I don\'t understand the process of double interpolation when solving for propert

Question

I don't understand the process of double interpolation when solving for properties. For this question I am given superheated water at a pressure P=150kPa and temperature T=280C. I need to find the specific volume (v) and internal energy (u) at this pressure and temperature. -------------------------------- At 100kPa: At 250C: v= 2.4062, u= 2733.9 At 300C: v= 2.6389, u=2810.7 -------------------------------- At 200kPa: At 250C: v= 1.9890, u=2731.4 At 300C: v=1.31623, u=2808.8 --------------------------------- The final answer is at P=150kPa and T=280C, v=1.9076 and u=2779. I do not know how these values were obtained.

Explanation / Answer

see 280 is in mid of 250 C and 350 C, so u can also use mean rather than interpolation. you will get a fairly close answer.

first we find at 100 Kpa :

u at 280 C = u(250) + u(300) /2 = (2733.9 + 2810.7)/2 = 2772.3

now at 200 Kpa:

u at 280 C = u(250) + u(300) /2 = ( 2731.4 + 2808.8)/2 = 2770.1 kJ/Kg

now 150 Kpa is in mid os 100 Kpa and 200 Kpa so instead of interpolation we find mean of uat 280 C for the Pressures 100 Kpa and 200 Kpa

so u at 150 Kpa = ( u at 100 Kpa + u at 200 Kpa)/2

u at 150 Kpa, 280 C =( 2772.3 + 2770.1) /2 = 2771.2 Kj/Kg

here the value which we have got 2771.2 is a bit far away from your answer of 2779 Kj/Kg. But you can use this value also with a good accuracy. The variation in result due to this will be very less. I have been doing the same for a very long time without any difficulty.

for (v) do the same steps

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