3.5 flow power - navapadol's blog · pdf file- “pump” is a tool to give power...
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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)
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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
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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.
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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)
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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)
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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.
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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
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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
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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)
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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.
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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
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- 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
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Summary
Flow rate Q = AV1
Resultant force F = 𝐹𝑥2 + 𝐹𝑦
2
Direction of resultant force 𝛼 = 𝑡𝑎𝑛−1𝐹𝑥
𝐹𝑦
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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.
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- 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
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- 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
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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
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- 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