Fluids In Rigid Body Motion Engineering Essay
11-38C A moving body of fluid can be treated as a rigid body when there are no shear stresses (i.e., no motion between fluid layers relative to each other) in the fluid body.
11-39C A glass of water is considered. The water pressure at the bottom surface will be the same since the acceleration for all four cases is zero.
11-40C The pressure at the bottom surface is constant when the glass is stationary. For a glass moving on a horizontal plane with constant acceleration, water will collect at the back but the water depth will remain constant at the center. Therefore, the pressure at the midpoint will be the same for both glasses. But the bottom pressure will be low at the front relative to the stationary glass, and high at the back (again relative to the stationary glass). Note that the pressure in all cases is the hydrostatic pressure, which is directly proportional to the fluid height.
11-41C When a vertical cylindrical container partially filled with water is rotated about its axis and rigid body motion is established, the fluid level will drop at the center and rise towards the edges. Noting that hydrostatic pressure is proportional to fluid depth, the pressure at the mid point will drop and the pressure at the edges of the bottom surface will rise due to rotation.
11-42 A water tank is being towed by a truck on a level road, and the angle the free surface makes with the horizontal is measured. The acceleration of the truck is to be determined.
ax
± = 15°
Water tank
Assumptions 1 The road is horizontal so that acceleration has no vertical component (az = 0). 2 Effects of splashing, breaking, driving over bumps, and climbing hills are assumed to be secondary, and are not considered. 3 The acceleration remains constant.
Analysis We take the x-axis to be the direction of motion, the z-axis to be the upward vertical direction. The tangent of the angle the free surface makes with the horizontal is
Solving for ax and substituting,
Discussion Note that the analysis is valid for any fluid with constant density since we used no information that pertains to fluid properties in the solution.
11-43 Two water tanks filled with water, one stationary and the other moving upwards at constant acceleration. The tank with the higher pressure at the bottom is to be determined.
Tank A
8 m
Water
az = 5 m/s2
Tank B
2 m
Water
g
z
·
2
·
1
·
2
·
1
Assumptions 1 The acceleration remains constant. 2 Water is an incompressible substance.
Properties We take the density of water to be 1000 kg/m3.
Analysis The pressure difference between two points 1 and 2 in an incompressible fluid is given by
or
since ax = 0. Taking point 2 at the free surface and point 1 at the tank bottom, we have and and thus
Tank A: We have az = 0, and thus the pressure at the bottom is
Tank B: We have az = +5 m/s2, and thus the pressure at the bottom is
Therefore, tank A has a higher pressure at the bottom.
Discussion We can also solve this problem quickly by examining the relation . Acceleration for tank B is about 1.5 times that of Tank A (14.81 vs 9.81 m/s2), but the fluid depth for tank A is 4 times that of tank B (8 m vs 2 m). Therefore, the tank with the larger acceleration-fluid height product (tank A in this case) will have a higher pressure at the bottom.
11-44 A water tank is being towed on an uphill road at constant acceleration. The angle the free surface of water makes with the horizontal is to be determined, and the solution is to be repeated for the downhill motion case.
z
x
az
ax
¡ = 20°
-±
Downhill motion
Uphill motion
z
x
ax
Free
surface
az
Water tank
¡ = 20°
±
Horizontal
Assumptions 1 Effects of splashing, breaking, driving over bumps, and climbing hills are assumed to be secondary, and are not considered. 2 The acceleration remains constant.
Analysis We take the x- and z-axes as shown in the figure. From geometrical considerations, the horizontal and vertical components of acceleration are
The tangent of the angle the free surface makes with the horizontal is
® ± = 22.2°
When the direction of motion is reversed, both ax and az are in negative x- and z-direction, respectively, and thus become negative quantities,
Then the tangent of the angle the free surface makes with the horizontal becomes
® ± = – 30.1°
Discussion Note that the analysis is valid for any fluid with constant density, not just water, since we used no information that pertains to water in the solution.
11-45E A vertical cylindrical tank open to the atmosphere is rotated about the centerline. The angular velocity at which the bottom of the tank will first be exposed, and the maximum water height at this moment are to be determined.
·
2 ft
z
r
Assumptions 1 The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2 Water is an incompressible fluid.
Analysis Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0, z = 0), the equation for the free surface of the liquid is given as
where h0 = 1 ft is the original height of the liquid before rotation. Just before dry spot appear at the center of bottom surface, the height of the liquid at the center equals zero, and thus zs(0) = 0. Solving the equation above for · and substituting,
Noting that one complete revolution corresponds to 2° radians, the rotational speed of the container can also be expressed in terms of revolutions per minute (rpm) as
Therefore, the rotational speed of this container should be limited to 108 rpm to avoid any dry spots at the bottom surface of the tank.
The maximum vertical height of the liquid occurs a the edges of the tank (r = R = 1 ft), and it is
Discussion Note that the analysis is valid for any liquid since the result is independent of density or any other fluid property.
11-46 A cylindrical tank is being transported on a level road at constant acceleration. The allowable water height to avoid spill of water during acceleration is to be determined
D=40 cm
ax = 4 m/s2
±
htank =60 cm
„z
Water tank
Assumptions 1 The road is horizontal during acceleration so that acceleration has no vertical component (az = 0). 2 Effects of splashing, breaking, driving over bumps, and climbing hills are assumed to be secondary, and are not considered. 3 The acceleration remains constant.
Analysis We take the x-axis to be the direction of motion, the z-axis to be the upward vertical direction, and the origin to be the midpoint of the tank bottom. The tangent of the angle the free surface makes with the horizontal is
(and thus ± = 22.2°)
The maximum vertical rise of the free surface occurs at the back of the tank, and the vertical midplane experiences no rise or drop during acceleration. Then the maximum vertical rise at the back of the tank relative to the midplane is
Therefore, the maximum initial water height in the tank to avoid spilling is
Discussion Note that the analysis is valid for any fluid with constant density, not just water, since we used no information that pertains to water in the solution.
11-47 A vertical cylindrical container partially filled with a liquid is rotated at constant speed. The drop in the liquid level at the center of the cylinder is to be determined.
z
r
·
zs
R = 20 cm
Free
surface
ho = 60 cm
g
Assumptions 1 The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2 The bottom surface of the container remains covered with liquid during rotation (no dry spots).
Analysis Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0, z = 0), the equation for the free surface of the liquid is given as
where h0 = 0.6 m is the original height of the liquid before rotation, and
Then the vertical height of the liquid at the center of the container where r = 0 becomes
Therefore, the drop in the liquid level at the center of the cylinder is
Discussion Note that the analysis is valid for any liquid since the result is independent of density or any other fluid property. Also, our assumption of no dry spots is validated since z0(0) is positive.
11-48 The motion of a fish tank in the cabin of an elevator is considered. The pressure at the bottom of the tank when the elevator is stationary, moving up with a specified acceleration, and moving down with a specified acceleration is to be determined.
Fish Tank
·
2
az = 3 m/s2
h = 40 cm
g
z
Water
·
1
Assumptions 1 The acceleration remains constant. 2 Water is an incompressible substance.
Properties We take the density of water to be 1000 kg/m3.
Analysis The pressure difference between two points 1 and 2 in an incompressible fluid is given by
or
since ax = 0. Taking point 2 at the free surface and point 1 at the tank bottom, we have and and thus
(a) Tank stationary: We have az = 0, and thus the gage pressure at the tank bottom is
(b) Tank moving up: We have az = +3 m/s2, and thus the gage pressure at the tank bottom is
(c) Tank moving down: We have az = -3 m/s2, and thus the gage pressure at the tank bottom is
Discussion Note that the pressure at the tank bottom while moving up in an elevator is almost twice that while moving down, and thus the tank is under much greater stress during upward acceleration.
11-49 vertical cylindrical milk tank is rotated at constant speed, and the pressure at the center of the bottom surface is measured. The pressure at the edge of the bottom surface is to be determined.
z
r
·
zs
R = 1.50 m
Free
surface
g
ho
Assumptions 1 The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2 Milk is an incompressible substance.
Properties The density of the milk is given to be 1030 kg/m3.
Analysis Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0, z = 0), the equation for the free surface of the liquid is given as
where R = 1.5 m is the radius, and
The fluid rise at the edge relative to the center of the tank is
The pressure difference corresponding to this fluid height difference is
Then the pressure at the edge of the bottom surface becomes
Discussion Note that the pressure is 14% higher at the edge relative to the center of the tank, and there is a fluid level difference of nearly 2 m between the edge and center of the tank, and these large differences should be considered when designing rotating fluid tanks.
11-50 Milk is transported in a completely filled horizontal cylindrical tank accelerating at a specified rate. The maximum pressure difference in the tanker is to be determined. ƒ-EES
ax = – 3 m/s2
·
1
z
x
g
·
2
Assumptions 1 The acceleration remains constant. 2 Milk is an incompressible substance.
Properties The density of the milk is given to be 1020 kg/m3.
Analysis We take the x- and z- axes as shown. The horizontal acceleration is in the negative x direction, and thus ax is negative. Also, there is no acceleration in the vertical direction, and thus az = 0. The pressure difference between two points 1 and 2 in an incompressible fluid in linear rigid body motion is given by
®
The first term is due to acceleration in the horizontal direction and the resulting compression effect towards the back of the tanker, while the second term is simply the hydrostatic pressure that increases with depth. Therefore, we reason that the lowest pressure in the tank will occur at point 1 (upper front corner), and the higher pressure at point 2 (the lower rear corner). Therefore, the maximum pressure difference in the tank is
since x1 = 0, x2 = 7 m, z1 = 3 m, and z2 = 0.
Discussion Note that the variation of pressure along a horizontal line is due to acceleration in the horizontal direction while the variation of pressure in the vertical direction is due to the effects of gravity and acceleration in the vertical direction (which is zero in this case).
11-51 Milk is transported in a completely filled horizontal cylindrical tank decelerating at a specified rate. The maximum pressure difference in the tanker is to be determined. ƒ-EES
z
x
·
2
·
1
g
ax = 3 m/s2
Assumptions 1 The acceleration remains constant. 2 Milk is an incompressible substance.
Properties The density of the milk is given to be 1020 kg/m3.
Analysis We take the x- and z- axes as shown. The horizontal deceleration is in the x direction, and thus ax is positive. Also, there is no acceleration in the vertical direction, and thus az = 0. The pressure difference between two points 1 and 2 in an incompressible fluid in linear rigid body motion is given by
®
The first term is due to deceleration in the horizontal direction and the resulting compression effect towards the front of the tanker, while the second term is simply the hydrostatic pressure that increases with depth. Therefore, we reason that the lowest pressure in the tank will occur at point 1 (upper front corner), and the higher pressure at point 2 (the lower rear corner). Therefore, the maximum pressure difference in the tank is
since x1 = 7 m, x2 = 0, z1 = 3 m, and z2 = 0.
Discussion Note that the variation of pressure along a horizontal line is due to acceleration in the horizontal direction while the variation of pressure in the vertical direction is due to the effects of gravity and acceleration in the vertical direction (which is zero in this case).
11-52 A vertical U-tube partially filled with alcohol is rotated at a specified rate about one of its arms. The elevation difference between the fluid levels in the two arms is to be determined.
z
r
h0 = 20 cm
R = 25 cm
Assumptions 1 Alcohol is an incompressible fluid.
Analysis Taking the base of the left arm of the U-tube as the origin (r = 0, z = 0), the equation for the free surface of the liquid is given as
where h0 = 0.20 m is the original height of the liquid before rotation, and · = 4.2 rad/s. The fluid rise at the right arm relative to the fluid level in the left arm (the center of rotation) is
Discussion Note that the analysis is valid for any liquid since the result is independent of density or any other fluid property.
11-53 A vertical cylindrical tank is completely filled with gasoline, and the tank is rotated about its vertical axis at a specified rate. The pressures difference between the centers of the bottom and top surfaces, and the pressures difference between the center and the edge of the bottom surface are to be determined. ƒ-EES
h = 3 m
D = 1.20 m
z
r
Assumptions 1 The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2 Gasoline is an incompressible substance.
Properties The density of the gasoline is given to be 740 kg/m3.
Analysis The pressure difference between two points 1 and 2 in an incompressible fluid rotating in rigid body motion is given by
where R = 0.60 m is the radius, and
(a) Taking points 1 and 2 to be the centers of the bottom and top surfaces, respectively, we have and . Then,
(b) Taking points 1 and 2 to be the center and edge of the bottom surface, respectively, we have , , and . Then,
Discussion Note that the rotation of the tank does not affect the pressure difference along the axis of the tank. But the pressure difference between the edge and the center of the bottom surface (or any other horizontal plane) is due entirely to the rotation of the tank.
11-54 Problem 11-53 is reconsidered. The effect of rotational speed on the pressure difference between the center and the edge of the bottom surface of the cylinder as the rotational speed varies from 0 to 500 rpm in increments of 50 rpm is to be investigated.
g=9.81 “m/s2”
rho=740 “kg/m3”
R=0.6 “m”
h=3 “m”
omega=2*pi*n_dot/60 “rad/s”
DeltaP_axis=rho*g*h/1000 “kPa”
DeltaP_bottom=rho*omega^2*R^2/2000 “kPa”
Rotation rate
, rpm
Angular speed
·, rad/s
„Pcenter-edge
kPa
50
100
150
200
250
300
350
400
450
500
0.0
5.2
10.5
15.7
20.9
26.2
31.4
36.7
41.9
47.1
52.4
0.0
3.7
14.6
32.9
58.4
91.3
131.5
178.9
233.7
295.8
365.2
11-55E A water tank partially filled with water is being towed by a truck on a level road. The maximum acceleration (or deceleration) of the truck to avoid spilling is to be determined.
ax
„h = 2 ft
±
Water tank
hw = 6 ft
z
x
L=20 ft
Assumptions 1 The road is horizontal so that acceleration has no vertical component (az = 0). 2 Effects of splashing, breaking, driving over bumps, and climbing hills are assumed to be secondary, and are not considered. 3 The acceleration remains constant.
Analysis We take the x-axis to be the direction of motion, the z-axis to be the upward vertical direction. The shape of the free surface just before spilling is shown in figure. The tangent of the angle the free surface makes with the horizontal is given by
®
where az = 0 and, from geometric considerations, tan± is
Substituting,
The solution can be repeated for deceleration by replacing ax by – ax. We obtain ax = -6.44 m/s2.
Discussion Note that the analysis is valid for any fluid with constant density since we used no information that pertains to fluid properties in the solution.
11-56E A water tank partially filled with water is being towed by a truck on a level road. The maximum acceleration (or deceleration) of the truck to avoid spilling is to be determined.
ax
„h = 0.5 ft
±
Water tank
hw = 3 ft
z
x
L= 8 ft
Assumptions 1 The road is horizontal so that deceleration has no vertical component (az = 0). 2 Effects of splashing and driving over bumps are assumed to be secondary, and are not considered. 3 The deceleration remains constant.
Analysis We take the x-axis to be the direction of motion, the z-axis to be the upward vertical direction. The shape of the free surface just before spilling is shown in figure. The tangent of the angle the free surface makes with the horizontal is given by
®
where az = 0 and, from geometric considerations, tan± is
Substituting,
Discussion Note that the analysis is valid for any fluid with constant density since we used no information that pertains to fluid properties in the solution.
11-57 Water is transported in a completely filled horizontal cylindrical tanker accelerating at a specified rate. The pressure difference between the front and back ends of the tank along a horizontal line when the truck accelerates and decelerates at specified rates. ƒ-EES
ax = – 3 m/s2
z
x
1
·
2
g
·
Assumptions 1 The acceleration remains constant. 2 Water is an incompressible substance.
Properties We take the density of the water to be 1000 kg/m3.
Analysis (a) We take the x- and z- axes as shown. The horizontal acceleration is in the negative x direction, and thus ax is negative. Also, there is no acceleration in the vertical direction, and thus az = 0. The pressure difference between two points 1 and 2 in an incompressible fluid in linear rigid body motion is given by
®
since z2 – z1 = 0 along a horizontal line. Therefore, the pressure difference between the front and back of the tank is due to acceleration in the horizontal direction and the resulting compression effect towards the back of the tank. Then the pressure difference along a horizontal line becomes
since x1 = 0 and x2 = 7 m.
(b) The pressure difference during deceleration is determined the way, but ax = 4 m/s2 in this case,
Discussion Note that the pressure is higher at the back end of the tank during acceleration, but at the front end during deceleration (during breaking, for example) as expected.
Review Problems
11-58 The density of a wood log is to be measured by tying lead weights to it until both the log and the weights are completely submerged, and then weighing them separately in air. The average density of a given log is to be determined by this approach.
Properties The density of lead weights is given to be 11,300 kg/m3. We take the density of water to be 1000 kg/m3.
Analysis The weight of a body is equal to the buoyant force when the body is floating in a fluid while being completely submerged in it (a consequence of vertical force balance from static equilibrium). In this case the average density of the body must be equal to the density of the fluid since
Lead, 34 kg
Log, 1540 N
FB
Water
Therefore,
where
Substituting, the volume and density of the log are determined to be
Discussion Note that the log must be completely submerged for this analysis to be valid. Ideally, the lead weights must also be completely submerged, but this is not very critical because of the small volume of the lead weights.
11-59 A rectangular gate that leans against the floor with an angle of 45° with the horizontal is to be opened from its lower edge by applying a normal force at its center. The minimum force F required to open the water gate is to be determined.
Assumptions 1 The atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. 2 Friction at the hinge is negligible.
Properties We take the density of water to be 1000 kg/m3 throughout.
Analysis The length of the gate and the distance of the upper edge of the gate (point B) from the free surface in the plane of the gate are
FR
F
45°
B
0.5 m
3 m
A
The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the plate area gives the resultant hydrostatic on the surface,
The distance of the pressure center from the free surface of water along the plane of the gate is
The distance of the pressure center from the hinge at point B is
Taking the moment about point B and setting it equal to zero gives
Solving for F and substituting, the required force is determined to be
Discussion The applied force is inversely proportional to the distance of the point of application from the hinge, and the required force can be reduced by applying the force at a lower point on the gate.
11-60 A rectangular gate that leans against the floor with an angle of 45° with the horizontal is to be opened from its lower edge by applying a normal force at its center. The minimum force F required to open the water gate is to be determined.
Assumptions 1 The atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. 2 Friction at the hinge is negligible.
Properties We take the density of water to be 1000 kg/m3 throughout.
FR
F
45°
B
1.2 m
3 m
AAnalysis The length of the gate and the distance of the upper edge of the gate (point B) from the free surface in the plane of the gate are
The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the plate area gives the resultant hydrostatic on the surface,
The distance of the pressure center from the free surface of water along the plane of the gate is
The distance of the pressure center from the hinge at point B is
Taking the moment about point B and setting it equal to zero gives
Solving for F and substituting, the required force is determined to be
Discussion The applied force is inversely proportional to the distance of the point of application from the hinge, and the required force can be reduced by applying the force at a lower point on the gate.
11-61 A rectangular gate hinged about a horizontal axis along its upper edge is restrained by a fixed ridge at point B. The force exerted to the plate by the ridge is to be determined.
Assumptions The atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience.
FR
3 m
A
2 m
ypProperties We take the density of water to be 1000 kg/m3 throughout.
Analysis The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the plate area gives the resultant hydrostatic force on the gate,
The vertical distance of the pressure center from the free surface of water is
11-62 A rectangular gate hinged about a horizontal axis along its upper edge is restrained by a fixed ridge at point B. The force exerted to the plate by the ridge is to be determined.
Assumptions The atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience.
FR
3 m
yP
h = 2 m
A
Properties We take the density of water to be 1000 kg/m3 throughout.
Analysis The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the wetted plate area gives the resultant hydrostatic force on the gate,
The vertical distance of the pressure center from the free surface of water is
11-63E A semicircular tunnel is to be built under a lake. The total hydrostatic force acting on the roof of the tunnel is to be determined.
Assumptions The atmospheric pressure acts on both sides of the tunnel, and thus it can be ignored in calculations for convenience.
Properties We take the density of water to be 62.4 lbm/ft3 throughout.
Analysis We consider the free body diagram of the liquid block enclosed by the circular surface of the tunnel and its vertical (on both sides) and horizontal projections. The hydrostatic forces acting on the vertical and horizontal plane surfaces as well as the weight of the liquid block are determined as follows:
Horizontal force on vertical surface (each side):
Fy
W
Fx
Fx
Vertical force on horizontal surface (downward):
R = 15 ft
Weight of fluid block on each side within the control volume (downward):
Therefore, the net downward vertical force is
This is also the net force acting on the tunnel since the horizontal forces acting on the right and left side of the tunnel cancel each other since they are equal ad opposite.
11-64 A hemispherical dome on a level surface filled with water is to be lifted by attaching a long tube to the top and filling it with water. The required height of water in the tube to lift the dome is to be determined.
Assumptions 1 The atmospheric pressure acts on both sides of the dome, and thus it can be ignored in calculations for convenience. 2 The weight of the tube and the water in it is negligible.
Properties We take the density of water to be 1000 kg/m3 throughout.
Analysis We take the dome and the water in it as the system.  When the dome is about to rise, the reaction force between the dome and the ground becomes zero. Then the free body diagram of this system involves the weights of the dome and the water, balanced by the hydrostatic pressure force from below. Setting these forces equal to each other gives
FV
R = 3 m
h
W
Solving for h gives
Substituting,
Therefore, this dome can be lifted by attaching a tube which is 2.02 m long.
Discussion This problem can also be solved without finding FR by finding the lines of action of the horizontal hydrostatic force and the weight.
11-65 The water in a reservoir is restrained by a triangular wall. The total force (hydrostatic + atmospheric) acting on the inner surface of the wall and the horizontal component of this force are to be determined.
FR
h = 25 m
ypAssumptions 1 The atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. 2 Friction at the hinge is negligible.
Properties We take the density of water to be 1000 kg/m3 throughout.
Analysis The length of the wall surface underwater is
The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the plate area gives the resultant hydrostatic force on the surface,
Noting that
the distance of the pressure center from the free surface of water along the wall surface is
The magnitude of the horizontal component of the hydrostatic force is simply FRsin ±,
Discussion The atmospheric pressure is usually ignored in the analysis for convenience since it acts on both sides of the walls.
11-66 A U-tube that contains water in right arm and another liquid in the left is rotated about an axis closer to the left arm. For a known rotation rate at which the liquid levels in both arms are the same, the density of the fluid in the left arm is to be determined.
1*
·
·
1
Fluid
Water
h = 10 cm
z
·
2
r
R1 = 5 cm
R2 = 15 cm
Assumptions 1 Both the fluid and the water are incompressible fluids. 2 The two fluids meet at the axis of rotation, and thus there is only water to the right of the axis of rotation.
Properties We take the density of water to be 1000 kg/m3.
Analysis The pressure difference between two points 1 and 2 in an incompressible fluid rotating in rigid body motion (the same fluid) is given by
where (for both arms of the U-tube).
Pressure at point 2 is the same for both fluids, so are the pressures at points 1 and 1* (P1 = P1* = Patm). Therefore, is the same for both fluids. Noting that for both fluids and expressing for each fluid,
Water:
Fluid:
Setting them equal to each other and solving for ²f gives
Discussion Note that this device can be used to determine relative densities, though it wouldn’t be a very practical.
11-67 A vertical cylindrical tank is completely filled with gasoline, and the tank is rotated about its vertical axis at a specified rate while being accelerated upward. The pressures difference between the centers of the bottom and top surfaces, and the pressures difference between the center and the edge of the bottom surface are to be determined. ƒ-EES
5 m/s2
h = 2 m
D = 1 m
z
r
Assumptions 1 The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2 Gasoline is an incompressible substance.
Properties The density of the gasoline is given to be 740 kg/m3.
Analysis The pressure difference between two points 1 and 2 in an incompressible fluid rotating in rigid body motion is given by . The effect of linear acceleration in the vertical direction is accounted for by replacing g by . Then,
where R = 0.50 m is the radius, and
(a) Taking points 1 and 2 to be the centers of the bottom and top surfaces, respectively, we have and . Then,
(b) Taking points 1 and 2 to be the center and edge of the bottom surface, respectively, we have , , and . Then,
Discussion Note that the rotation of the tank does not affect the pressure difference along the axis of the tank. Likewise, the vertical acceleration does not affect the pressure difference between the edge and the center of the bottom surface (or any other horizontal plane).
11-68 A rectangular water tank open to the atmosphere is accelerated to the right on a level surface at a specified rate. The maximum pressure in the tank above the atmospheric level is to be determined. ƒ-EES
L =5 m
ax = 2 m/s2
h0 =2.5 m
1.5 m
Water tank
Vent
±
·
1
·
2
Assumptions 1 The road is horizontal during acceleration so that acceleration has no vertical component (az = 0). 2 Effects of splashing, breaking and driving over bumps are assumed to be secondary, and are not considered. 3 The vent is never blocked, and thus the minimum pressure is the atmospheric pressure.
Properties We take the density of water to be 1000 kg/m3.
Analysis We take the x-axis to be the direction of motion, the z-axis to be the upward vertical direction. The tangent of the angle the free surface makes with the horizontal is
(and thus ± = 11.5°)
The maximum vertical rise of the free surface occurs at the back of the tank, and the vertical midsection experiences no rise or drop during acceleration. Then the maximum vertical rise at the back of the tank relative to the neutral midplane is
which is less than 1.5 m high air space. Therefore, water never reaches the ceiling, and the maximum water height and the corresponding maximum pressure are
Discussion It can be shown that the gage pressure at the bottom of the tank varies from 29.5 kPa at the back of the tank to 24.5 kPa at the midsection and 19.5 kPa at the front of the tank.
11-69 Problem 11-68 is reconsidered. The effect of acceleration on the slope of the free surface of water in the tank as the acceleration varies from 0 to 5 m/s2 in increments of 0.5 m/s2 is to be investigated.
g=9.81 “m/s2”
rho=1000 “kg/m3”
L=5 “m”
h0=2.5 “m”
a_z=0
tan(theta)=a_x/(g+a_z)
h_max=h0+(L/2)*tan(theta)
P_max=rho*g*h_max/1000 “kPa”
Acceleration
ax, m/s2
Free surface angle, ±°
Maximum height
hmax, m
Maximum pressure
Pmax, kPa
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
0.0
2.9
5.8
8.7
11.5
14.3
17.0
19.6
22.2
24.6
27.0
2.50
2.63
2.75
2.88
3.01
3.14
3.26
3.39
3.52
3.65
3.77
24.5
25.8
27.0
28.3
29.5
30.8
32.0
33.3
34.5
35.8
37.0
Note that water never reaches the ceiling, and a full free surface is formed in the tank.
11-70 An elastic air balloon submerged in water is attached to the base of the tank. The change in the tension force of the cable is to be determined when the tank pressure is increased and the balloon diameter is decreased in accordance with the relation P = CD-2.
Water
P1=100 kPa
D1=30 cm
Assumptions 1 The atmospheric pressure acts on all surfaces, and thus it can be ignored in calculations for convenience. 2 Water is an incompressible fluid. 3 The weight of the balloon and the air in it is negligible.
Properties We take the density of water to be 1000 kg/m3.
Analysis The tension force on the cable holding the balloon is determined from a force balance on the balloon to be
The boyuancy force acting on the ballon initially is
The variation of pressure with diameter is given as , which is equivalent to . Then the final diameter of the ball becomes
The bouyancy force acting on the balloon in this case is
Then the percent change in the cable for becomes
.
Therefore, increasing the tank pressure in this case results in 98.4% reduction in cable tension.
Discussion We can obtain a relation for the change in cable tension as follows:
.
11-71 Problem 11-70 is reconsidered. The effect of air pressure above water on the cable force as the pressure varies from 0.1 MPa to 20 MPa is to be investigated.
P1=0.1 “MPa”
Change=100*(1-(P1/P2)^1.5)
Tank pressure
P2, MPa
%Change in cable tension
0.1
0.2
0.3
0.4
0.6
0.8
1
2
3
4
5
6
7
8
9
10
0.0
64.6
80.8
87.5
93.2
95.6
96.8
98.9
99.4
99.6
99.7
99.8
99.8
99.9
99.9
99.9
11-72 An iceberg floating in seawater is considered. The volume fraction of iceberg submerged in seawater is to be determined, and the reason for their turnover is to be explained.
Assumptions 1 The buoyancy force in air is negligible. 2 The density of iceberg and seawater are uniform.
Properties The densities of iceberg and seawater are given to be 917 kg/m3 and 1042 kg/m3, respectively.
Analysis (a) The weight of a body floating in a fluid is equal to the buoyant force acting on it (a consequence of vertical force balance from static equilibrium). Therefore,
Sea
FB
W
Iceberg
W = FB
Therefore, 88% of the volume of the iceberg is submerged in this case.
(b) Heat transfer to the iceberg due to the temperature difference between the seawater and an iceberg causes uneven melting of the irregularly shaped iceberg. The resulting shift in the center of mass causes turn over.
Discussion Note that the submerged fraction depends on the density of seawater, and this fraction can different in different seas.
11-73 A cylindrical container equipped with a manometer is inverted and pressed into water. The differential height of the manometer and the force needed to hold the container in place are to be determined. ƒ-
air
h
Manometer fluid
SG=2.1
20 cm
D = 30 cm
F
water
A
B
·
·
Assumptions 1 The atmospheric pressure acts on all surfaces, and thus it can be ignored in calculations for convenience. 2 The variation of air pressure inside cylinder is negligible.
Properties We take the density of water to be 1000 kg/m3. The density of the manometer fluid is
Analysis The pressures at point A and B must be the same since they are on the same horizontal line in the same fluid. Then the gage pressure in the cylinder becomes
The manometer also indicates the gage pressure in the cylinder. Therefore,
A force balance on the cylinder in the vertical direction yields
Solving for F and substituting,
Discussion We could also solve this problem by considering the atmospheric pressure, but we would obtain the same result since atmospheric pressure would cancel out.
11-74 … 11-75 Design and Essay Problems
11-75 The volume of a rock can be determined without using any volume measurement devices as follows: We weigh the rock in the air and then in the water. The difference between the two weights is due to the buoyancy force, which is equal to . Solving this relation for Vbody gives the volume of the rock.
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