ENGR 207 Fluid Mechanics: Applications of Momentum Principle PDF

# ENGR 207 Fluid Mechanics ## *Chapter Four: Finite Control Volume Analysis* ### **4. Application on Momentum Principle** - Safwat Mahmoud, PhD, PMP® - Associate Professor of Sanitary & Environmental Engineering, Cairo University - Adjunct Faculty in the Environmental Engineering Program - Zewail City ## **CONTENT** - Law of conservation of momentum - Control volume concept - Applications - Free Jet Strikes Plate - Sudden Expansion and Contraction - Bends ## **The Law of Conservation of Momentum** - States that a body in motion cannot gain or lose momentum unless some external force(s) is applied. - The classical statement of this law is Newton's Second Law of Motion, i.e. - **force(s) = rate of change of momentum** ## **Principle of Momentum** - Momentum *M* = *mv* - Rate of Momentum: $$\frac{dM}{dt}=\frac{d}{dt}(mv)=\frac{d}{dt}(ma)=F$$ - **The rate of change of momentum is equal to the force applied.** - $$F=\frac{d}{dt}(ρV v)=\frac{d}{dt}(ρV)v$$ - *ρ* is constant for incompressible fluid. - $$F=ρQv$$ ## **Principle of the Momentum (Cont.)** - For this stream tube, the change of momentum of fluid ABCD: - = momentum of fluid A'AB'B - momentum of fluid CC'D'D - = *mv1- mv2* = *ρV*v1- *ρV*v2* - = *ρQ*v1*dt - *ρQ*v2*dt - Rate of change of momentum between AB and CD = *ρQ*(v1-v2*) - (F)from fluid = *ρQ*[vin-vout]* - Remember: **ρQ=m** ## **The Control Volume (CV) Concept** - Equations for flow, forces, and the energy will be written for **a fixed control volume of fluid** (e.g. a section of pipe). - **A properly drawn and labeled CV is an immense help in problem-solving.** - (F)from fluid = *ρQ*[vin-vout]* - **Forces resulting from fluid and affecting adjacent bodies.** - (F)on fluid = *ρQ*[vout-vin]* - **Forces on the fluid from adjacent bodies.** - The difference in velocities (*vin- vout*) represents the change in momentum of the fluid. This change in momentum results in a force being exerted on the surrounding bodies. ## **Applications: Free Jet Strikes Plate** ### Assumptions; - 1. **The plate is smooth(frictionless)** - the force in the plate direction equals zero. - 2. **There is no impact** - *Q1 = Q2+Q3* - 3. **Water strikes the plate and leaves it with the same velocity.** ## **Free Jet on Flat & Curved Plates** ## **Applications: Flat Plate** ### **1) Fixed plate (Stationary)** - *V* is the jet velocity. - In the X-Direction: - *Fx* = *ρQ*[vin -vout]* - Fx = *ρa1*V[V-0]= [*ρa1*[V2] - In the Y-Direction: - *Fy* = *ρQ*[vin-vout]* = 0 - *Fy* = 0 - [*ρQ2*v - *ρQ3*v] = 0 - *Q2* = *Q3*= *Q*/2* - *Q* = *a*V* ## **Example 1** - Given a tank discharging 0.4 m³/s to a plate, as shown in the figure. Calculate the required horizontal force *F* to hold the plate in position. - **Solution:** - Applying Bernoulli's equation between points *A* and *B*: - *PA*/γ + *zA*/g + *VA2*/(2g) = *PB*/γ + *zB*/g + *VB2*/(2g) - *VB* = √2*PA*/ρ = √2*75000/1000 = 12.3 m/s - *F* = *ρQ*V = 1000 * 0.4 * 12.3 = **4.9 kN** ## **Applications: (a) Flat Plate** ### **2) Single moving plate** - *V* is the jet velocity. - *v* is the velocity of the moving plate. - *vr* = *V-v* = relative velocity = velocity at which water hits the plate. - *Fx* = *ρQ*[vin - vout]* - *Fx* = *ρa1*vr*[vr-0]* - *Fx* = *ρa1*[V - v]2* - **Pay attention:** *Q1* not *Q* ### **3) Series of moving plates** - *Fn* = *ρQ*[vin -vout,n]* - = *ρQ*[vr - 0]* - = *Q*[v-v]* / *g* - *Q* = *Q* for total vanes - Note: Work done (W.D) = *F*d - Power = W.D/t = Rate of Work Done = *F*d/t - Output power = W.D/t = *Fx*v = *ρQ*[V -v]*v ## **Applications: (a) Flat Plate (Cont.)** - Output power = W.D/t = *Fx*v = *ρQ*[V-v]*v - Input power = W.D/t = (1/2)*mv2= (1/2)*ρQV2= *ρQ*V2/2 - Efficiency (η) = (O.P)/(I.P)= 2[V-v]v/V2 - To get maximum efficiency, differentiate with respect to *v*: - For ηmax: - *dn/dv* = 2[V-2v]v/V2=0 - *V*=2*v* - ηmax= 2[2*v*-v]v/4v2= 0.5 = 50% ## **Summary** | Parameter | Flat Plate | Curved Plate | | ------------- | ------------- | ------------- | | Fixed Plate | *Fx* = *ρQ*V | *Fx*= *ρQ*V*(1-cosθ) * *Fy* = *ρQ*(-v*sinθ)* | |. |. |. | | Moving Plate | *Fx* = *ρQ*vr | *Fx*= *ρQ*vr*(1-cosθ) * *Fy* = *ρQ*(-vr*sinθ)*| |. |. |. | | Series of Moving Plates | *Fn* = *ρQ*vr * η= 2V-v/V2* | *Fn*= *ρQ*v*(1-cosθ) * η= 2V-v/V2*(1-cosθ)* | ## **Applications: (b) Curved Plate** ### **1) Fixed plate (Stationary) ** - *V* is the jet velocity - *Fx*= *ρQ*[vin,x - vout,x]* - *Fx*= *ρQ*(V - Vcosθ) = *ρa1*V(V - Vcosθ) - *Fy*= *ρQ*(0 - Vsindθ) = *ρa1*V(0 - Vsindθ) ### **2) Single moving curved plate** - *v* is the velocity of the moving plate. - *vr* = *V-v* = relative velocity = velocity at which water hits the plate. - *Fx* = *ρQ*[vin,x - vout,x]* - *Fx*= *ρa1*vr*[vr- vrcosθ]* - *Fy*= *ρa1*vr*[0-vrsinθ]* - W.D/t = *Fx* = *Fx*v ## **Application: (b) Curved Plate** ### **3) Series of moving curved plates** - *Fx* = *ρQ*[vr - vrcosθ]* - *Fx* = *ρQ*[V-v]*(1-cosθ) - W.D/t = *O.P* = *ρQ*[V-v]*(1-cosθ)*v - *I.P* = W.D/t = *ρQ*V2/2 - η = (O.P)/(I.P) = 2[V-v]*(1-cosθ)v/V2 ## **Applications: (b) Curved Plate (Cont.)** - For ηmax: - 2[V-2v]*(1-cosθ)v/V2=0 - (1-cosθ)≠0 - V - 2v =0 - V=2v - ηmax= 2[2v-v]*(1-cosθ)v/4v2 = 0.5(1-cosθ) ## **Summary** | Parameter | Flat Plate | Curved Plate | | ------------- | ------------- | ------------- | | Fixed Plate | *Fx* = *ρQ*V | *Fx*= *ρQ*V*(1-cosθ) * *Fy* = *ρQ*(-v*sinθ)* | |. |. |. | | Moving Plate | *Fx* = *ρQ*vr | *Fx*= *ρQ*vr*(1-cosθ) * *Fy* = *ρQ*(-vr*sinθ)*| |. |. |. | | Series of Moving Plates | *Fn* = *ρQ*vr * η= 2V-v/V2* | *Fn*= *ρQ*v*(1-cosθ) * η= 2V-v/V2*(1-cosθ)* | ## **Sudden Enlargement and Contraction** ### **Losses in Sudden Enlargement** - To calculate the losses due to sudden enlargement: - Applying B.E between points 1, 2 - *P1*/γ + *V12*/(2g) = *P2*/γ + *V22*/(2g) + *h1* - *h1* = (*P1* - *P2*)/γ + (*V12*/(2g) - *V22*/(2g) - **(F)on fluid** = **ρQ*[vout - vin]*** - **Forces on the fluid from adjacent bodies** - *F1 - F2* = *Q*[vout-vin]*/g - *P1*A1 + *p*(A2 - A1)- *P2*A2 = *Q*[v2 - v1]*/g - **but p'=P1** (from experiments) - *P1*A1 + *P1*A2 -*P1*A1 - *P2*A2 = *Q*[v2 - v1]*/g - (*P1* - *P2*)A2 = *Q*A2[v22 - v12]*/g - *P1*/γ - *P2*/γ = *Q*V22*/g - *QV12*/g - **From (1) & (2)** - *hl* = *V12*/g - *V1V2*/g + *V22*/(2g) - *V22*/(2g) - *hl* = (*V1* - *V2*)2/(2g) ### **Losses in Sudden Contraction** - Losses are mainly due to sudden enlargement between Ac and A2: - *hl* = (*Vc -*V2*)2/(2g) - *hl* = [(*A2*/Ac -1) * V22/(2g)]= [(*Cc* -1)* *V22*/(2g)] = k*V22*/(2g) - *k* = *φ*(D1/D2)<1.0 - AcK = A2V2 - Cc = Ac/A2 ## **Bends** - Sharp bends result in separation downstream of the bend. - The turbulence in the separation zone causes flow resistance. - Greater radius of bend reduces flow resistance. ## **Vertical Bend** - Control Volume = abcd - **For x direction:** - *P1*A1 - *P2*A2cosθ - *Fx* = *ρQ*(vout,x - vin,x) - **(Fx)on fluid** = *P1*A1 -*P2*A2cosθ - *Q/g*[v2cosθ - v1] - **For z direction:** - -*P2*A2sinθ - W + *Fz* = *ρQ*(vout,z - vin,z) - **(Fz)on fluid** = - *P2*A2sinθ + W + *Q/g*[v2sinθ - 0.0] - *FT* = √*Fx2* + *Fz2* - tanα = *Fz*/ *Fx* - **(Fx) on bend** = - *P1*A1 + *P2*A2cosθ + *Q/g*[v2cosθ - v1] - **(Fz) on bend** = -*P2*A2sinθ - W - *Q/g*[v2sinθ - 0.0] - **If the bend is horizontal, the weight W is not included.** ## **Example** - Find the x and y components of the force required to hold the bend shown below in place. Pressure p1 and p2 are gauge pressures. - **Solution:** - **Momentum Equation:** - ∑Fx = m(-V2-V1) - *P1*A1 + Fx + *P2*A2 = *m*(-V2 - V1) - Fx = *m*(-V2 - V1) - *P1*A1 - *P2*A2 - **Mass Flow Rate** - *m* = *ρAV1* = 998*π/4*0.22*10 = 314.16kg/sec - **Horizontal Force** - *Fx* = *m*(-V2 - V1) - *P1*A1 - *P2*A2 - *Fx*= 314.16(-10-10) - 140000*π/2*0.22- 120000*π/2*0.22 - *Fx* = -14451 N - **Vertical Force:** - *Fy*= 0 - **Fx** = -14451 N - force acting by bend on fluid. - **Fx** = 14451 N- force acting by bend on fluid. - **Fx** = 14451 N - force acting on bend by fluid. - **Fx**= 14451 N - force required to hold bend in place. ## **Thank You** Any Questions?

ENGR 207 Fluid Mechanics: Applications of Momentum Principle PDF

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