Different units in Bernoulli's equation's terms?

In summary, the units in Bernoulli's equation must match on both sides of the equation, with the first term representing pressure with units of force/area and the second term representing the product of density and velocity squared with units of force. If the units do not match, it is likely due to incorrect units of mass or force being used.
  • #1
sinueus
4
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
 
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  • #2
A lb isn't a unit of force, it's a units of mass.
 
  • #3
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.
 
  • #4
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.
 

1. What are the units for pressure in Bernoulli's equation?

The units for pressure in Bernoulli's equation are typically given in pascals (Pa) or newtons per square meter (N/m2). However, any unit of pressure can be used as long as it is consistent throughout the equation.

2. What units are used for velocity in Bernoulli's equation?

The units for velocity in Bernoulli's equation are typically given in meters per second (m/s) or feet per second (ft/s). Again, any consistent unit of velocity can be used.

3. How do you convert between different units in Bernoulli's equation?

To convert between units in Bernoulli's equation, you can use conversion factors or unit conversions. For example, to convert from feet per second to meters per second, you would multiply by 0.3048 since 1 ft/s is equal to 0.3048 m/s.

4. What units are used for density in Bernoulli's equation?

The units for density in Bernoulli's equation are typically given in kilograms per cubic meter (kg/m3) or pounds per cubic foot (lb/ft3). As long as the units are consistent throughout the equation, any unit of density can be used.

5. Do all terms in Bernoulli's equation have the same units?

No, the terms in Bernoulli's equation do not have to have the same units. However, it is important to note that the units must be consistent within each term. For example, if one term has units of pressure and another has units of velocity, the units must be converted to be consistent before the equation can be solved.

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