1 if a back pressure turbine is used under the same inlet steam conditions, exit pressure is 1.4 bar and turbine efficiency is 75%. If the exhaust steam is used in the heating process and heating capacity needed is 70x103 MJ/h, calculate the steam mass flow rate, the power generated and the heat added in the boiler. Assume that the boiler efficiency is 84%, and the heating condensate is returned back to the boiler as saturated liquid.

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**h1=3000kJ/kg h2s=2333kJ/kg h2=2500kJ/kg h3=458.5kJ/kg**

H=70 x 106/3600== ms (h2 –h3) = ms= 9.52 kg/s Power(E)= ms(h1-h2)=4760 kW Neglecting the pump power Qadd= ms(h1-h3)/ηb =24195 kW

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Extraction Turbine The extraction turbine has opening(s) in its casing for extraction of a portion of the steam at some intermediate pressure before condensing the remaining steam

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Extraction Turbine At steam extraction locations there are usually steam flow control valves that control system flow rates and cost.

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**Mixed Pressure Turbine**

used in conjunction with a multistage turbine in cases where the exhaust is not sufficient to develop the power required

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**Mixed Pressure Turbine**

A heat accumulator in its simple form, is a cylindrical vessel where exhaust steam is led to submerged orifices, The steam condenses by direct contact with water. The water absorbs its latent heat, thus its sensible heat increases, and the pressure of steam above it rises. When the turbine consumes more steam than is coming, there is a slight reduction of pressure in the steam space which causes the water to evaporate, thus causing the pressure to fall gradually and the water looses sensible heat so on…….. The first operation is called storage period while the second is called the generation period. The steam consumption and the pressure of the steam in the accumulator is shown next

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**Mixed Pressure Turbine**

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**Mixed Pressure Turbine**

-Case (1) Power from H.P. turbine work only, Power1 = m x Δ hHP where m is the steam consumption in H.P. turbine work only -Case (2) Power from L.P. turbine work only Power2 = m1 x Δ hLP Where m1 is the steam consumption in L.P. turbine work only

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**Case (3) Power from both L.P & H.P. turbine work**

h3M(m+m1) =m h2+m1h3

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Example Steam enters a turbine at 1200 kPa and 350°C and it exits at 100 kPa, 150°C. The water mass flow rate through the turbine is 2 kg/s. Determine.

Example Steam enters a turbine at 1200 kPa and 350°C and it exits at 100 kPa, 150°C. The water mass flow rate through the turbine is 2 kg/s. Determine.

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