Results 1 to 11 of 11

Math Help - Linear interpolation

  1. #1
    Super Member
    Joined
    Sep 2008
    Posts
    607

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Sep 2008
    Posts
    607
    Quote Originally Posted by Ackbeet View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Sep 2008
    Posts
    607
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Great. And the value is closer to the 100C value than the 95C value, right?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,373
    Thanks
    1315
    (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)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Jun 2009
    Posts
    660
    Thanks
    133
    \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
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Sep 2008
    Posts
    607
    Both ways give a different answer, which one is more accurate? And can you please explain why you add the value of 'b' ?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Jun 2009
    Posts
    660
    Thanks
    133
    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.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,682
    Thanks
    614
    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

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. linear interpolation
    Posted in the Statistics Forum
    Replies: 0
    Last Post: December 12th 2010, 10:51 AM
  2. Linear interpolation
    Posted in the Advanced Math Topics Forum
    Replies: 1
    Last Post: November 9th 2009, 05:26 AM
  3. Linear Interpolation
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: February 27th 2009, 01:14 PM
  4. linear interpolation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 19th 2009, 07:57 AM
  5. Linear interpolation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 13th 2008, 09:49 AM

Search Tags


/mathhelpforum @mathhelpforum