Buoyancy and stability in fluid mechanics | Summaries Fluid Mechanics

UNIT 5

BUOYANCY AND STABILITY

Whenever an object is floating in a fluid or when it is completely submerged in the fluid, it is subjected

to a buoyant force that tends to lift it upward, helping to support it. Buoyancy is the tendency of a fluid

to exert a supporting force on a body placed in the fluid. There is need to understand the concept of

buoyancy and make calculations to determine the net forces exerted on objects immersed in fluids or

the position of an object when it is floating. Also, the need to learn about the stability of floating or

submerged bodies to ensure that they will stay in the preferred orientation even when subjected to

external forces that tend to tip them over.

Stability the ability of a body in a fluid to return to its original position after being tilted about a

horizontal axis.

BUOYANCY

A body in a fluid, whether floating or submerged, is buoyed up by a force equal to the weight of the

fluid displaced. The buoyant force acts vertically upward through the centroid of the displaced volume.

𝑭𝒃= 𝜸𝒇𝒗𝒅 (1)

Where

𝐹𝑏= Buoyant force

𝛾𝑓= Specific weight of the fluid

𝑣𝑑= Displaced volume of the fluid

When a body is floating freely, it displaces a sufficient volume of fluid to just balance its own weight.

The analysis of problems dealing with buoyancy requires the application of the equation of static

equilibrium in the vertical direction, ∑𝑭𝒗 = 0, assuming the object is at rest in the fluid. The following

procedure is recommended for all problems, whether they involve floating or submerged bodies.

Procedure for Solving Buoyancy Problems

1. Determine the objective of the problem solution. Do you want to determine a force, a weight,

a volume, or a specific weight?

2. Draw a free-body diagram of the object in the fluid. Show all forces that act on the free body

in the vertical direction, including the weight of the body, the buoyant force, and all external

forces. If the direction of some force is not known, assume the most probable direction and

show it on the free body.

3. Write the equation of static equilibrium in the vertical direction, ∑𝑭𝒗 = 0, assuming the positive

direction to be upward.

4. Solve for the desired force, weight, volume, or specific weight, remembering the following

concepts:

a. The buoyant force is calculated from Fb = 𝜸fVd.

b. The weight of a solid object is the product of its total volume and its specific weight;

that is, w = 𝜸V.

c. An object with an average specific weight less than that of the fluid will tend to float

because w < Fb with the object submerged.

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UNIT 5

BUOYANCY AND STABILITY

Whenever an object is floating in a fluid or when it is completely submerged in the fluid, it is subjected to a buoyant force that tends to lift it upward, helping to support it. Buoyancy is the tendency of a fluid to exert a supporting force on a body placed in the fluid. There is need to understand the concept of buoyancy and make calculations to determine the net forces exerted on objects immersed in fluids or the position of an object when it is floating. Also, the need to learn about the stability of floating or submerged bodies to ensure that they will stay in the preferred orientation even when subjected to external forces that tend to tip them over. Stability the ability of a body in a fluid to return to its original position after being tilted about a horizontal axis. BUOYANCY A body in a fluid, whether floating or submerged, is buoyed up by a force equal to the weight of the fluid displaced. The buoyant force acts vertically upward through the centroid of the displaced volume. 𝑭𝒃 = 𝜸𝒇𝒗𝒅 (1) Where 𝐹𝑏= Buoyant force 𝛾𝑓= Specific weight of the fluid 𝑣𝑑= Displaced volume of the fluid When a body is floating freely, it displaces a sufficient volume of fluid to just balance its own weight. The analysis of problems dealing with buoyancy requires the application of the equation of static equilibrium in the vertical direction, ∑^ 𝑭𝒗 = 0 , assuming the object is at rest in the fluid. The following procedure is recommended for all problems, whether they involve floating or submerged bodies. Procedure for Solving Buoyancy Problems

Determine the objective of the problem solution. Do you want to determine a force, a weight, a volume, or a specific weight?
Draw a free-body diagram of the object in the fluid. Show all forces that act on the free body in the vertical direction, including the weight of the body, the buoyant force, and all external forces. If the direction of some force is not known, assume the most probable direction and show it on the free body.
Write the equation of static equilibrium in the vertical direction, ∑^ 𝑭𝒗 = 0 , assuming the positive direction to be upward.
Solve for the desired force, weight, volume, or specific weight, remembering the following concepts: a. The buoyant force is calculated from Fb = 𝜸 fVd. b. The weight of a solid object is the product of its total volume and its specific weight; that is, w = 𝜸 V. c. An object with an average specific weight less than that of the fluid will tend to float because w < Fb with the object submerged.

d. An object with an average specific weight greater than that of the fluid will tend to sink because w > Fb with the object submerged. e. Neutral buoyancy occurs when a body stays in a given position wherever it is submerged in a fluid. An object whose average specific weight is equal to that of the fluid is neutrally buoyant. Example Problem 1: A cube 0. 6 0 m on a side is made of bronze having a specific weight of 8 8. kN/m^3. Determine the magnitude and direction of the force required to hold the cube in equilibrium when completely submerged (a) in water and (b) in mercury. The specific gravity of mercury is 13.54. Solution Consider part (a) first. Imagine the cube of bronze submerged in water do Step 1 of the procedure Assuming the bronze cube will not stay in equilibrium by itself, some external force is required. The reason is to find the magnitude of this force and the direction in which it would act that is, up or down. Step 2 Draw a free body diagram of the cube, showing the three forces acting on the cube in the vertical direction as shown in fig. 1

The weight of the cube w, acting downward through its center of gravity
The buoyant force Fb, acting upward through the centroid of the displaced volume
The externally applied supporting force Fe (a)Force acting on the cube (b) Two-dimensional body-free diagram Figure 1 Free-body diagram of a cube Step 3 Assuming positive force is acting upward, then the equation will be ∑ (^) 𝐹𝑣 = 0 𝐹𝑏 + 𝐹𝑒 − 𝑤 = 0 (2)

The following are the correct equations of equilibrium. Notice the differences and relate them to the figures: 𝐹𝑏 + 𝐹𝑒 − 𝑤 = 0 ∕ 𝐹𝑏 −𝐹𝑒 − 𝑤 = 0 Now, solve for Fe. 𝐹𝑒 = 𝑤 − 𝐹𝑏 ∕ 𝐹𝑒 = 𝐹𝑏 − 𝑤 w =19.20 kN Fb = 𝛾mVd 𝛾m= 13. 54 x 9.81 =13 2. 83 kN/m^3 Fb = 𝛾mVd = 132.83 x 10^3 x 0.216 =28.69 kN 𝐹𝑒 = 𝑤 − 𝐹𝑏 ∕ 𝐹𝑒 = 𝐹𝑏 − 𝑤 Fe = 19.20 − 28. 69 = − 9. 49 𝑘𝑁 ∕ Fe =28.69 – 19.20 = +9.49 kN Notice that both solutions yield the same numerical value, but they have opposite signs. The negative sign for the solution on the left means that the assumed direction for Fe in Fig. 2 (a) was wrong. Therefore, both approaches give the same result. The required external force is a downward force of 9.49 kN Because the specific weight of the cube is less than that of the mercury, it would tend to float without an external force. Therefore, a downward force, as pictured in Fig. 2 (b), would be required to hold it in equilibrium under the surface of the mercury. Problem 2: A brass cube 0. 8 m on a side weighs 900 N. We want to hold this cube in equilibrium under water by attaching a light foam buoy to it. If the foam weighs 3. 64 kN/m^3 , what is the minimum required volume of the buoy? Solution Figure 3 Free-body diagram for brass and foam together

Calculate the minimum volume of foam to hold the brass cube in equilibrium The equilibrium equation is ∑ (^) 𝐹𝑣 = 0 𝑭𝒃𝑩 + 𝑭𝒃𝑭 − 𝒘𝑩 − 𝒘𝑭 = 𝟎 𝑤𝐵 = 900 N 𝛾𝐹 = 3. 64 𝑘𝑁/𝑚^3 𝛾𝑤 = 9. 81 𝑘𝑁/𝑚^3 𝑉𝑑𝐵 = 0. 8 × 0. 8 × 0. 8 = 0. 512 𝑚^3 𝐹𝑏𝐵 = 𝛾𝑤𝑉𝑑𝐵 = 9.81 x 10^3 x 0 .512 = 5022.7 = 5 .02 kN 𝑤𝐹 = 𝛾𝐹 𝑉𝑑𝐹 𝐹𝑏𝐹 = 𝛾𝑤𝑉𝑑𝐹 FBb + FBf − wB − wF = 0

7 + 𝛾𝑤𝑉𝑑𝐹 − 900 − 𝛾𝐹 𝑉𝑑𝐹 = 0 𝛾𝑤𝑉𝑑𝐹 − 𝛾𝐹𝑉𝑑𝐹 = 5022. 7 − 900 (𝛾𝑤 − 𝛾𝐹 )𝑉𝑑𝐹 = 4122. 7 𝑉𝑑𝐹 =
7 (𝛾𝑤−𝛾𝐹) =^
7 ( 9. 81 − 3. 64 ) 103 =^ 0.^668 m 3 This means that if 0. 668 m^3 of foam were attached to the brass cube, the combination would be in equilibrium in water without any external force. It would be neutrally buoyant. BUOYANCY MATERIALS The design of floating bodies often requires the use of Lightweight materials that offer a high degree of buoyancy. In addition, when a relatively heavy object must be moved while submerged in a fluid, it is often desirable to add buoyancy to facilitate mobility. The buoyancy material should typically have the following properties: a. Low specific weight and density b. Little or no tendency to absorb the fluid c. Compatibility with the fluid in which it will operate d. Ability to be formed to appropriate shapes e. Ability to withstand fluid pressures to which it will be subjected f. resistance and damage tolerance g. Attractive appearance Foam materials are popular for buoyancy applications. They are made up of a continuous network of closed, hollow cells that contain air or other light gases to yield the low specific weight. The closed cells also ensure that the fluid is not absorbed. The following tests are performed to evaluate the performance of foams: density, tensile strength, tensile elongation, tear strength, compression set, compressive deflection, thermal stability, thermal conductivity, and water absorption. The details of the tests are prescribed in ASTM D 3575, Standard Test Methods for Flexible Cellular Materials Made from Olefin Polymers. Other standards apply to other materials.

body. Figure 5(b) shows that if the body is rotated slightly, the center of buoyancy shifts to a new position because the geometry of the displaced volume has changed. The buoyant force and the weight now produce a righting couple that tends to return the body to its original orientation. Thus, the body is stable. To state the condition for stability of a floating body, we must define a new term, metacenter. The metacenter ( mc ) is defined as the intersection of the vertical axis of a body when in its equilibrium position and a vertical line through the new position of the center of buoyancy when the body is rotated slightly. This is illustrated in Fig. 5(b). (a) Original position (b) Tilted position Figure 5 Method of finding the metacenter Condition of Stability for Floating Bodies A floating body is stable if its center of gravity is below the metacenter It is possible to determine analytically if a floating body is stable by calculating the location of its metacenter. The distance to the metacenter from the center of buoyancy is called MB and is calculated from MB= 𝐼 𝑉𝑑^ (4) In this equation, Vd is the displaced volume of fluid and I is the least moment of inertia of a horizontal section of the body taken at the surface of the fluid. If the distance MB places the metacenter above the center of gravity, the body is stable. Procedure for Evaluating the Stability of Floating Bodies

Determine the position of the floating body, using the principles of buoyancy.
Locate the center of buoyancy, cb ; compute the distance from some reference axis to cb , called ycb · Usually, the bottom of the object is taken as the reference axis.
Locate the center of gravity, cg ; compute ycg measured from the same reference axis.
Determine the shape of the area at the fluid surface and compute the smallest moment of inertia I for that shape.
Compute the displaced volume Vd.
Compute MB = 𝐼 𝑉𝑑

Compute ymc = ycb + MB.
If ymc > ycg,(mc greater than cg) the body is stable.
If ymc < ycg, (mc less than cg) the body is unstable.

Problem 3: A solid cylinder is 6 m in diameter, 10 m high, and weighs 1500 kN. If the cylinder is placed in oil (sg = 0.90) with its axis vertical, would it be stable or unstable? Solution Position of cylinder in oil 𝑉𝑑 = 𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑 𝑣𝑜𝑙𝑢𝑚𝑒 = 𝐴𝑋 =

𝜋𝐷^2

Equilibrium equation. ∑ (^) 𝐹𝑣 = 0 𝑤 = 𝐹𝑏 = 𝛾𝑜𝑣𝑑=𝛾𝑜 𝜋𝐷^2 4

𝜋𝐷^2 𝛾𝑜

4 × 1500 × 103

𝜋 62 × 0. 90 × 9. 81 × 103

The center of buoyancy cb is at a distance 𝑋 2 from the bottom of the cylinder. 𝑐𝑏 =

01 2 =^3.^01 𝑚 The center of gravity cg is at 𝐻 2 = 5 m from the bottom of the cylinder, assuming the material of the cylinder is of uniform specific weight. The position of the metacenter mc, MB = 𝐼 𝑉𝑑 𝐼 =

Buoyancy and stability in fluid mechanics, Summaries of Fluid Mechanics

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UNIT 5

BUOYANCY AND STABILITY

𝜋𝐷^2

𝜋𝐷^2 𝛾𝑜

4 × 1500 × 103

𝜋 62 × 0. 90 × 9. 81 × 103

𝜋𝐷^4

= 63. 62 𝑚^4

𝜋𝐷^2

× 6. 01 = 28. 27 × 6. 01 = 169. 93 𝑚^3

MB =