1 Linear Interpolation
Property tables such as steam tables are tabulated at discrete values of the specific properties. When referring
to the tables to find the specific properties, very often the property which we are interested in lies between the
tabulated values. In such cases, interpolation is required to obtain the correct value. The easiest method is to
use Linear Interpolation (Note: Interpolation is approximation)
Problem1 : Find values of P, h and v at
(a) T = 210
(b) T = 225.
T P h v
200 (T1) 100 (P1) 3490 0.2150
220 (T2) 140 (P2) 3541 0.2340
240 190 3615 0.2453
SOLUTION: Part a)
For this, we are required to find the values at T = 210 which lies in between 200 (Smaller Value (T1)) and 220
(Higher Value, (T2))
Before we begin, lets use the following convention while finding the interpolated value:
Smaller value - The First value in the table (200 for part (a) for T)
Higher value - The second value in the table (220 for part (a) for T)
Given Value - The value at which properties are to be found (210 for part (a) for T))
Given Property - Property that is known (T for this problem)
Required Property - Properties to be found (P, h and v for this problem)
Adopt the following procedure:
1. Find the difference (T2-T1) for the Given Property (Temperature) i.e. (Higher - Smaller)
2. Find the difference (T3-T1) for Given property (Temperature) i.e. (Given Value - Smaller)
3. Find the difference (P2-P1) for the Required Property (Pressure) i.e. (Higher - Smaller)
Then put the above values in the following equation
Preq =(T3−T1)
(T2−T1)∗(P2−P1) + P1
In other words,
Preq =(Given Value - Smaller, for T)
(Higher - Smaller, for T) ∗(Higher - Smaller, for P) + Smaller P
For part (b) for P,
T1= 220, T2= 240, T3= 225
P1= 140, P2= 190
ANSWER:
T P h v
200 100 3490 0.2150
210 120 3515.5 0.2245
220 140 3541 0.2340
225 152.5 3559.5 0.2368
240 190 3615 0.2453
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