1. ## Linear interpolation

Hello,

I need to find the pressure of steam at $98C^{o}$

however my steam table does not give a value for 98C, however does give a value for 95C and 100C,

I am told in order to find the pressure of steam at 98C I have interpolate from the tables,

I am not sure how to do this?

at 95C the pressure is 0.08453MPa
and at 100C the pressure is 0.1013MPa

and since 98C is not half way between, how would I interpolate?

is it $\frac{3}{5} ( 0.08453 + 0.1013) ?$ Multiplying by 3/5 as its 3 units away from 95 and after that another 5 units to 100C

Is this correct?

Thanks.

2. Your expression is incorrect. Find the equation of the line of pressure as a function of temperature. Then just plug in 98. What do you get?

3. Originally Posted by Ackbeet
Your expression is incorrect. Find the equation of the line of pressure as a function of temperature. Then just plug in 98. What do you get?

Okay, but I am not sure how to do that? I dont have any variables to work with?

Thank you.

4. Well, use P for pressure, and T for temperature. Then postulate that P = m T + b. Plug in two data points where you know both P and T in order to get two equations. Use those two equations to solve for m and b. Then plug in T = 98. How does that sound?

5. Thanks I get it, I got 0.093182, which seems about right, as it should be higher than the pressure at 95 but lower than the pressure for 100C.

6. Great. And the value is closer to the 100C value than the 95C value, right?

7. (3/5) of the way from a to b is NOT (3/5)(a+ b). The distance from a to b is b- a and 3/5 of that is (3/5)(a- b). Now add that to b: b+ (3/5)(a- b)= (3/5)a+ (2/5)b. (notice that 2+ 3= 5)

8. $\begin{array}{ccc}\mbox{Temp(T)} & \mbox{Pressure(P)} & \Delta \\ 95 &0.08453 & \mbox{ } \\ \mbox{ } & \mbox{ } & 0.01677 \\ 100 & 0.1013 & \mbox{ } \end{array}$

$P(98) \approx 0.08453 + (3/5)0.01677 = 0.09460$

9. Both ways give a different answer, which one is more accurate? And can you please explain why you add the value of 'b' ?

10. The two methods are equivalent. If, instead of the numerical values, you call the pressure at 95, $a$, and the pressure at 100, $b$, then, using my notation, $\Delta =b-a$.
Then,

$P(98)\approx a + (3/5)(b-a) = (2/5)a+(3/5)b,$

(and now substituting the numbers),

$=(2/5)0.08453+(3/5)0.1013=0.094592.$

11. Hello, Tweety!

$\text{At }95^oC\text{ the pressure is 0.08453 MPa}$
$\text{and at }100^oC\text{ the pressure is 0.1013 MPa.}$

$\text{Use linear interpolation to find the pressure at }98^oC.$

We are given two points: . $(95,\,0.08453)\,\text{ and }\,(100,\,0.1013)$

Find the equation of the line through those two points.

. . The slope is: . $m \:=\:\frac{0.1013 - 0.08453}{100-95} \:=\:\frac{0.01677}{5} \:=\:0.003354$

. . Then: . $y - 0.08453 \:=\:0.003354(x - 95)$

. . The equation of the line is: . $y \:=\:0.003354x - 0.2341$

If $x = 95\text{, then: }\:y \:=\:0.003354(95) - 0.2341 \:=\:0.094592$