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show that the units for all 3 terms in bernouliis equation are the same

 
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Jan9-10, 05:14 AM   #1
 

show that the units for all 3 terms in bernouliis equation are the same


1. The problem statement, all variables and given/known data

bernouliis equaton

P + 1/2 Rho v^2 + Rho g h = constant

find the units of the three terms on the left hand side & show they are the same and then the units of the constant on the right

3. The attempt at a solution

all i can think to do here is check the SI units for each term

units for pressure = Pa or NM^-2
units for density = kg/m^3
units for velocity =m/s
units for g = m/s
units for h = m

but i cant see how i combine them or how they all come out the same ?
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Jan9-10, 07:26 AM   #2
 
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Pressure P = NM^-2 = kg*M*s^-2*M^-2 = kg*M^-1*s^-2.
Similarly work out for the other two expression. You can see that all the tree terms have the same dimensions.
Jan9-10, 07:35 AM   #3
 
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Quote by victoriafello View Post
but i cant see how i combine them or how they all come out the same ?
Express each factor in terms of the basic units: m, kg, s. Then work out the units of each term in the equation by multiplying the units of each factor.
units for pressure = Pa or NM^-2
Hint: Express the pressure in terms of those three basic units.
units for g = m/s
Those units should be ms^-2.
Jan11-10, 06:37 AM   #4
 

show that the units for all 3 terms in bernouliis equation are the same


Ok so pressure becomes kgm^-1 s^-2, then the other terms expressed in
M , kg, s are

½ pv^2 is Kg m^-3 m/s^-1
Pgh is kg m^-3 m/2^2

I think I am still having a problem with the powers thou, I am not combining the units correctly
Jan11-10, 07:04 AM   #5
 
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Quote by victoriafello View Post
Ok so pressure becomes kgm^-1 s^-2,
Good.
then the other terms expressed in
M , kg, s are

½ pv^2 is Kg m^-3 m/s^-1
Pgh is kg m^-3 m/2^2
Show how you got these.

What are the units for density? For v?
Jan11-10, 11:03 AM   #6
 
density is mass / volume so its units are Kg m^-3
for velocity units are m/s^-1

so combined you get Kg m^-3 m/s^-1

and for the second part
density - Kg m^-3
units for g = m/s^-2
units for h - m

combines to give Kg m^-3 m/s^-2

i must be getting this wrong but i have re read my text book and i cant see where to correct it,
Jan11-10, 11:09 AM   #7
 
The dimensions of velocity are [tex][L]^1\times [T]^{-1}[/tex]

The dimensions of velocity squared are [tex][L]^2 \times [T]^{-2}[/tex]
Jan11-10, 11:27 AM   #8
 
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Quote by victoriafello View Post
density is mass / volume so its units are Kg m^-3
for velocity units are m/s^-1
The density units are correct, but velocity has units of m/s or m*s^-1.

so combined you get Kg m^-3 m/s^-1
Not quite. The units for ½ρv² would be:
[kg*m^-3]*[m*s^-1]² = [kg*m^-3]*[m*s^-1]*[m*s^-1]

See if you can simplify that by collecting all the powers of m and s. (For example, what's m^-3*m*m simplify to?)
Jan12-10, 05:07 AM   #9
 
Ok I think I see it now,

So m^-3*m*m simplifies to m^-1
And s^-1*s^-1 simplifies to s^-2

So the units for ½ρv² would be kg m^-1s^-2

Then for the second term
Units for density are kg*m^-3
Units for g are m*s^-2
Units for h are m

This is

[kg*m^-3]*[m*s^-2]*[m]

Collecting the m terms gives m^-1 so the units are

Kg m^-1s^-2

Then all I need is the units for the constant at the end, if the three terms all have the same units then the constant must be in the same units to the power 3, like if you have a volume then m*m*m the result is m^3 ?

if so how do I do this for kg m^-1s^-2 ?
Jan12-10, 05:12 AM   #10
 
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Quote by victoriafello View Post
Ok I think I see it now,

So m^-3*m*m simplifies to m^-1
And s^-1*s^-1 simplifies to s^-2

So the units for ½ρv² would be kg m^-1s^-2

Then for the second term
Units for density are kg*m^-3
Units for g are m*s^-2
Units for h are m

This is

[kg*m^-3]*[m*s^-2]*[m]

Collecting the m terms gives m^-1 so the units are

Kg m^-1s^-2
Good.

Then all I need is the units for the constant at the end, if the three terms all have the same units then the constant must be in the same units to the power 3, like if you have a volume then m*m*m the result is m^3 ?
Don't confuse three terms (which add) with three factors (which multiply). All the terms have the same units, including the constant. So you're done.
Jan13-10, 06:38 AM   #11
 
thanks so much for your help !!
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