Different units in Bernoulli's equation's terms?

AI Thread Summary
The discussion focuses on the unit consistency in Bernoulli's equation, specifically addressing the mismatch between pressure and kinetic energy terms. Initially, there was confusion regarding the units of pressure (force per unit area) and the kinetic energy term, which appeared to differ. After clarification, it was established that in the SI system, pressure is expressed in units of slug/(ft sec^2), aligning with the kinetic energy term's units of slug/(ft sec^2) when properly calculated. This resolution underscores the importance of using correct units in fluid dynamics equations. The conversation highlights the necessity of understanding fundamental physics concepts to avoid unit discrepancies.
sinueus
Messages
4
Reaction score
0
When I check units in the terms in Bernoulli's equation, they do not match:
Given:
p1 + 1/2 (row) v1^2 = p2 + 1/2 (row) v2^2

On the left hand side of the equation, the first term, pressure must have units of force/unit area; say lb/ft^2. The units in the second term are the product of

row, the density in, say, lb/ft^3
v1^2, the square of speed, say, ft^2/sec^2,

or [lb/ft^3][ft^2/sec^2] = [ lb/(ft sec^2) ] and this is different from the [lb/ft^2] of the first term. Why?

Thanks
 
Physics news on Phys.org
A lb isn't a unit of force, it's a units of mass.
 
I thought so!

Thanks,
I haven't done physics in a log time and I am rusty on the basics. I thoght I had the problem that you mention but was not sure. I'll try again using proper force units and I'll come back.
 
I am back!

After reviewing my units, I found that in the SI system, the unit of force is the pound (lb), the unit of mass is the slug and that of acceleration is ft/s^2. I remember that 1lb = 1 slug ft/sec^2. Armed with this knowledge, I atacked the problem as follows:

pressure (p) has units of lb/ft^2 = (slug ft)/(sec^2 ft^2) =
slug/(ft sec^2)

velocity has units of ft/sec

row, the density has units of slug/ft^3

Then, the second term in the equation must have units of:

(row)v^2 = [slug/ft^3][ft^2/sec^2] = slug/(ft sec^2)] which matches the pressure units of the first term.

Thanks for the reply. It motivated me to arrive at the right answer.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top