Saturated water vapor undergoes a throttling process from 1bar to a 0.35bar. What is the change in temperature for this process?
1 answer:
Answer:
-25.63°C.
Explanation:
We know that throttling is a constant enthalpy process
From steal table
We know that if we know only one property in side the dome then we will find the other property by using steam property table.
Temperature at saturation pressure 1 bar is 99.63°C and Temperature at saturation pressure 0.35 bar is about 74°C .
So from above we can say that change in temperature is -25.63°C.
But there is no any option for that .
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Answer:
Explanation:
The adiabatic throttling process is modelled after the First Law of Thermodynamics:
Properties of water at inlet and outlet are obtained from steam tables:
State 1 - Inlet (Liquid-Vapor Mixture)
State 2 - Outlet (Superheated Vapor)
The change of entropy of the steam is derived of the Second Law of Thermodynamics:
Answer:
The inside temperature, is approximately 248 °C.
Explanation:
The parameters given are;
Temperature of the air = 20°C
Carbon steel surface temperature 250°C
Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²
Convection heat transfer coefficient = 25 W/(m²·K)
Heat lost by radiation = 300 W
Assumption,
Air temperature = 20 °C
Hot plate temperature = 250 °C
Thermal conductivity K = 65.2 W/(m·K)
Steady state heat transfer process
One dimensional heat conduction
We have;
Newton's law of cooling;
q = h×A×( -
) + Heat loss by radiation
= 25×0.325×(250 - 20) + 300
= 2456.25 W
The rate of energy transfer per second is given by the following relation;
Thermal conductivity K = 65.2 W/(m·K)
Therefore;
The inside temperature, = 247.99 °C ≈ 248 °C.