Tuesday, August 9, 2011

MrsDrPoe: Conservation of Momentum, Part II

It's that time again folks! For this Thesis Tuesday, we will continue our discussion on the conservation of momentum equation by looking at a brief example.

Consider water flowing through a 20 deg bend shown below at a rate of 0.025 m3/s. The flow is frictionless, gravitational effects are negligible, and the pressure at section 1 is 150 kPa and section 2 is 74.5 kPa. Determine the x and y components of force required to hold the bend in place.

Alright...let's organize the problem from the given information in the statement, the original figure (not shown), and a fluid property table:

Given: theta = 20 deg, Q1 = 0.025 m3/s, P1 = 150 kPa, P2 = 74.5 kPa, rho = 999 kg/m3, D1 = 0.1 m, D2 = 0.05 m
Fx, Fy
Assume: steady, laminar, incompressible, uniform flow, frictionless, neglect gravity

Now onto the solution (employing the process we outlined last week):

Step 1: We must first choose a CV, keeping in mind the criteria for the "best" CV. In the figure above, the CV we will use is represented by the dashed line- note that it is perpendicular to flow into and out of the bend and that it cuts through the bend support.

Step 2: Next, we must apply the continuity equation:

d/dt(int(rho)dV) + int(rho*(V.n))dA = 0

For steady (d/dt(...) = 0), incompressible (rho = constant), uniform (V = constant) flow with 2 in/out surfaces:

rho*int(V1.n1)dA1 + rho*int(V2.n2)dA2 = rho*|V2|*A2 - rho*|V1|*A1 = 0

Since we know the volumetric flowrate (Q1) and bend diameter at 1, we can find the velocity at 1:

|V1|= Q1/(0.25*pi*D1*D1)

Similarly, since we know the diameters at 1 and 2, we can determine the areas:

A1 = 0.25*pi*D1*D1 (etc.)

Thus, we can solve for the velocity at surface 2:

|V2|= |V1|*D1*D1/(D2*D2) = 12.732 m/s

Step 3: Then we look at the external forces acting on the bend. In this case, we have two pressure forces and the support force of the bend (which we would not "see" if our CV did not intersect the support apparatus):

Fp1 = (p1*A1 0 0)
Fp2 = (Fp2x Fp2y 0) = (-p2*cos(theta)*A2 p2*sin(theta)*A2 0)
Fs = (Fsx Fsy Fsz)

Step 4: Next, we must look at the left side of the linear momentum equation for steady, uniform, incompressible flow with 2 surfaces:

d/dt(int(rho*V)dV + int(rho*V*(V.n))dA

rho*int(V1*(V1.n1))dA1 + rho*int(V2*(V2.n2))dA2

-rho*int(V1*|V1|)dA1 + rho*int(V2*|V2|)dA2

-rho*V1*|V1|*A1 + rho*V2*|V2|*A2

Step 5: Finally, we combine the right and left sides of the momentum equation, splitting it into scalar components (remember: this equation is a vector equation!), and solve for the unknown support forces:

Vector Eqn:
-rho*V1*|V1|*A1 + rho*V2*|V2|*A2 = Fs + Fp1 + Fp2

-rho*|V1|*|V1|*A1 + rho*|V2|*|V2|*cos(theta)*A2 = Fsx + rho*A1 -p2*cos(theta)*A2

Fsx = -821.322 N

-rho*|V2|*|V2|*sin(theta)*A2 = Fsy + p2*sin(theta)*A2

Fsy = -158.79 N

0 = Fsz

While it may seem a bit complicated at first, I can assure you that the more problems you work with this outlined process, the easier it becomes. Please be sure to comment with any questions you may have, and stay tuned for more momentum madness next week!


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