Upload
adjei-baldan
View
216
Download
1
Embed Size (px)
DESCRIPTION
FLUID MECHANICS
Citation preview
CHE 204: TRANSPORT PHENOMENA ITERM 102ALSHAMI
Chapter 2: MOMENTUM BALANCESDefinition Momentum (M) can simply be defined as the power residing in a moving object:
Therefore, the dimensions of momentum is mass* length/time (ML/T).Since we are about to perform momentum BALANCES in accordance with the conservation laws introduced in chapter one, we must ask: How can momentum be transferred?The answer is in TWO ways: 1) by force, and 2) by convection.1. By ForceIn fluid mechanics, the most frequently occurring forces are those due to pressure (normal to surface), shear stress (tangential), and gravity (body force)
A force is readily seen to be equivalent to a rate of change of momentum by examining its dimensions:
This is a direct application of Newtons 2nd law of motion:
2. By ConvectionThe convective transfer of momentum by flow is due to the convective action of the moving fluid.Momentum Balance:
equation (5) is a vector equation and can be represented by three equations for each coordinate direction:x-direction:
y-direction:
z-direction:
Where,Fp = Forces due to pressure FR = Forces due to the solid containing the fluid (Reaction forces)Fb = Forces due to gravity (body forces)For the case shown in the figure below:CV(uA)1
(uA)2
P1A1P2A2FRxy
x-direction:
For horizontal flow (no elevation): gravity force (mg) = 0
Solving for the reaction force (force required to hold the conduit in place), yields:
Also, since
Equation (13), can be rewritten as
The same procedure is repeated to find the Reaction forces in y-direction (FRy):Momentum entering the CV is only in the x-direction => y-momentum in = 0Momentum leaving the CV the y-direction = Force due to pressure in the y-direction = P2A2 sin
Finally, the magnitude of the resultant force is given by
And it makes an angle
ExampleWater with a density of 1000 Kg/m3 flows in a right angle bend .The pressure at the outlet is 1 atm. The volumetric flow rate is 0.025 m3/s. The inlet diameter is 0.05 m and the outlet diameter is 0.025 m. The friction losses and the gravity forces are neglected. Calculate the force needed to hold the bend stationary (magnitude and direction )
21
Solution
Volumetric flow rate =Q
The energy balance should be used to find the unknown pressure
P2 = 1 atm (0 gauge)P1 = unknown = (negligible frictional losses)W = 0 (no shaft work)Z2 = Z1 (no elevation)Substituting gives:
P1-P2 =1.2 * 107 Pa = 118.8 atmP1= 119.8 abs =118.8 atm gauge P2= 1 atm abs = 0 gauge
Applying the momentum equation, mtotg = 0
A1= =1.96*10-3 m2A2= 4.91*10-4m2
By substituting all the numbers in FRx to get
FRx = 2.5x 104 N (acting in the opposite direction)
FRy= 1.27x103 N
The angle is in the second quarter because X is negative and Y is positive
1