Interpolation Tutorial Example: The internal energy, 2950 kJ/kg, is between 2875.2 and 2957.3 kJ/kg. Therefore the temperature is between 350 and 400 °C. To determine exact temperature we need to INTERPOLATE. INTERPOLATION Firstly find the ratio y, by dividing distance AB by distance AC: (all of these values will be known) y= AB AC = ab ac = B−A C−A Rearranging the previous formula gives the unknown value as, b = a + y(c − a) y= 2950 − 2875.2 2957.3 – 2875.2 = 0.91 Therefore the temperature is, T = 350 + (0.91)(50) = 395.5 °C x y C B A a b c
Interpolation tutorial using thermodynamics steam tables
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Interpolation Tutorial Example:
Theinternalenergy,2950kJ/kg,isbetween2875.2and2957.3kJ/kg.Thereforethe
temperatureisbetween350and400C.Todetermineexacttemperatureweneedto
INTERPOLATE. INTERPOLATION Firstly find the ratio y, by dividing
distance AB bydistance AC: (all of these values will be known) y
=ABAC=abac =BACA Rearranging the previous formula gives theunknown
value as, b = a +y(c a) y =2950 2875.22957.3 2875.2 = 0.91
Therefore the temperature is,T = 350 + (0.91)(50) = 395.5 C x y C
BA a b c
