Answer:hmmmmmmmmmm give an hour.
Explanation:
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Answer:
The level of the service is loss and the density is 34.2248 pc/mi/ln
Explanation:
the solution is attached in the Word file
Answer:
The temperature of the first exit (feed to water heater) is at 330.15ºC. The second exit (exit of the turbine) is at 141ºC. The turbine Power output (if efficiency is %100) is 3165.46 KW
Explanation:
If we are talking of a steam turbine, the work done by the steam is done in an adiabatic process. To determine the temperature of the 2 exits, we have to find at which temperature of the steam with 1000KPa and 200KPa we have the same entropy of the steam entrance.
In this case for steam at 3000 kPa, 500°C, s= 7.2345Kj/kg K. i=3456.18 KJ/Kg
For steam at 1000 kPa and s= 7.2345Kj/kg K → T= 330.15ºC i=3116.48KJ/Kg
For steam at 200 kPa and s= 7.2345Kj/kg K → T= 141ºC i=2749.74KJ/Kg
For the power output, we have to multiply the steam flow with the enthalpic jump.
The addition of the 2 jumps is the total power output.
Answer:
The pressure in the tank is 70.183 k Pa
The volume of the tank is 4.73 m³
Explanation:
Volume of the liquid phase
v(f) = m(f).v(f)
= 8 kg . 0.001036 m³/kg
= 0.008288 m³
Volume of the vapor phase
v(g) = m(g).v(g)
=2 kg . 2.3593 m³/kg
= 4.7186 m³
Volume of the tank = Volume of the liquid phase + Volume of the vapor phase
Volume of the tank
= 0.008288 m³ + 4.7186 m³ = 4.73 m³
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