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

[Amr719]

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

 

[Chestermiller]

Linearly interpolate.

Chet

 

[Amr719]

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

 

[Chestermiller]

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

 

[Amr719]

I don't know exactly but if there is a faster way , I should go with it

 

[Chestermiller]

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

 

[Amr719]

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 :)

 

[Chestermiller]

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

 

[Amr719]

I can't understand . what I know that it is at the superheated vapour Tables

 

[Chestermiller]

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

 

[Amr719]

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

 

[Chestermiller]

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

 

[russ_watters]

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?

 

[Chestermiller]

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

 

[SteamKing]

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.

 

[Amr719]

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

 

[Amr719]

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

 

[Chestermiller]

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

 

[Amr719]

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

 

[Chestermiller]

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

 

[russ_watters]

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?

 

[Amr719]

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 ?

 

[Chestermiller]

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

 

[Amr719]

now I know if I didn't find a table for a specific pressure I should interpolate .Do I understand correctly ?

 

[Chestermiller]

now I know if I didn't find a table for a specific pressure I should interpolate .Do I understand correctly ?

Yes