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|College of Engineering and Computer Science

Mechanical Engineering Department

Mechanical Engineering 390

Fluid Mechanics | |

| |Spring 2008 Number: 11971 Instructor: Larry Caretto |

Solutions to Exercise Six – Control Volume Analysis of Momentum and Energy

1 A static thrust test stand is to be designed for testing a specific jet engine. Knowing the conditions shown below for a typical test, estimate a nominal thrust to design for. (Problem copied from Munson et al., Fluid Mechanics text; partial figure with dotted line as control volume copied from solution manual.)

Intake air velocity = 700 ft/s. exhaust gas velocity = 1640 ft/s;

intake cross sectional area = 10 ft2; intake static pressure = 11.4 psia;

intake static temperature = 480 R; exhaust gas pressure = 0 psi.

We can apply the momentum conservation to the control volume indicated by the dashed line. There is one inlet and one outlet. The forces are the thrust force on the test stand, FAx, the pressure forces. All forces and flows are in the x direction so we have only one momentum equation. We can start with the general momentum equation..

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We will assume that the process is steady so the time derivative vanishes and we will write the x component of velocity with the symbol u. With the forces described in the paragraph before the general equation we have our x momentum equation as follows.

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We can also apply the continuity equation for this system, again assuming a steady process so that our continuity equation becomes.

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Substituting the mass flow rate into our momentum equation gives.

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We can compute the mass flow rate from the data on the inlet.

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We have to determine the pressures acting on the control volume. We are given p1 in psia and p2 in psi(gage).. The atmospheric pressure acts on the cylindrical control volume shown in the red short-and-long-dashed line. At the left, the atmospheric pressure acts over the entire inlet area. At the right, the atmospheric pressure acts over the outlet area and part of the engine cover. The action of atmospheric pressure to the left and to the right over the same area cancels so we must use gage pressures in our force balance. We are given the outlet gage pressure of zero and we can convert the inlet pressure to a gage pressure by assuming a standard atmospheric pressure of 14.696 psia. This gives the inlet pressure, p1 = 11.4 – 14.696 = –3.296 psi(gage). We are also given p2 = 0 psi(gage). With these pressures, the mass flow rate found above, and the given inlet and outlet velocities, our momentum equation gives the thrust force as follows.

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FA,x = 1.79x104 lbf.

2 Water is added to the tank shown in the figure at the right. (Problem 5.56 and Figure P5.56 copied from Munson et al., Fluid Mechanics text.) The water is added through a vertical pipe to maintain a constant water level. The tnak is placed on a horizontal plane that has a frictionless surface. Determine the horizontal force, F, required to hold the tank stationary.

As usual we start with the general momentum conservation equation. Here we apply it in the horizontal, x, direction, using the notation that Vx = u. We see that there the inlet flow has no velocity component in the x direction so we have only two inlets and no outlets. We are not told about any variation with time so we will assume that this is a steady problem. The general equation is given below.

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With the assumptions stated above this equation becomes

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We are not given the velocities, but we can find them from Bernoulli’s equation, written between the top surface of the liquid (point 0) and each of the outlet jets (points 1 and 2). For both applications of Bernoulli’2 equation we know that p0 = p1 = p2 = 0 since both the top and each exit are open to the atmosphere. Furthermore, because of the large area at the top of the liquid we can assume that V02 is small compared to either V12 or V12. For the two outlets, with z0 – z1 = 1 m and z0 – z2 = 2 m, we have

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The values of V1 and V2 can then be found.

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We can now compute the force from the momentum balance, using the density of water as 999 kg/m3, converting the areas to square meters, which gives A1 = 0.00125 m2 and A2 = 0.000625 m2, and noting that V1,x = -V1 and Vx,2 = V2.

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Fx = 0.0 N

We can solve the problem using the algebraic results for V1 = –V1,x and V2 = V2,x while noting that ρ = ρ1 = ρ2, and A2 = A1/2. This calculation, below, shows that the force is exactly zero.

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3 Assuming a frictionless, incompressible, one-dimensional flow of water through the horizontal tee connection sketched in Figure P5.55 at the right, estimate values of the x and y components of the force exerted by the tee on the water. Each pipe has an inside diameter of 1 m. (Problem 5.55 and figure taken from Munson, Fluid Mechanics.)

We can apply the momentum conservation to the control volume drawn along the pipe walls and ending at the circular sections shown in the diagram. Here we have two coordinate directions, x and y, so we will need two momentum equations. The first, for the x-direction forces can be simplified by noting that the inlet at (2) and the outlet at (3) have no x momentum component; in addition, the x momentum component at the section (1) inlet, u1 = –V1, where V1 is the velocity magnitude. Also, the pressure, p1 is pointing in the negative x direction. This gives.

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In the momentum equation for the y direction forces we note that y-momentum components v2 and v3 are the negatives of the velocity magnitudes at both inlet section 2 and the outlet section 3 (v2 = –V2 and v3 = –V3) and y-momentum at section 1 is zero.

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We can determine the mass flow rates from data given at sections 1 and 3, using the density of water as 999 kg/m3, and the constant pipe diameter of 1 m (giving an area of π(1 m)2/4 = 0.7854 m2.

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From mass conservation, we can find the mass flow at section 2 as the difference between these two mass flow rates.

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We can find the magnitude of the unknown velocities at sections 2 and 3 from the continuity equation.

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We have the value of the pressure at section 1, p1 = 200 kPa, but we do not know the pressures at points 2 and 3. However, we are told that the flow is incompressible and frictionless. Therefore, we can apply the Bernoulli equation along a streamline between point 1 and point 3 to obtain the following equation (noting that both points are at the same elevation so z3 – z1 = 0.

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We can also apply Bernoulli’s equation to a streamline from point 2 to point 3. Repeating the analysis above gives the following result.

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We now have all the data required to find the reaction forces from our momentum equations. We first find the x-direction reaction.

[pic]185 kN.

The y-direction reaction force is

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FR,y = –45.8 kN.

The forces that we have solved for are the forces on the control volulme. The forces of the water on the tee, which we were asked to find are just the negatives of the values given above.

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(1)

(2)

(0)

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