9/12/23, 11:54 AM
Interpolation Thermodynamics: Finding Temperature
Notes on Linear Interpolation in Thermodynamics
Objective:
To find the desired thermodynamic property (temperature or internal energy) at a given specific volume V (r)
using linear interpolation from a property table.
Formula:
The general linear interpolation formula is given by:
(x−x1 )(y2 −y1 )
x 2 −x 1
y = y1 +
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Here:
y : the property value we want to find (temperature or internal energy).
x: the specific volume V (r) given in your problem.
x1 and x2 : the specific volume values surrounding x in your table.
y1 and y2 : the property values corresponding to x1 and x2 , respectively.
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Procedure:
Identify Values: Locate and note down the values of x1 , x2 , y1 , and y2 from the nearest entries in your table
to the specific volume x you have.
Substitute Values: Substitute the values you have into the linear interpolation formula.
Calculate: Solve the equation step by step to find the value of y .
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Examples:
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9/12/23, 11:54 AM
Interpolation Thermodynamics: Finding Temperature
1. For Temperature:
Given a specific volume of 0.25 m³/kg, and a table with values:
V (m3 /kg)
0.2
0.3
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T (°C)
150
200
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The interpolation gives:
T = 150°C +
(0.25−0.2)(200−150)
0.3−0.2
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= 175°C
2. For Internal Energy:
Given a specific volume of 0.25 m³/kg, and a table with values:
V (m3 /kg)
0.2
0.3
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U (kJ/kg)
1500
2000
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The interpolation gives:
U = 1500 kJ/kg +
(0.25−0.2)(2000−1500)
0.3−0.2
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= 1750 kJ/kg
Conclusion:
Linear interpolation helps in estimating the property values (like temperature and internal energy) at specific
volumes not directly listed in the property tables.
Note:
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9/12/23, 11:54 AM
Interpolation Thermodynamics: Finding Temperature
Ensure x is between x1 and x2 .
Double-check the values from your table to avoid errors.
Perform calculations carefully to ensure accuracy.
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Best wishes for your exam!
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