What should I do if didn't find the given pressure?

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Discussion Overview

The discussion revolves around the challenge of finding enthalpy values in superheated steam tables when the given pressure does not match any entries in the tables. Participants explore methods for obtaining the required values, including linear interpolation and potential alternative approaches.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant expresses difficulty in finding the enthalpy value for a pressure of 1.1 MPa in the superheated tables.
  • Another participant suggests using linear interpolation as a standard method for obtaining the value.
  • Some participants inquire about faster methods that may have been taught in class, indicating uncertainty about the interpolation process.
  • There are repeated requests for clarification on how to perform interpolation, with specific values from the tables being discussed.
  • Participants share values from their tables for pressures of 1.0 MPa and 1.5 MPa, which are necessary for the interpolation process.
  • One participant expresses a lack of familiarity with interpolation, while others emphasize its importance and standard use in engineering contexts.
  • There is a discussion about the need to understand the reasoning behind interpolation rather than just applying it mechanically.

Areas of Agreement / Disagreement

Participants do not reach a consensus on alternative methods to interpolation, and there is ongoing uncertainty about the best approach. However, there is agreement on the necessity of interpolation when specific pressure values are not available in the tables.

Contextual Notes

Limitations include the absence of specific tables for the pressure of 1.1 MPa and the reliance on values from surrounding pressures (1.0 MPa and 1.5 MPa) for interpolation. Some participants express confusion about the interpolation process and its application.

Amr719
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I want to know what should I do if I don't find the given pressure in a problem in the super heated tables ?
I have P=1.1MPa and I want to get the " h " but I don't find this pressure in the tables
 
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Linearly interpolate.

Chet
 
Isn't there another way ? My teacher told me another way to do it that takes less time in the exam , but I can't remeber it
 
Amr719 said:
Isn't there another way ? My teacher told me another way to do it that takes less time in the exam , but I can't remeber it
How long does it take to linearly interpolate?

Chet
 
I don't know exactly but if there is a faster way , I should go with it
 
Amr719 said:
I don't know exactly but if there is a faster way , I should go with it
I can't think of a faster way. I bet I could do the linear interpolation in less than a minute. Interpolation is the standard way of working with tables.

Chet
 
Could you tell me how to do the interpolation for this : I have P=1.1MPa T=250°C and I want to get "h"
Sorry I am not familiar with this type of solution so I need your help :)
 
Amr719 said:
Could you tell me how to do the interpolation for this : I have P=1.1MPa T=250°C and I want to get "h"
Sorry I am not familiar with this type of solution so I need your help :)
At what values of the pressure does the table give values of h? (on either side of 1.1 MPa) What are those values of h at 250 C?

Chet
 
I can't understand . what I know that it is at the superheated vapour Tables
 
  • #10
Amr719 said:
I can't understand . what I know that it is at the superheated vapour Tables
Write down some of the numbers from your table. For example in the steam tables I have,

P = 10 bars, T = 240 C, h = 2920
P = 10 bars, T = 280 C, h = 3008
P = 15 bars, T = 240 C, h = 2899
P = 15 bars, T = 280 C, h = 2993

Your turn.

Chet
 
  • #11
That's the question. Which table ? I don't have a table for the pressure 1.1MPa . I have for 1.0 MPa and for 1.5 MPa
 
  • #12
Amr719 said:
That's the question. Which table ? I don't have a table for the pressure 1.1MPa . I have for 1.0 MPa and for 1.5 MPa
I'm acutely aware of that. What does your table give for h at 1.0 MPa and 1.5 MPa? After you provide those values, I will show you how to get the value at 1.1 MPa.

Chet
 
  • #13
Amr719, it doesn't seem like you know what it means to "interpolate". Did you look that up after Chestermiller said it is what is needed?
 
  • #14
russ_watters said:
Amr719, it doesn't seem like you know what it means to "interpolate". Did you look that up after Chestermiller said it is what is needed?
Shocking, huh? Don't they teach interpolation in high school algebra any more?

Chet
 
  • #15
Chestermiller said:
Shocking, huh? Don't they teach interpolation in high school algebra any more?

Chet
The problem is, "interpolate" is too big to fit on a calculator key. :oops:
 
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  • #16
russ_watters said:
Amr719, it doesn't seem like you know what it means to "interpolate". Did you look that up after Chestermiller said it is what is needed?
No I know it but like I said before I'm not familiar with it because my teacher doesn't use this method. That's all
 
  • #17
Chestermiller said:
I'm acutely aware of that. What does your table give for h at 1.0 MPa and 1.5 MPa? After you provide those values, I will show you how to get the value at 1.1 MPa.

Chet
Table for 1.0MPa : Table for 1.5MPa:
T=240,h=2920.4 T=240,h=2899.3
T=280,h=3008.2 T=280,h=2991.7
 
  • #18
Amr719 said:
Table for 1.0MPa : Table for 1.5MPa:
T=240,h=2920.4 T=240,h=2899.3
T=280,h=3008.2 T=280,h=2991.7
And, what, you never noticed that these are the same values I gave you from my table in post #10?

I am going to show you how to get the value of h at 250 C and 1.0 MPa. Then you are going to show me how you apply the same interpolation approach to get the value of h at 250 C and 1.5 MPa.

$$h(250 C,1 MPa) = 2920.4 + \frac{(250 - 240)}{(280-240)}(3008.2-2920.4)=2942.4$$

Now I want you to apply this same algorithm to get the value of h at 250 C and 1.5 MPa. Do you think you can do that?

Please don't tell me at this point that you need to have the value of h at 1.1 MPa. I know that. There will be another step after you complete this step.

Chet
 
  • #19
Chestermiller said:
And, what, you never noticed that these are the same values I gave you from my table in post #10?

I am going to show you how to get the value of h at 250 C and 1.0 MPa. Then you are going to show me how you apply the same interpolation approach to get the value of h at 250 C and 1.5 MPa.

$$h(250 C,1 MPa) = 2920.4 + \frac{(250 - 240)}{(280-240)}(3008.2-2920.4)=2942.4$$

Now I want you to apply this same algorithm to get the value of h at 250 C and 1.5 MPa. Do you think you can do that?

Please don't tell me at this point that you need to have the value of h at 1.1 MPa. I know that. There will be another step after you complete this step.

Chet
h(250C,1.5MPa)=2899.3+((250-240)/(280-240))*(2991.7-2899.3)=2922.4
 
  • #20
Looks good. Now, do you think you can take these results for h at 250 C for 1.0 MPa and 1.5 MPa and, by using this same kind of interpolation algorithm for pressure, find the value of h at 250 C and 1.1 MPa?

Chet
 
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  • #21
Amr719 said:
No I know it but like I said before I'm not familiar with it because my teacher doesn't use this method. That's all
Sometimes teachers show you an easier exmple, then give you a slightly harder example than what they showed you. An important part of being an engineer is being able to apply separate pieces of your knowledge to a problem you haven't seen before in order to solve it. You must learn to think beyond just learning the exact thing the teacher taught.

So -- now that you know interpolation works here, can you tell me why?
 
  • #22
russ_watters said:
Sometimes teachers show you an easier exmple, then give you a slightly harder example than what they showed you. An important part of being an engineer is being able to apply separate pieces of your knowledge to a problem you haven't seen before in order to solve it. You must learn to think beyond just learning the exact thing the teacher taught.

So -- now that you know interpolation works here, can you tell me why?

Why what ? Why it works here ?
 
  • #23
Amr719 said:
Why what ? Why it works here ?
Russ's question meant do now you understand why and how the linear interpolation formula that you used gives a good approximation to the enthalpy at 250 and 1.1 MPa, or are you just satisfied with applying it blindly and moving on? Be aware that understanding how and why it works is more important than getting the answer to a specific homework question.

Chet
 
  • #24
now I know if I didn't find a table for a specific pressure I should interpolate .Do I understand correctly ?
 
  • #25
Amr719 said:
now I know if I didn't find a table for a specific pressure I should interpolate .Do I understand correctly ?
Yes
 
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