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Pengalaman Belajar Fisika di SMAN BI 1 Banjar

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Experiment on Fluids:
Finding the Velocity of a Fluid in a Confined Container

SUBJECT: Aeronautics
TOPIC: Fluid Velocity
DESCRIPTION: A set of mathematics problems dealing with fluid velocity.
CONTRIBUTED BY: Carol Hodanbosi
EDITED BY: Jonathan G. Fairman - August 1996

Purpose:

To calculate the velocity of a confined fluid, given the cross-section area and velocity of another region.

Concept:

The drawing below is a cross-section of a circular cone attached to a circular cylinder.

When a fluid (liquid or gas) is in a confined space, with no change in pressure or temperature, one can use the equation of continuity to find the velocity of the fluid if one knows the cross-section area and velocity in one of the regions. The formula for this is A1*V1 = A2*V2, where A is the cross-section area of one location and V is the velocity for that location.

Analysis:

Given three different locations in a confined container, A, B, and C, all having different radii, can you find the other two velocities of the fluid, if the velocity at A is given?

1. If the cross-section at A has a radius of 6 meters, can you find the area of the slice through the cone
   (Area = pi * r ²)?
   (answer)
2. If the velocity of the fluid at location A is 10.0 m/s, and the radius at location B is 4.2 meters, can you find the velocity at location B?
   (answer)
3. If the velocity at location C is 8.6 m/s, can you find the radius at location C?
   (answer) Extension:
4. If the radius of a fourth location, D, is one-half the radius of A , how would the velocity at location D compare to the velocity of the fluid at A? If D had one-third the radius of A, compare the velocity of the fluid at D to A. Explain your reasoning showing calculations.
   (answer)

Sumber:

NASA

Fisika SMA

Pengalaman Belajar Fisika di SMAN BI 1 Banjar

Fluids Pressure and Depth

SUBJECT: Aeronautics
TOPIC: Hydrostatic Pressure
DESCRIPTION: A set of mathematics problems dealing with hydrostatics.
CONTRIBUTED BY: Carol Hodanbosi
EDITED BY: Jonathan G. Fairman - August 1996

A fluid is a substance that flows easily. Gases and liquids are fluids, although sometimes the dividing line between liquids and solids is not always clear. Because of their ability to flow, fluids can exert buoyant forces, multiply forces in a hydraulic systems, allow aircraft to fly and ships to float.

The topic that this page will explore will be pressure and depth. If a fluid is within a container then the depth of an object placed in that fluid can be measured. The deeper the object is placed in the fluid, the more pressure it experiences. This is because is the weight of the fluid above it. The more dense the fluid above it, the more pressure is exerted on the object that is submerged, due to the weight of the fluid.

The formula that gives the P pressure on an object submerged in a fluid is:

P = r * g * h

where

• r (rho) is the density of the fluid,
• g is the acceleration of gravity
• h is the height of the fluid above the object

If the container is open to the atmosphere above, the added pressure must be included if one is to find the total pressure on an object. The total pressure is the same as absolute pressure on pressure gauges readings, while the gauge pressure is the same as the fluid pressure alone, not including atmospheric pressure.

P_total = P_atmosphere + P_fluid

P_total = P_atmosphere + ( r * g * h )

A Pascal is the unit of pressure in the metric system. It represents 1 newton/m²

Example:
Find the pressure on a scuba diver when she is 12 meters below the surface of the ocean. Assume standard atmospheric conditions.

Solution:
The density of sea water is 1.03 X 10 ³ kg/m³ and the atmospheric pressure is 1.01 x 10⁵ N/m².

P_fluid = r g h = (1.03 x10 ³ kg/m³) (9.8 m/s²) (12 m) = 1.21 x 10⁵ Newtons/m² P_total = P_atmosphere + P_fluid = (1.01 x 10⁵) + (1.21 x 10⁵ ) Pa = 2.22 x 10 ² kPa (kilo Pascals)

Exercises :

What is the pressure experienced at a point on the bottom of a swimming pool 9 meters in depth? The density of water is 1.00 x 10³ kg/m³.
(answer)

The interior of a submarine located at a depth of 45 meters is maintained at normal atmospheric conditions. Find the total force exerted on a 20 cm by 20 cm square window. Use the density of sea water given above.
(answer)

How many atmospheres is a depth of 100 meters of ocean water?
(answer)

If the weight density of pure water is 62 pounds/ft³, find the weight of water in a swimming pool whose dimensions are 20 ft by 10 ft by 6 feet.
(answer)

An airplane in level flight whose mass is 20,000 kg has a wing area of 60 m². What is the pressure difference between the upper and lower surfaces of its wing? Express your answer in atmospheres.
(answer)

Sumber:

NASA

Fisika SMA

Pengalaman Belajar Fisika di SMAN BI 1 Banjar

Ada Apa Dengan Fisika?

Pascal's Principle and Hydraulics

SUBJECT: Physics
TOPIC: Hydraulics
DESCRIPTION: A set of mathematics problems dealing with hydraulics.
CONTRIBUTED BY: Carol Hodanbosi
EDITED BY: Jonathan G. Fairman - August 1996

Hydraulic systems use a incompressible fluid, such as oil or water, to transmit forces from one location to another within the fluid. Most aircraft use hydraulics in the braking systems and landing gear. Pneumatic systems use compressible fluid, such as air, in their operation. Some aircraft utilize pneumatic systems for their brakes, landing gear and movement of flaps.

Pascal's law states that when there is an increase in pressure at any point in a confined fluid, there is an equal increase at every other point in the container.

A container, as shown below, contains a fluid. There is an increase in pressure as the length of the column of liquid increases, due to the increased mass of the fluid above.

For example, in the figure below, P3 would be the highest value of the three pressure readings, because it has the highest level of fluid above it.

If the above container had an increase in overall pressure, that same added pressure would affect each of the gauges (and the liquid throughout) the same. For example P1, P2, P3 were originally 1, 3, 5 units of pressure, and 5 units of pressure were added to the system, the new readings would be 6, 8, and 10.

Applied to a more complex system below, such as a hydraulic car lift, Pascal's law allows forces to be multiplied. The cylinder on the left shows a cross-section area of 1 square inch, while the cylinder on the right shows a cross-section area of 10 square inches. The cylinder on the left has a weight (force) on 1 pound acting downward on the piston, which lowers the fluid 10 inches. As a result of this force, the piston on the right lifts a 10 pound weight a distance of 1 inch.

The 1 pound load on the 1 square inch area causes an increase in pressure on the fluid in the system. This pressure is distributed equally throughout and acts on every square inch of the 10 square inch area of the large piston. As a result, the larger piston lifts up a 10 pound weight. The larger the cross-section area of the second piston, the larger the mechanical advantage, and the more weight it lifts.

The formulas that relate to this are shown below:

P1 = P2 (since the pressures are equal throughout).

Since pressure equals force per unit area, then it follows that

F1/A1 = F2/A2

It can be shown by substitution that the values shown above are correct,

1 pound / 1 square inches = 10 pounds / 10 square inches

Because the volume of fluid pushed down on the left side equals the volume of fluid that is lifted up on the right side, the following formula is also true.

V1 = V2

by substitution,

A1 D1 = A2 D2

• A = cross sectional area
• D = the distance moved

or

A1/A2= D2/D1

This system can be thought of as a simple machine (lever), since force is multiplied.The mechanical advantage can be found by rearranging terms in the above equation to

Mechanical Advantage(IMA) = D1/D2 = A2/A1

For the sample problem above, the IMA would be 10:1 (10 inches/ 1 inch or 10 square inches / 1 square inch).

Given these simple formulas, try to answer the questions below.

Exercises:

1. A hydraulic press has an input cylinder 1 inch in diameter and an output cylinder 6 inches in diameter.
   1. Assuming 100% efficiency, find the force exerted by the output piston when a force of 10 pounds is applied to the input piston.
      (answer)
      
      
   2. If the input piston is moved through 4 inches, how far is the output piston moved?
      (answer)
      
      
2. A hydraulic system is said to have a mechanical advantage of 40. Mechanical advantage (MA) is FR (output) / FE (input). If the input piston, with a 12 inch radius, has a force of 65 pounds pushing downward a distance of 20 inches, find
   
   
   1. the volume of fluid that has been displaced
      (answer)
      
      
   2. the upward force on the output piston
      (answer)
      
      
   3. the radius of the output piston
      (answer)
      
      
   4. the distance the output piston moves
      (answer)
      
      
      
3. What pressure does a 130 pound woman exert on the floor when she balances on one of her heels? Her heels have an average radius of 0.5 inch.
   (answer)
   
   
4. A car has a weight of 2500 pounds and rests on four tires, each having a surface area of contact with the ground of 14 square inches. What is the pressure the ground experiences beneath the tires that is due to the car?
   (answer)
   
   Extension :
   
5. The input and output pistons of a hydraulic jack are respectively 1 cm and 4 cm in diameter. A lever with a mechanical advantage of 6 is used to apply force to the input piston. How much mass can the jack lift if a force of 180 N is applied to the lever and efficiency is 80%?
   (answer)

Sumber:

NASA