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CHEN 3200 Fluid Mechanics Spring 2011

Homework 3 solutions

1. An artery with an inner diameter of 15 mm contains blood flowing at a rate of 5000 mL/min. Further along the artery, arterial plaque has partially clogged the artery, reducing the area available for blood flow. When the person is lying down (the artery is horizontal), the pressure difference between the clean and clogged regions of the artery is 830 Pa. Determine how much (as a percentage) of the cross-‐sectional area is clogged by the plaque. (15 points)

Since !h=0, the governing equation becomes

The velocity at point 1 is givenby

Solving the governing equation for p1-p2gives

The velocities are related by the continuity equation, V2=(A1/A2)V2, thus

The SG of blood is approximately 1.06, so the equation abovebecomes

Therefore, A2=0.35A1, or 65% of the artery isblocked.

V2=(A1/A2)V1, thus


Since !h=0, the governing equation becomes

The velocity at point 1 is givenby

Solving the governing equation for p1-p2gives

The velocities are related by the continuity equation, V2=(A1/A2)V2, thus

The SG of blood is approximately 1.06, so the equation abovebecomes

Therefore, A2=0.35A1, or 65% of the artery isblocked.


2. Benezne flows through a circular tube with an inside diameter of 50 mm. A bar, with a smooth rounded end, has a diameter of 40 mm and partially plugs the end of the tube where the fluid is released into a tank at atmospheric pressure. Assume uniform velocity profiles throughout the system.

(a) What pressure is measured by the gage? (10 points)

(a)

V1 = 7 m/s

Since !h=0 and p2=0 (gage pressure), the governing equation becomes

or,

From the continutity equation, we can solve for V2=(A1/A2)V1=(D2/(D2-d2))V1, where D=50mm and d=40 mm. Plugging in these values,

The pressure at point 1 is then given by (using SGbenzene=0.876)

Since !h=0 and p2=0 (gage pressure), the governing equation becomes

or,

From the continutity equation, we can solve for V2=(A1/A2)V1=(D2/(D2-d2))V1, where D=50mm and d=40 mm. Plugging in these values,

The pressure at point 1 is then given by (using SGbenzene=0.876)


3. Two connected cylindrical chambers are filled with an incompressible fluid with SG=0.85. The chamber on the left is capped by a piston with a diameter of 10 cm. The chamber on the right has a diameter of 25 cm, and is also capped by a piston that supports the weight of a large stone block. (15 points)

(a) If the cylinder and block have a total weight of 9800 N, what force must be applied to the piston on the left to hold the block at the height shown? (10 points) (b) If the block is to be raised 6 cm, how far must the piston on the left be pushed down? (3 points) (c) If some of the fluid in the left cylinder was replaced by air, would the system be more or less efficient for raising the block? Explain your answer. (2 points)

F

3 m

The area of the left and right cylinders is givenby

The pressure just below the cylinder on the right is equal to the weight of the piston andblock, divided by the area

That pressure is also equal to the pressure applied by the piston on the left, plus the weight ofthe fluid, i.e.

Solving for the force on the leftgives

Thus, the force applied on the left is considerably less than the weight of the cylinder and blockon the right. This is the principle behind hydraulic lifts.

The change in volume on the right when the piston is raised 6 cm must equal the change involume in the left cylinder. The change in volume on the right is

The distance the piston on the left must move is givenby

So, the force required to lift the piston and block on the right is small, but the piston on theleft must be moved much further than the desired change in height on the right.

The system would be less efficient. Air is compressible, so some of the force applied by thepiston on the left would be used to compress the air, and less force would be transferred throughthe fluid to the piston on the right.


The area of the left and right cylinders is givenby

The pressure just below the cylinder on the right is equal to the weight of the piston andblock, divided by the area

That pressure is also equal to the pressure applied by the piston on the left, plus the weight ofthe fluid, i.e.

Solving for the force on the leftgives

Thus, the force applied on the left is considerably less than the weight of the cylinder and blockon the right. This is the principle behind hydraulic lifts.

The change in volume on the right when the piston is raised 6 cm must equal the change involume in the left cylinder. The change in volume on the right is

The distance the piston on the left must move is givenby

So, the force required to lift the piston and block on the right is small, but the piston on theleft must be moved much further than the desired change in height on the right.

The system would be less efficient. Air is compressible, so some of the force applied by thepiston on the left would be used to compress the air, and less force would be transferred throughthe fluid to the piston on the right.


4. A cylindrical tank with diameter D is initially filled with fluid to a height of H+h0. The fluid then exits the tank through a hole of diameter d, located a distance H above the bottom of the tank. (25 points)

(a) Derive an expression for the dimensionless fluid height, h/h0, as a function of time. (10 points) (b) Make a plot of h/h0 versus t using D/d = 10, for values of h0 = 0.2, 0.5 and 1 m. (3 points) (c) Make a plot of h/h0 versus t using h0 = 1 m, for values of D/d = 2, 5 and 10. (3 points) (d) Now, assume that the tank is continuously refilled to maintain a constant fluid height. If the

fluid is ethanol, and H = 0.3 m and h0 = 0.1 m, what will be the fluid velocity leaving the tank? (5 points)

(e) For the same conditions as in (d), what is the horizontal distance from the edge of the tank that the jet of ethanol will strike the ground? (4 points)

(a)

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Applying the Bernoulli equation between point 1 (the fluid at the top of the tank) and point 2(just outside the hole in the tank), and making the assumptions/simplifications that p1=p2 sinceboth points are at atmospheric pressure, and that V1=0 since the fluid at the top of the tank isheld constant, we have

parallel to the ground

We can use the classical equations of motion to calculate how far the fluid will travel before ithits the ground.

In the downward z direction, the acceleration due to gravity is defined as

Integrating this equation twice gives us

where C1 and C2 are constants of integration. As boundary conditions, we can say that theinitial velocity in the downward (z) direction is 0, implying that C1=0. Also, at t=0, the fluid isat the height of the hole. If we assign the ground to be z=0, then C2=0.3 m.

The time taken for the fluid to reach the ground is then

In this time, the fluid will have travelled to the right a distance of V2t,or

-‐g


5. Water flows from a large tank and is expelled from the horizontal pipe to the right. Calculate the velocity and flow rate in the pipe. (15 points)

Problem 6.52 [2]

H = h1 =

(h2)

Given: Flow through tank-pipe system

Find: Velocity in pipe; Rate of discharge

Solution:

Basic equation p!

V2

2g z const "p ! g "h Q V A

Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline

Hence, applying Bernoulli between the free surface and the manometer location

patm!

p! g H

V2

2where we assume VSurface <<, and H = 4 m

Hence p patm ! g H ! V2

2

For the manometer p patm SGHg ! g h2 ! g h1 Note that we have water on one side and mercury onthe other of the manometer

Combining equations ! g H ! V2

2SGHg ! g h2 ! g h1 or V 2 g H SGHg h2 h2

Hence V 2 9.81m

s24 13.6 0.15 0.75( ) m V 7.29

ms

The flow rate is Q V# D2

4Q

#4

7.29ms

0.05 m( )2 Q 0.0143m3

s

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