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Fundamental Question about Dynamic Pressure.

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PhilipFry

Mechanical
Aug 3, 2001
56
Alright, I can't believe I need to ask this question, but I'm looking at the term in Bernoulli's equation for dynamic pressure (1/2*rho*velocity^2). Can someone explain to me the units on this this? When I look closely, it looks like it gives the answer as a unit of force, not pressure.

rho=density=kg/m^3
velocity=m/s^2

rho*velocity^2=kg/m^2*(m^2/s^2)=kg/ms^2=1 Newton!

What am I missing here?
 
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Oops, my fingers got ahead of me.

rho=density=kg/m^3
velocity=m/s

rho*velocity^2=kg/m^3*(m^2/s^2)=kg/ms^2=1 Newton!
 
Ah, yes, I'm retared. I should turn in my degree.
 
still not sure if you are clear. The units work out to (Kg-m)/(s-m)^2, or a Newton/m^2 --- pressure
 

To PhilipFry,

As BronYrAur says 1 kg/(ms2) is a pressure unit = 1 Pascal = 1 Pa = 1 N/m2. And as blackwed rightly says the unit of force (=mass[×]acceleration) 1 N = kg.m/s2.

When [ρ]V2/2 (kinetic energy per unit mass of the Bernoulli equation) is divided by [ρ]g, it results in V2/2g, and we obtain velocity head, with dimensions of length, m.
 
The unit what you got doesn't indicate either force or pressure, per se.

Force is either expressed in Newtons or Kgf
Pressure is either expressed in Newtons/m2 or kgf/m2

What you got is a simplified unit and is kgm/(m.s2) and here kgm is kg mass (later on I will just use kg for kg mass).

Multiply and divide this with meters and you will get

kg x m/ (m x m x s2) = kg x m/(m2 x s2)
by rearranging the terms,

(kg x m/s2)x(1/m2) = N/m2, which is a pressure unit.

Fluid mechanics text books, use a constant gc when dealing with FPS system of units and express the dynamic head as rho*v2/2gc. Here, gc is 32.17 ft.lbm/lbf.sec2

This constant is not required for metric system as Newton is directly expressed as kg.m/sec2. However, you can derive the pressure units for dynamic head by using the constant gc with units kg.m/N.sec2


 
Yeah, I got it. I've really got to cut down on smoking crack. ;-)
 
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