3.5 flow power - navapadol's blog · pdf file- “pump” is a tool to give power...

3.5 Flow Power

- Naturally, fluid flows form high to low position.

- If we would like flow from low to high position, we need to give power to fluid.

- “Pump” is a tool to give power to fluid.

𝑃1𝛾+𝑉12

2𝑔+ 𝑧1 + ℎ𝑃 =

𝑃2𝛾+𝑉22

2𝑔+ 𝑧2 + ℎ𝐿

When hP = Energy head of pump (m)hL = Energy head of friction loss (m)

Flow power

PF = 𝛾QH When PF = flow power (kW)𝛾 = Specific weight of fluid (kN/m3)

Q = Flow rate (m3/s)H = Total energy head (m)

𝑃

𝛾+

𝑉2

2𝑔+ z = H

The total energy head can be found by

Ex. Water surface is higher than referent point for 20 m. It has 5 m3/sof flow rate. Find the flow power caused by the water at surface.

3.6 Momentum Equation

The force caused by fluid movement which reacts with obstruction need

to study because it impacts to the equipment. The impact of fluid needs to analyze the direction and amount of its force.

The momentum of fluid mass needs to consider with the impacted Structure. It relates to the 2nd law of Newton.

F = ma

The change of momentum = ma (kg. m/s2 or N)

The change of momentum of fluid mass = ρQdv when dv = v2 – v1

The unit of the change of momentum of fluid mass can be found by

ρQdv = kg/m3 × m3/s × m/s = kg.m/s = N

Thus, the equation of force that reacts with obstruction called “momentum equation”

F = ρQdv F = Force of fluid (N)ρ = Fluid density (kg/m3)Q = Flow rate (m3/s)dv = v2-v1 = different velocity (m/s)

Application of momentum equation

1. Fluid force reacts the planar object surface- Fluid impacts the planar object surface V1, and it makes plain object

Move as V2. The cross section of fluid is A

V1

A

V2Injector

F = ρQdv = ρQd(v2 – v1) It can divided to 2 consideration.

1.1 static planar object

Thus V2 = 0 m/s we have F = -ρQdv1

(symbol “ - ” means the reaction force of planar object.)

The flow rate (Q) of static planar object: Q = Av1

V1

A

Injector

V2 = 0 m/s

1.2 planar object moves along the flow direction

The momentum equation is F = ρQdv = ρQd(v2 – v1)

Flow rate (Q): Q = A.dv (Use dv = IdvI )

V1

A

V2Injector

1.3 planar object moves opposite the flow direction

V1

A

V2

Injector

The velocity of planar object is –v2 , and velocity of fluid is v1

According to momentum equation F = ρQdv= ρQd(v2 – v1)

Replace v2 as –v2 = ρQd(-v2 – v1)So F = -ρQd(v2 + v1)

Flow rate (Q): Q = A(v2+v1)

2. Fluid force reacts the curve object surface- Fluid impacts the curve object surface V1, and it makes

fluid direction changed along the curve surface.

- The angle occurs in x, y plain as shown in following figure

V1

A

Injector

V1

θ

V1cosθ

V1sinθ

curve object

α

Fy

Fx

F

- Fluid jet which has V1 impacts to curve Surface (no friction force).

- The direction of fluid jet will change along the curve made the θ to y-axis.

F and θ can be separately considered by

2.1 Static curve surfaceF and θ have to consider in x-axis and y-axis

- In x-axis (Fx) From momentum equation

Fx= ρQ(V2-V1)

Consider only force in x-axis. Thus velocity in x-axis after impact curve surfacecan be found by V1sinθ Then

Fx= ρQ(V1sinθ-V1)

Fx= -ρQV1 (1-sinθ) and Q = AV1

- In y-axis (Fy) From momentum equation

Fy= ρQ(V2-V1)

Consider only force in y-axis. Thus velocity in y-axis after impact curve surfacecan be found by V1cosθ Then

*** no velocity of V1 in y-axis then V1 = 0

Fy= ρQ(V1cosθ-0)

Fy= ρQV1cosθ

Summary

Flow rate Q = AV1

Resultant force F = 𝐹𝑥2 + 𝐹𝑦

2

Direction of resultant force 𝛼 = 𝑡𝑎𝑛−1𝐹𝑥

𝐹𝑦

2.2 Moving curve surface (fluid velocity < 90 degree)

V1-V2

A

Injector

V1-V2

θ

(V1-V2)cosθ

(V1-V2)sinθ

curve object

α

Fy

Fx

F

-Fluid flows from injector, and it impact to curve surface. Then the curve surface moves with V2

-To determine F and α, it has to calculate in X axis and Y axis separately.

- In x-axis (Fx) From momentum equation

Fx= ρQ(V2-V1)

V2 = out coming velocity = (V1-V2)sinθV1 = incoming velocity = (V1-V2)

Then

V1-V2

A

Injector

V1-V2

θ

(V1-V2)cosθ

(V1-V2)sinθ

curve object

α

Fy

Fx

F

Fx= ρQ[(V1-V2)sin θ – (V1-V2)]

Fx= ρQ(V1-V2)(sin θ - 1)

Fx= -ρQ(V1-V2)(1 - sin θ)

Flow rate => Q = A(V1 – V2) when V1, V2 are Vfluid and Vcurve object Respectively

Power => P = Fx.V2

- In y-axis (Fy) From momentum equation

Fy= ρQ(V2-V1)

V2 = out coming velocity = (V1-V2)cosθ

V1 = incoming velocity = (V1-V2) = “0”

Then Fy= ρQ[(V1-V2)cos θ – (V1-V2)]

Fy= ρQ[(V1-V2)cos θ – 0]

Fy= ρQ(V1-V2)cos θ

V1-V2

A

Injector

V1-V2

θ

(V1-V2)cosθ

(V1-V2)sinθ

curve object

α

Fy

Fx

F

0

Flow rate => Q = A(V1 – V2) when V1, V2 are Vfluid and Vcurve object Respectively

Resultant Force => 𝐹 = 𝐹𝑋2 − 𝐹𝑦

2 α = 𝑡𝑎𝑛−1𝐹𝑥𝐹𝑦

2.3 Moving curve surface (fluid velocity > 90 degree)

V1-V2Injector

-(V1-V2)

θ

-(V1-V2)cosθ

-(V1-V2)sinθ

curve object

α

Fy

Fx

F

x

y

- In x-axis (Fx) From momentum equation

Fx= ρQ(V2-V1)

V2 = out coming velocity = -(V1-V2)sinθV1 = incoming velocity = (V1-V2)

Then Fx= ρQ[-(V1-V2)sin θ – (V1-V2)]

Fx= -ρQ[(V1-V2)sin θ + (V1-V2)]

Fx= -ρQ(V1-V2)(sin θ + 1)

Flow rate => Q = A(V1 – V2) when V1, V2 are Vfluid and Vcurve object Respectively

Power => P = Fx.V2

- In y-axis (Fy) From momentum equation

Fy= ρQ(V2-V1)

V2 = out coming velocity = -(V1-V2)cosθ

V1 = incoming velocity = (V1-V2) = “0”

Then Fy= ρQ[[-(V1-V2)cos θ – (V1-V2)]

Fy= ρQ[-(V1-V2)cos θ – 0]

Fy= -ρQ(V1-V2)cos θ

V1-V2Injector

-(V1-V2)

θ

-(V1-V2)cosθ

-(V1-V2)sinθ

curve object

α

Fy

Fx

F

x

y

0

Flow rate => Q = A(V1 – V2) when V1, V2 are Vfluid and Vcurve object Respectively

Resultant Force => 𝐹 = 𝐹𝑋2 − 𝐹𝑦

2 α = 𝑡𝑎𝑛−1𝐹𝑥𝐹𝑦