Hi, Im a mature student studying a distance learning HNC so I don't have anybody to turn to help with a problem very easily. I don't expect anybody to do the work for me, but I'm wondering if anybody can help with this question?

You are testing a prototype loudspeaker for your Audio division, which may possibly be used in smoke-filled environments when nuclear reactor emergencies occur. A senior colleague has asked you to use dimensional analysis to predict how the speed of sound (u) in a gas may be influenced by the gas pressure (p), gas density (r) and acceleration due to gravity (g).

I have:

pressure P = ML^-1T^-2
density r = ML^-3
acceleration g = LT^-2
Speed v = LT^-1

LT^-1 = (ML^-1T^-2)^A (ML^-3)^B (LT^-2)^C

LT^-1 = M^A+B L^C-A-3B T^-2A-2C

M 0=A+B
L 1=C-A-3B
T -1=-2A-2C

Ive gotten this far and now I'm stumped on whether I am right or wrong, where I'm wrong, and how to solve for ABC.

any help greatly appreciated.

Lloyd
Original post by lwood10
Hi, Im a mature student studying a distance learning HNC so I don't have anybody to turn to help with a problem very easily. I don't expect anybody to do the work for me, but I'm wondering if anybody can help with this question?

You are testing a prototype loudspeaker for your Audio division, which may possibly be used in smoke-filled environments when nuclear reactor emergencies occur. A senior colleague has asked you to use dimensional analysis to predict how the speed of sound (u) in a gas may be influenced by the gas pressure (p), gas density (r) and acceleration due to gravity (g).

I have:

pressure P = ML^-1T^-2
density r = ML^-3
acceleration g = LT^-2
Speed v = LT^-1

LT^-1 = (ML^-1T^-2)^A (ML^-3)^B (LT^-2)^C

LT^-1 = M^A+B L^C-A-3B T^-2A-2C

M 0=A+B
L 1=C-A-3B
T -1=-2A-2C

Ive gotten this far and now I'm stumped on whether I am right or wrong, where I'm wrong, and how to solve for ABC.

any help greatly appreciated.

Lloyd

Correct. You just need to solve the simultaneous system of three equations in three variables.

E.g. if you rearrange the first one for A=-B then you can substitute this into the the second and third eqns, hence be leftover with two equations in two variables B,C to solve (which should be more familiar territory for you)
Original post by RDKGames
Correct. You just need to solve the simultaneous system of three equations in three variables.

E.g. if you rearrange the first one for A=-B then you can substitute this into the the second and third eqns, hence be leftover with two equations in two variables B,C to solve (which should be more familiar territory for you)

Thank you very much. Sorry if I am digging too much, but if I dont get this correct on the first submission I can only get a Pass and not a merit or distinction.

So would this be correct?

1 = B-3B+c
1 = C-1B Call A

-1 = -2(-B)-2C
-1 = 2B-2C
-2B = -2C Call B

B into A
1 = C+1-2C
C = 0

C = 0 INTO A
1 = 0-2B
B = -1/2

0 = A-1/2
A = 1/2

Back to original u = p^1/2 r^-1/2 g^0
u = SQRp/r

Thanks,
Lloyd

Also I am new here, if correct am I to delete this thread to deter copying?
Original post by lwood10
Thank you very much. Sorry if I am digging too much, but if I dont get this correct on the first submission I can only get a Pass and not a merit or distinction.

So would this be correct?

1 = B-3B+c
1 = C-1B Call A

-1 = -2(-B)-2C
-1 = 2B-2C
-2B = -2C Call B

B into A
1 = C+1-2C
C = 0

C = 0 INTO A
1 = 0-2B
B = -1/2

0 = A-1/2
A = 1/2

Back to original u = p^1/2 r^-1/2 g^0
u = SQRp/r

Thanks,
Lloyd

Also I am new here, if correct am I to delete this thread to deter copying?

Dont confuse yourself (and others) by naming equations A and B when you have variables in the problem as A and B already.

Anyway, your equation "A" is incorrect; B-3B is not equal to -B.

Your equation "B" is incorrect; where does the -1 go?

That said, when it comes to you substituting things in, you get this correct. So I assume you just made typos rather than algebraic errors.

No need to delete anything, though you can if you want.
(edited 3 years ago)
Original post by RDKGames
Dont confuse yourself (and others) by naming equations A and B when you have variables in the problem as A and B already.

Anyway, your equation "A" is incorrect; B-3B is not equal to -B.

Your equation "B" is incorrect; where does the -1 go?

That said, when it comes to you substituting things in, you get this correct. So I assume you just made typos rather than algebraic errors.

No need to delete anything, though you can if you want.

Yes sorry I just checked my notepad and "A" was 1=C-2B and "B" was -2B=1-2C.
I had never considered calling them anything else, thank you for the tip.

Lloyd
Hi, Is anybody able to show how you get the values of A,B & C from the following:-

[L] -a-3b+c = 1
[T] -2a-2c = -1
[M] a+b=0

I don't understand how people are getting A=0.5 B=-0.5 C=0 and I can't find any advice or help to get me to the right answer. Been trying to figure it out for some time now.

Any help is appreciated, thanks.
(edited 2 years ago)
Original post by James2233
Hi, Is anybody able to show how you get the values of A,B & C from the following:-

[L] -a-3b+c = 1
[T] -2a-2c = -1
[M] a+b=0

I don't understand how people are getting A=0.5 B=-0.5 C=0 and I can't find any advice or help to get me to the right answer. Been trying to figure it out for some time now.

Any help is appreciated, thanks.

This thread is a year old. However, you can rearrange your third equation to give b = -a, then substitute that result into your first equation. Your first two equations will both be in terms of (only) a and c, which enables you to solve them simultaneously.