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
*********************************************************************
[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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