Dimensional analysis of the equation u=prg

[dtfw]

Homework Statement

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).

Relevant Equations

u=LT^-1
p=MLT^-2/L2
r=ML^-3
g=LT^-2

u=prg
LT^-1 = MLT^-2 ML^-3 LT^-2 = (ML^-2T^-2)^a (ML^-3)^b (LT^-2)c
L^2

LT^-1 = M^a L^-2a T^-2a M^b L^-3b L^c T^-2c
=M^(a+b) L^(-2a+3b+c) T(-2a+2c)
M; a+b=0
L;-2a+3b+c=1 + T^-1

Either the formula is mixed up or my method because there seem to be more unknown quantities for me to work anything out, please advise

 

[haruspex]

ML^-2T^-2

Check that term.

 

[dtfw]

sorry that's a type error it MLT^-2/L^2 from p=f/a

rearranging is ML^-2T^-2

Is that correct or do you mean it is the wrong formula to insert ?

 

[haruspex]

sorry that's a type error it MLT^-2/L^2 from p=f/a

rearranging is ML^-2T^-2

Still wrong. Look at the exponent on L.

 

[dtfw]

so it is ML^-1T^-2 . thanks for your help however, this still honestly doesn't help where I am stuck at, it is still going to be as follows

M^(a+b) L^(-1a+3b+c) T^(-2a+2c)

M; a+b=0
L; -1a+3b+c=1 +T^-1and still there are too many unknowns to do anything with

 

[haruspex]

so it is ML^-1T^-2 . thanks for your help however, this still honestly doesn't help where I am stuck at, it is still going to be as follows

M^(a+b) L^(-1a+3b+c) T^(-2a+2c)

M; a+b=0
L; -1a+3b+c=1 +T^-1and still there are too many unknowns to do anything with

I can’t decipher the last equation above.
You have three unknowns, a, b and c, and one equation for each of M, L and T. You seem to have mangled the L and T equations into one meaningless expression.

 

[dtfw]

dimensional analysis is completely new to me and I am going through the steps correctly , its literally how it ends up with a T^_1 being on one side. The only thing I can determine is that the equation doesn't work at all. If not I am utterly clueless, basically for the equation to work , the L should equal zero but the T^-1 which is unknown still ends up onto the side of the L; value

 

[haruspex]

dimensional analysis is completely new to me and I am going through the steps correctly , its literally how it ends up with a T^_1 being on one side. The only thing I can determine is that the equation doesn't work at all. If not I am utterly clueless, basically for the equation to work , the L should equal zero but the T^-1 which is unknown still ends up onto the side of the L; value

The way it works is that you take your whole equation
$LT^{-1} = M^a L^{-a }T^{-2a }M^b L^{-3b }L^c T^{-2c}$
and split it out into three separate equations, one for each of M, L and T.
The principle is that no amount of time can be equivalent to an amount of mass, etc., so the exponents must balance for each physical attribute separately.

 

[dtfw]

so LT^-1=M^a+b L^-a-3b+c T-2a-2c

M; a+b=0
L; -a-3b+c=?
T; -2a-2c=?

Is that correct other than the = part?

 

[haruspex]

so LT^-1=M^a+b L^-a-3b+c T-2a-2c

M; a+b=0
L; -a-3b+c=?
T; -2a-2c=?

Is that correct other than the = part?

Right, and for the =?, you need to plug in the corresponding exponent on the left (i.e. in the velocity).

 

[dtfw]

M; a+b=0
L; -a-3b+c=1T^-1
T; -2a-2c= 1L

correct?

 

[haruspex]

M; a+b=0
L; -a-3b+c=1T^-1
T; -2a-2c= 1L

correct?

No. I don’t even know what "1T^-1" means.
The form of each equation should be
Sum of exponents of X on the left = sum of exponents of X on the right
where X is respectively M, L, T.
The resulting equations should not have M, L or T in them.

 

[dtfw]

right M; a+b=0 because there is no number value
L;-a-3b+c however from the left there is still LT^-1 therefore there is 1L T^-1
T; -2a-2c 1L left over

if I am doing this wrong please advise , the example I have used before was a lot simpler than this where the L=0 as well so it was a straight forward substitution , this is a lot more complicated and I honestly can't suss it

 

[haruspex]

from the left there is still LT^-1 therefore there is 1L T^-1

Untangle the L and T terms, like you did with M.
LT^-1 is an L¹ and a T^-1. The L¹ has to match the L^-a-3b+c on the right and the T^-1 has to match the T^-2a-2c on the right.

 

[Tim86]

Untangle the L and T terms, like you did with M.
LT^-1 is an L¹ and a T^-1. The L¹ has to match the L^-a-3b+c on the right and the T^-1 has to match the T^-2a-2c on the right.

Sorry to revive old thread.

So it would be;

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

Then you solve equation? For some reason this doesn't work out for me.

 

[haruspex]

Sorry to revive old thread.

So it would be;

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

Then you solve equation? For some reason this doesn't work out for me.

In what way does it not work out?
Please post your working.

 

[Tim86]

In what way does it not work out?
Please post your working.

Well I get;

A=0.5
B=-0.5
C=0

but that doesn't seem right.

 

[mjc123]

Why not? It is right.

 

[Jimmy23]

Hi,
Ive been trying to get past this last step i can't work out how people get A=0.5 B=-0.5 C=0 from
[L] -a-3b+c = 1
[T] -2a-2c = -1
[M] a+b=0

I can't find help anywhere. Can someone please explain how it's done

 

[Jimmy23]

From [M] b=-a
plug into [L] 2a+c=1
add to [T] -c=0 →c=0
plug into [L] →a=0.5
plug into [M] →b=-0.5

Hi,

thanks for your reply. I'm a little confused on the following.

1) am I right in saying a=-b and b=-a?

2) when you plug are both of these true 1=c-b-3b and 1=c-a-3a?

3) if I were to pick one of the above. Let's say, 1=c-a-3a. Why is it 1=c-2a and not 1=c-4a?

4) on the next step (add to [T] -c=0 →c=0). Does this come from 1+(-1)=(2a+c)+(-2a-2c) which then becomes 0=-c?

5)why is 0 equal to -c and c?

6)when you plug a=0.5 into 0=a+b you get 0=0.5+b. Then take 0.5 to both sides you get b=-0.5. Instead of taking 0.5, can you not take b to get -b=0.5?

Sorry for all the question lol.

Thanks,
James

 

[mjc123]

1. Yes.
2. No, neither is true. What do you get if you substitute a = -b?
3. Do 2 correctly and you'll see.
4. Yes.
5. If -c = 0, what is c?
6. Yes. It amounts to the same thing. (You do realize that b = -0.5 means -b = 0.5, don't you? You seem to show some confusion in handling negative numbers.)

 

[Jimmy23]

1. Yes.
2. No, neither is true. What do you get if you substitute a = -b?
3. Do 2 correctly and you'll see.
4. Yes.
5. If -c = 0, what is c?
6. Yes. It amounts to the same thing. (You do realize that b = -0.5 means -b = 0.5, don't you? You seem to show some confusion in handling negative numbers.)

Hi,

You say my first question correct (a=-b and b=-a are both correct) if you then plug these into [L] -a-3b+c = 1 separately. You end up with both 1=c-b-3b and 1=c-a-3a respectively. But then you answered question 2 as 'No, neither is true' what am I doing wrong?

if you were to substitute
a for -b. then you'd re-write [L] -a-3b+c = 1 as -b-3b+c=1, correct me if I'm wrong?

 

[Jimmy23]

You plugged them in incorrectly. What is your educational background?

Ahh I think I see what I've done. [L] -a-3b+c = 1
Substitute a for -b. Therefore, -(-b)-3b+c=1.
-(-b) is just b. So you get b-3b which is -2b. -2b+c=1. Is this correct?

This is all new to me. I'm a student. Thanks

 

[Jimmy23]

What is the highest level math class that you have taken?

No. b-3b=-2b

Yeah sorry that's what I meant

 

[Jimmy23]

If you were to substitute b with -a. Would 2a+c=1 be true?

 

[Jimmy23]

Yes.

Thanks for your help. Sometimes it's easier to talk it through with someone who knows what they're doing. Its ard to teach yourself something you don't know.

So
A=0.5 or 0=0.5-A or -0.5=-A
B=-0.5 or 0=-0.5-b or 0.5=-b
C=0 or -C=0

What level of education do you think I'm at? What is your highest level of maths?

 

[jedishrfu]

@Jimmy23 We ask about education level so that we can provide a level appropriate answer.

If you were doing an advanced calculus problem while still at a precalculus math level then we have some work to do to explain things clearly. However, if you're clearly well versed in Calculus then we can leave out some stuff in our explanations.

To ask us about our math level doesn't help you much.

Jedi

 

[Mark44]

So
A=0.5 or 0=0.5-A or -0.5=-A
B=-0.5 or 0=-0.5-b or 0.5=-b
C=0 or -C=0

None of these is really helpful. All you're doing is replacing one equation by two or three equivalent equations, starting from the solution and going around in circles.

Of course A = 0.5 is equivalent to 0 = 0.5 - A or to -0.5 = -A, and of course C = 0 is equivalent to -C = 0 as well. Writing the equations above does you no good if you're trying to find the solution of the system as shown in post #19.
[L] -a-3b+c = 1
[T] -2a-2c = -1
[M] a+b=0

One way to solve this system is to replace a in equations 1 and 2 to get a system of two equations involving b and c. Then you could solve one of those equations for, say, b, and substitute its value for b in the remaining equation.

What level of education do you think I'm at? What is your highest level of maths?

It's not up to us to guess what level you're at -- you tell us. If I had to guess, I wouldn't guess a very high grade.

 

[Jimmy23]

Thanks you for your help guys, I appreciate it.
I think the part that tripped me up was when you substitute one for the other I overlooked the fact the negative is still there and so a negative plus a negative is a positive.I was asking A=0.5 or 0=0.5-A or -0.5=-A more so for reassurance that I know what I'm doing, not to solve it lol. And same with the a=-b b=-a, which then leads to -(-b)-3b+c=1 or
-a-(3a)+c=1. Its more to get my head around the way its done.

My level of education isn't great. Hence why I'm trying to better myself. Thank you for being patient.

Out of curiosity, I was wondering what level you guys are at.

Thanks,
James

 

[haruspex]

a negative plus a negative is a positive.

I think you mean a negative times a negative is a positive.