entc 370prof. alvarado1 thermodynamics lab properties of pure substances entc - 370
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ENTC 370 PROF. ALVARADO 1
THERMODYNAMICS LABProperties of Pure Substances
ENTC - 370
Pure Substances
• It can exist in three different phases- solid, liquid, gas
• Under certain conditions, two phases can co-exist
• Ex: liquid and gas co-existing in the evaporator and condenser of a refrigerator
• An Extensive property can be converted to an Intensive property by specifying the property per unit mass, such as specific properties.
Property of a substance is independent of the path travelled
Pure Substances• A state of a substances can be specified by
two independent properties
• In single phase, pressure and temperature are the two common independent properties
• In two phase, pressure and temperature are no longer independent. In this case we have to use another independent property (mostly “quality”) along with pressure or temperature to determine the state of the substance.
Quality• The quality of a fluid is the percentage of
mass that is vapor– Saturated vapor has a "quality" of 100%– Saturated liquid has a "quality" of 0%
• X = 1, Saturated Vapor
X = 0, Saturated Liquid
0 < X < 1, Mixture
total
g
m
mx
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Example
• Suppose that we want to complete the following table of properties for R-134a (‘Refrigerant found in air conditioning systems’):
T [°C] P [kPa] V [m3/kg]Phase
Description
-8 320
180Saturated vapor
(x = 1.0)
Note: Remember that in order to specify a thermodynamic state, you just need two independent properties.
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Thermodynamics Properties in EES
• EES provides thermophysical property data on a wide variety of fluids that are found in engineering applications
• To access this option, in the menu Options, select Function info and The following window will appear
• Select the fluid and the property of interest, and paste it to the equations window
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Properties
• Remember EES uses the following representations for the most common properties:– T= temperature– P= Pressure– x= Quality– u= Internal Energy– h= Enthalpy– v= Specific Volume– s= Entropy
• Use EES Property Calculator instead of the EES function to fill out the previous table
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Example• In order to find the missing properties of the first row, we just input
the equations, press F2 and get the solutions
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Example
• The second line of the table gives us the pressure and tell us that the refrigerant is in the form of saturated vapor (quality=1).
• If we paste the temperature function in the equations window; we will see that the default properties used to calculate the temperature are Pressure and enthalpy– TEMPERATURE(R134a,h=h1,P=P1)
• It is possible to change the enthalpy for quality and obtain the temperature
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Example
• The completed table looks like this:
T [°C] P [kPa] V [m3/kg]Phase
Description
-8 320 7.569 * 10-4 Compressed Liquid
-12.73 180 0.1104Saturated
vapor
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Problem 1
• A piston-cylinder device contains 0.3 m3 of liquid water and 0.7 m3 of water vapor in equilibrium at 600 kPa. Heat is transferred at constant pressure until the temperature reaches 400 °C.
– What is the initial temperature of the water?
– Determine the total mass of the water.
– Calculate the final volume
– Let the final temperature vary from 180 to 1080 °C. Determine the impact on final volume in the tank. Plot the final volume vs. final temperature. Discuss results.
– Let the pressure vary from 150 to 1050 kPa. Determine the impact on the final volume of water. Plot the total volume of water vs. Pressure. Discuss results.
H2O
P=600kPa
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Lab Report – Problem 1 only
• Discussion: State the problem with your own words
• Equation: Summary of the equations used in the experiment with explanation.
• Findings: Summary of the experimental data, you could use tables, figures, graphs.
• Conclusions: Interpretation of the findings.