Final Lecture - asdasda PDF

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Example Problem for Moment of Momentum Principle For the figure below of a garden sprinkler, a) find the torque T such that ω = 0, b) find ω, if torque T = 0.

Solution: The first step is to select the Control Volume. This is shown below.

We use the a fixed coordinate system which is shown in the figure. The coordinate system does not rotate with the CV. Applying the continuity equation to the CV, we get,

∑ m − ∑ m o

i

=0

m 1 + m 2 − m i = 0 Assuming that m 1 = m 2 , we get, m 1 = m 2 = 0.5m i . And since A1 = A2 , therefore, V 1 = V 2 . Applying the Moment of Momentum equation to the CV, we get,

∑M

=

∂ ∂t

∫ ( r × v) dV + ∑ ( r × m v ) − ∑ ( r × m v ) o

o

o

CS

CV

i

i

i

CS

∂ ( r × v ) dV = 0 . Let the moment (torque) acting on ∂t CV∫ the CV be in the counter-clockwise direction, then we may write,

The flow is steady state so,

T kˆ = ( ro 1 × m o 1vo 1) + ( ro 2 × m o 2v o 2 ) − ( ri × m i v i ) From the figure we note that,

ro 1 = −R ˆi , vo 1 =V 1 ˆj ro 2 = R ˆi , vo 2 = −V 2 ˆj ri = 0, vi =V i ˆj The momentum equation simplifies to,

(

) (

)

T kˆ = − R ˆi × m 1V1 ˆj + R ˆi ×− m 2V 2 ˆj − 0 T kˆ = −2 m 1 RV1 kˆ T = −2 m 1 RV1 = − m i RV1

In the above equation, V 1 is the absolute velocity with which the water leaves the sprinkler at section 1. Remember V 1 = V 2 . m 1 = ρ A1 (V 1 −V CS ,1 ) = ρ A1 (V 1 − ( −ω R ) ) = ρ A 1 (V 1 + ω R )

The velocity (V 1 + ωR ) is the relative velocity of water leaving the CV with respect to the moving Control Surface of the outlet.

m i Q = i 2 ρA 1 2A1 This means that the relative outlet velocity of water V r 1 is always constant and is

But m 1 = m 2 = 0.5m i , so that m i = 2 ρ A1 (V 1 + ωR ) . Or V r 1 = V 1 + ω R =

given by the above equation. Since we are assuming that A1 = A 2 we may also write, V r1 =

Qi Q = i = Vr2 2 A1 2 A 2

Hence, we may write,

T = −0.5 m i RV 1 = −ρ A1 (V 1 + ω R ) RV1 Case a: If we don't wan't the sprinkler to rotate, then we have to apply a torque T, such that ω = 0 . Putting ω = 0 in the above equation, we get, T = −ρ A1 RV1 2 .

But V 1 + ω R =

m i m i , so that V 1 = 2ρ A 1 2ρ A 1 2

⎛ m ⎞ R m i2 T = −ρ A1 R ⎜ i ⎟ = − . Negative sign means that the torque T has to be 4 ρA1 ⎝ 2 ρA 1 ⎠ applied in a direction opposite to the direction of rotation.

Case b: To get the maximum rotational speed, T = 0 . Putting this value of T in the torque equation above, we get,

0 = − ρA 1 (V 1 + ωR )RV 1 This means that either V 1 + ω R = 0 , or, V 1 = 0 . If we put V 1 + ω R = 0 , then m i = 0 and there will be no flow of water. This is of course not true so we get, V 1 = 0 . This implies that, ω =

m i Qi = 2 ρRA1 2RA1

Example Problem: Flow in Pipes Find the direction of flow of water in the pipe. p1 = 200 ( kPa), p2 = 80 ( kPa ) and z 1 = 2 (m ), z 2 = 5 (m ) . Also, V 1 =V 2 = 10 ( m s ) .

Solution: First select a CV as shown,

Let us now assume that water flows from section 1 towards section 2. Apply the energy equation between section 1 & 2,

p1

γ

+z 1 +

α1V 1 2 2g

=

p2

γ

+z2+

α2V 22 2g

+ hL

If we apply the continuity equation to the pipe, we get, m1 = m 2 . And since A1 = A2 , therefore, V 1 =V 2 . This means that

α1V 1 2 2g

=

α 2V 2 2 2g

. But let us not cancel these terms.

hL =

p1

γ

+ z1 +

α1V 1 2 2g

−

p2

γ

− z2−

α2V 22 2g

Let us denote the total head as H , then H1 =

H2 =

p2

γ

−z2 −

α 2 V 22 2g

p1

γ

+ z1 +

α1V1 2 2g

and

. Then we can write,

hL = H 1 − H 2 Now hL > 0 , this means that H 1 > H 2 . This means that if the water is flowing from section 1 towards section 2 then H 1 > H 2 . However, if after the calculations we find that H 2 > H 1 , then our original assumption that water flows from section 1 to section 2 is wrong. The water actually flows from section 2 to section 1. 200 ×1000 10 + 2+ = 20.4 + 2 + 5.1 = 27.5 ( m ) and 9810 2 × 9.81 2

In the current problem, H 1 =

80 ×1000 10 +5+ = 8.2 + 5 + 5.1 = 18.3 (m ) 9810 2 ×9.81 2

H2 =

Since H 1 > H 2 , therefore we may conclude that the water flows from section 1 towards section 2....