Partial Differential Equations in Two or More Dimensions



5.4 Conservation of Momentum

5.4a Macroscopic Momentum Balance. The equation for the conservation of momentum with respect to a control volume (CV) can be written as follows:

[pic]

[pic](m[pic])cv = [pic]( [pic] + [pic]

The total force acting on the control volume consists both of surface forces and body forces. The surface force is due to interaction between the control volume and its surrounding through direct contact at the boundary. The body force is due to the location of the control volume in a force field. The gravitational field and its resultant force is the most common example of the body force that will act on the entire control volume and not just at the control surface.

Example 5.4-1. ----------------------------------------------------------------------------------

[pic]

Figure 5.4-1 Gravity-flow tank

The above figure shows a tank into which an in compressible liquid is pumped at a volumetric flow rate Qo (ft3/s). The height of liquid in the vertical tank is h (ft). The flow rate out of the tank is Q (ft3/s).The length of the exit line is L (ft) and its inside diameter is Dp (ft). The cylindrical tank has an inside diameter Dt (ft). The liquid level h is determined as a function of time given an initial height ho and initial velocity Vo of the liquid in the pipe.

Solution ------------------------------------------------------------------------------------------

Step #1: Define the system.

We assume plug-flow conditions and incompressible liquid for the liquid flowing through the pipe. Therefore all the liquid is moving at the same velocity, more or less like a solid rod. The velocity V (ft/s) is equal to the volumetric flow Q divided by the cross-sectional area Ap of the pipe.

[pic]

Figure 5.4-2 Exit line of the gravity-flow tank

Let control volume (CV) be the fluid inside the exit line then the mass m of fluid inside the pipe is ApL(.

Step #2: Find equation that contains the liquid level h.

The liquid level can be obtained from the mass balance however the volumetric flow rate must be obtained from the momentum balance.

Step #3: Apply the momentum balance on system.

[pic] = [pic]( ApL(V)

[pic] ( [pic] = Q(V

[pic] = Fo ( FL ( Ff , where

Fo = Ap(hg (hydraulic force) + ApPatm (static force due to surrounding pressure)

FL = ApPatm (static force due to surrounding pressure)

Ff = KfLV2 is the frictional force due to the viscosity of the liquid pushing in the opposite direction from right to left and opposing the flow.

The momentum equation for the fluid inside the exit line becomes

[pic]( ApL(V) = Fo ( FL ( Ff = Ap(hg ( KfLV2

or

[pic] = [pic]h ( [pic]V2 (E-1)

To describe the system completely a total continuity equation (mass balance) on the liquid in the tank is also needed.

At[pic] = Qo ( Q = Qo ( ApV

or

[pic] = [pic] ( [pic]V (E-2)

Step #4: Specify the boundary conditions for the differential equations.

We need to solve two coupled ordinary differential equations (E-1) and (E-2). Physical dimensions, parameter values, and initial flow rate and liquid height are given in Table 5.4-1

Table 5.4-1 Gravity-flow tank data

_______________________________________________________________________

Pipe: ID = 3 ft Area = 7.06 ft2 Length = 3000 ft

Tank: ID = 12 ft Area = 113 ft2 Height = 7 ft

Initial values h = 3.0 ft V = 2.0 ft/s

Parameters density = 62.4 lb/ft3 Kf = 0.0281 lbf(s2/ft3

_______________________________________________________________________

Step #5: Solve the resulting equations and verify the solution.

Substituting the numerical values of parameters into Eqs. (E-1) and (E-2) give

[pic] = 0.0107h ( 0.00205V2 (E-3)

[pic] = [pic] ( 0.06248V (E-4)

The above equations can be solved by Matlab with initial values of V and h at a given Qo. Table 5.4-2 lists the Matlab programs and results with Qo = 35.1 ft3/s. Figure 5.4-3 shows the results in graphical forms.

Table 5.4-2 Gravity-flow tank results

% Example 5.4-1, gravity flow tank

% Main program

xspan=0:10:500;

[x,y]=ode45('ex541',xspan,[2 3]);

plot(x,y(:,1),'-',x,y(:,2),':')

grid on

xlabel('t(s)')

ylabel('V,h')

legend('V(ft/s)','h(ft)')

% Example 5.4-1, gravity flow tank

function yy = ex541(x,y)

% y(1)=V, y(2)=h

yy(1,1)=0.0107*y(2) - 0.00205*y(1)^2;

yy(2,1)=35.1/113- 0.06248*y(1);

t (sec) V (ft/s) h (ft)

0 2.0000 3.0000

10.0000 2.3225 4.7643

20.0000 2.7830 6.2814

30.0000 3.3314 7.4812

40.0000 3.9112 8.3254

50.0000 4.4704 8.8100

60.0000 4.9660 8.9622

70.0000 5.3712 8.8320

80.0000 5.6730 8.4842

90.0000 5.8703 7.9809

100.0000 5.9718 7.3821

110.0000 5.9925 6.7461

120.0000 5.9491 6.1199

130.0000 5.8571 5.5362

140.0000 5.7314 5.0198

150.0000 5.5878 4.5889

160.0000 5.4381 4.2515

170.0000 5.2896 4.0072

180.0000 5.1502 3.8529

190.0000 5.0272 3.7803

200.0000 4.9246 3.7785

210.0000 4.8437 3.8344

220.0000 4.7843 3.9347

230.0000 4.7467 4.0652

240.0000 4.7295 4.2115

250.0000 4.7300 4.3626

260.0000 4.7450 4.5096

270.0000 4.7711 4.6440

280.0000 4.8045 4.7591

290.0000 4.8418 4.8514

300.0000 4.8804 4.9201

310.0000 4.9177 4.9653

320.0000 4.9514 4.9883

330.0000 4.9799 4.9917

340.0000 5.0023 4.9790

350.0000 5.0190 4.9540

360.0000 5.0297 4.9203

370.0000 5.0349 4.8817

380.0000 5.0354 4.8420

390.0000 5.0319 4.8034

400.0000 5.0254 4.7677

410.0000 5.0168 4.7364

420.0000 5.0072 4.7109

430.0000 4.9973 4.6918

440.0000 4.9877 4.6789

450.0000 4.9789 4.6717

460.0000 4.9712 4.6697

470.0000 4.9649 4.6720

480.0000 4.9602 4.6776

490.0000 4.9572 4.6856

500.0000 4.9556 4.6951

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[pic]

Figure 5.4-3 Velocity and liquid height in an unsteady-state gravity flow tank.

The following example presents a system where energy and momentum balance must be applied to obtain the results.

Example 5.4-21 ----------------------------------------------------------------------------------

Advertised is a small toy, aerocket, that will send up a signal flare and the operation “is so simple that it is amazing” (Fig. 5.4-4).

[pic]

Figure 5.4-4 Aerocket

Our examination of this device indicates that it is a sheet metal tube 7 ft long and 1 in.2 area. A plug shaped into the form of a piston fits into the tube and a mechanical trigger holds it in place 2 ft above the bottom. The mass of the piston is 3.46 lb. To operate the device, the volume below the piston is pumped up to a pressure of about 4 atm absolute with a small hand pump, and then the trigger is depressed allowing the piston to fly out the top. The pyrotechnic and parachute devices contained within the piston are actuated by the acceleration force during ejection.

When we operated this toy last summer, the ambient temperature is 90oF. Assuming no friction in the piston and no heat transfer or other irreversibilities in the operation, how high would you expect the piston to go? What would be the time required from the start to attain this height?

Solution ------------------------------------------------------------------------------------------

Step #1: Define the system.

System: The piston.

Step #2: Find equation that contains h, the height the piston will attain.

The momentum balance will provide the velocity of the piston as it leaves the metal tube. An energy balance can then be applied to determine h.

Step #3: Apply the momentum balance on the system.

[pic]

Let the positive direction be pointing upward, the momentum balance on the piston becomes

m [pic] = (P ( Patm) Ap ( mg

where

m = mass of the piston

Vvel = velocity of the piston

P = pressure of the compressed air within the metal tube

Patm = surrounding atmospheric pressure

Ap = cross-sectional area of the piston

g = acceleration of gravity

We need to obtain the velocity of the piston as a function of the vertical distance z. First, the relation between the pressure P and the volume V of the moles N of compressed air within the metal tube must be determined. For the adiabatic expansion of air

dU = dW => NCvdT = ( PdV

From ideal gas law: PV = NRT => d(PV) = NR dT

N dT = [pic] (PdV + VdP)

[pic]PdV + [pic]VdP = ( PdV => CvVdP = ( (R + Cv)P dV = ( CpPdV

[pic] = ([pic] [pic]

Integrating the equation from the initial conditions (Pi, Vi) to (P, V) gives

ln (P/Pi) = ( ( ln(V/Vi) , where ( = [pic]

Therefore

P = Pi[pic]= Pi[pic] = Pi[pic]

From the momentum equation

m [pic] = (P ( Patm) Ap ( mg

m [pic] = Pi Ap[pic] ( Patm Ap ( mg

[pic] = [pic]z-( ( [pic]

Let a = [pic] , and b = [pic]

The momentum equation becomes

[pic] = az-( ( b => [pic][pic] = az-( ( b => Vvel [pic] = az-( ( b

Step #4: Specify the boundary conditions for the differential equation.

At z = zi = 2 ft, Vvel = 0

At z = 7 ft, Vvel = ?

Step #5: Solve the resulting equation and verify the solution.

How high would you expect the piston to go?

Integrate the momentum equation to obtain

[pic][pic] = [pic] z1-( ( bz + C1

At z = zi = 2 ft, Vvel = 0 => C1 = bzi ( [pic] zi1-(

The numerical values of a, b, and C1 can be evaluated

a = [pic] = [pic] = 1.4441(103 ft2.4/s2

Note the conversion factor from lbf to lb

b = [pic] = [pic] + 32.2 = 169 ft/s2

C1 = bzi ( [pic] zi1-( = (169)(2) + [pic]2-0.4 = 3,074 ft2/s2

Therefore

Vvel = 20.5(3,074 ( 3,610z-0.4 ( 169z)0.5

At z = 7 ft, Vvel = 21.6 ft/s

The height h attained by the rocket when it leaves the cylinder is evaluated from the conservation of energy. At the maximum height, Vvel = 0, all the kinetic energy becomes the potential energy.

[pic](Vvel2 = (gh => h = [pic][pic] = [pic][pic] = 7.2447 ft

The total height is 14.24 ft from the ground.

What would be the time required from the start to attain this height?

The total time consists of the time the piston spent within the cylinder and the time spent outside the cylinder.

Time spent within the cylinder can be obtained by integrating

Vvel = [pic] = 20.5(3,074 ( 3,610z-0.4 ( 169z)0.5

from z = 2 ft to z = 7 ft. This equation is rearranged and integrated numerically using Simpson’ formula with 5 points.

t = [pic] = [pic]

|z(ft) |f(z) |

|2.00 |f1 = 0.047909 |

|3.25 |f2 = 0.026712 |

|4.50 |f3 = 0.022637 |

|5.75 |f4 = 0.020751 |

|7.00 |f5 = 0.019641 |

t = [pic] (f1 + 4f2 + 2f3 + 4f4 + f5) = 0.126 sec

Time spent outside the cylinder is obtained by applying Newton’s second law to the pistion with the positive direction upward.

m [pic] = ( mg => [pic] = ( g => Vvel = [pic] = ( gt + Vvel,0

at t = 0 , Vvel = Vvel,0 = 21.6 ft/s

Integrate the equation again

z = ( 0.5gt2 + Vvel,0t + zo

at t = 0, z = zo = 0

z = ( 0.5gt2 + Vvel,0t = ( 16.1t2 + 21.6t

at z = 7.2447 ft = ( 16.1t2 + 21.6t

Solve this equation for t => t = 0.67 sec

Time to attain a height of 14.24 ft is: 0.126 + 0.67 = 0.796 sec

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