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nata0808 [166]
3 years ago
7

Superheated water vapor at a pressure of 20 MPa, a temperature of 500oC, and a flow rate of 10 kg/s is to be brought to a satura

ted vapor state at 10 MPa in an open feedwater heater. This process is accomplished by mixing this stream with a stream of liquid water at 20oC and 10 MPa. What flow rate is needed for the liquid stream?
Engineering
1 answer:
katrin2010 [14]3 years ago
7 0

Answer:

1.96 kg/s.

Explanation:

So, we are given the following data or parameters or information which we are going to use in solving this question effectively and these data are;

=> Superheated water vapor at a pressure = 20 MPa,

=> temperature = 500°C,

=> " flow rate of 10 kg/s is to be brought to a saturated vapor state at 10 MPa in an open feedwater heater."

=> "mixing this stream with a stream of liquid water at 20°C and 10 MPa."

K1 = 3241.18, k2 = 93.28 and 2725.47.

Therefore, m1 + m2= m3.

10(3241.18) + m2 (93.28) = (10 + m3) 2725.47.

=> 1.96 kg/s.

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A steam power plant operating on a simple ideal Rankine cycle maintains the boiler at 6000 kPa, the turbine inlet at 600 °C, and
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Answer:

ηa=0.349

ηb=0.345

Explanation:

The enthalpy and entropy at state 3 are determined from the given pressure and temperature with data from table:

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 q_{4} =\frac{s_{4}-s_{liq50}  }{s_{evap,50} } \\=\frac{7.1693-1.0912}{6.5019}=0.935

The enthalpy at state 4 then is:  

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h_{1}=340.54kJ/kg\\ \alpha _{1}=0.00103m^3/kg

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=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.349

Part B  

In the case when the water is at a lower temperature than the saturation temperature at the condenser pressure we look into table and see the water is in a compressed liquid state. Then we take the enthalpy and specific volume for that temperature with data from  and the saturated liquid values:  

h_{1}=293.7kJ/kg\\ \alpha _{1}=0.001023m^3/kg

The enthalpy at state 2 is then determined from an energy balance on the pump:  

h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1}  )

    =299.79 kJ/kg  

The thermal efficiency in this case then is:  

=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.345

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