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CHE 204: TRANSPORT PHENOMENA I TERM 102 ALSHAMI CHAPTER 2: MOMENTUM BALANCES Definition – Momentum (M) can simply be defined as ‘the power residing in a moving object’: M=mu( 1) 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 Force In 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: Force≡ ML T 2 = ML T T = Momentum Time ( 2) This is a direct application of Newton’s 2 nd law of motion: F=ma=m du dt = d ( mu ) dt = ˙ M ( 3) 2. By Convection The convective transfer of momentum by flow is due to the convective action of the moving fluid. Momentum Balance: 1

Chapter 2_Lect Notes_Momentum Eqn

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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 )

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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

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