Uploaded by nethravathi sasikumar

fluid mechanics problems

advertisement
Fluid Mechanic
Ammar Ali Abd
Email: [email protected]
Hydrostatic pressure
Consider a tank of fluid which contains a very thin plate of (neutrally buoyant)
material with area A. This situation is shown in Figure below. If the plate is in
equilibrium (it does not start to move), then the forces on it must be balanced. On the
top side of the plate acts the weight of the fluid, with density ρ, above it in the
downward direction. Because it does not accelerate, there must be a force acting
upwards on the underside of the plate - this is the effect of the hydrostatic pressure
acting on the plate.
The weight of the fluid above the plate is:
𝐹𝑀 = 𝑑𝑒𝑛𝑠𝑖𝑑𝑦 βˆ— π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ 𝑓𝑙𝑒𝑖𝑑 βˆ— π‘Žπ‘π‘π‘’π‘™π‘Žπ‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›
Balancing this is the hydrostatic pressure, p, acting over the area, A, of the
underside of the plate depicted in Figure. Balancing the two forces yields:
𝑃𝐴 = πœŒπ‘”π΄β„Ž
Cancelling A gives the expression for the hydrostatic pressure as:
𝑃𝐴 = πœŒπ‘”π΄β„Ž , 𝑖𝑛
𝑁
(π‘ƒπ‘Žπ‘ π‘π‘Žπ‘™π‘ ) β†’ πΈπ‘ž. 1
π‘š2
Noting that h is the vertical depth. You can now see the relationship between
hydrostatic pressure and gravitational forces.
If we evaluated the expression above, we would not find the total pressure acting on
the plate. This is because, using the same arguments, there is a column of atmospheric
air above the water in the tank and this presses down on the water which, in turn,
transmits the additional pressure onto the plate. If we denote the air pressure at the
liquid surface PA (again, in N/m2), then the total pressure at a depth h is:
π‘ƒπ‘Žπ‘π‘  = πœŒπ‘”β„Ž + 𝑃𝐴 β†’ πΈπ‘ž. 2
This is called the absolute pressure whilst the pressure of Eq. 1 is called the gauge
pressure, i.e. the pressure relative to atmospheric pressure.
Water Resources Engineering College – Al-Qasim Green University
Fluid Mechanic
Ammar Ali Abd
Email: [email protected]
Pressure modes
If a vessel were to be completely empty, containing no molecules whatsoever the
pressure would be zero. When you are using this zero as the pressure reference point
it is called absolute pressure because there is no lower pressure than the absence of all
molecules.
There are many applications were measuring pressure is not really dependent on the
absolute pressure but the difference between it and the pressure of the atmosphere.
When using the atmospheric pressure as the reference point we call the mode gauge
pressure. The classic example is a tire, on a typical tire the pressure to be 30 psi (207
kPa) above atmospheric pressure as 0 gauge pressure would mean a flat tire, even
though there is technically still atmospheric air pressure in it. The difference between
the absolute pressure and gauge pressure value is the variable value of atmospheric
pressure:
π΄π‘π‘ π‘œπ‘™π‘’π‘‘π‘’ π‘ƒπ‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’ = π‘”π‘Žπ‘’π‘”π‘’ π‘ƒπ‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’ + π‘Žπ‘‘π‘šπ‘œπ‘ π‘β„Žπ‘’π‘Ÿπ‘–π‘ π‘ƒπ‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’
And the vacuum pressure can be shown by:
π΄π‘π‘ π‘œπ‘™π‘’π‘‘π‘’ π‘ƒπ‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’ = π‘”π‘Žπ‘’π‘”π‘’ π‘ƒπ‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’ βˆ’ π‘£π‘Žπ‘π‘’π‘’π‘š π‘ƒπ‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’
Where gauge pressure can be illustrated by:
π‘ƒπ‘”π‘Žπ‘’π‘”π‘’ = πœŒπ‘”β„Ž
Pressure Measurements:
1- The Manometer
It is commonly used to measure small and moderate pressure differences.
Consider the manometer shown above that is used to measure the pressure in the
tank. Since the gravitational effects of gases are negligible, the pressure anywhere in
the tank and at position 1 has the same value. Furthermore, since pressure in a fluid
does not vary in the horizontal direction within a fluid, the pressure at point 2 is the
Water Resources Engineering College – Al-Qasim Green University
Fluid Mechanic
Ammar Ali Abd
Email: [email protected]
same as the pressure at point 1, P2 = P1. The differential fluid column of height h is in
static equilibrium, and it is open to the atmosphere. Then the pressure at point 2 is
determined directly from:
𝑃2 = πœŒπ‘”β„Ž + π‘ƒπ‘Žπ‘‘π‘š
where 𝜌 is the density of the fluid in the tube. Note that the cross-sectional area of the
tube has no effect on the differential height h, and thus the pressure exerted by the
fluid. However, the diameter of the tube should be large enough (more than a few
millimeters) to ensure that the surface tension effect and thus the capillary rise is
negligible.
Example 1: Measuring Pressure with a Manometer. A manometer is used to measure
the pressure in a tank. The fluid used has a specific gravity of 0.85, and the manometer
column height is 55 cm, as shown in Figure below. If the local atmospheric pressure is
96 kPa, determine the absolute pressure within the tank.
Solution:
Now with the specific gravity it easy to find the density of the manometer fluid to be:
𝑆𝑃𝑔 =
πœŒπ‘“
𝐹𝑙𝑒𝑖𝑑 𝑑𝑒𝑛𝑠𝑖𝑑𝑦
π‘˜π‘”
β†’ 0.85 =
β†’ πœŒπ‘“ = 850 3
π‘€π‘Žπ‘‘π‘’π‘Ÿ 𝑑𝑒𝑛𝑠𝑖𝑑𝑦
1000
π‘š
And the pressure can be calculated by:
𝑃 = πœŒπ‘”β„Ž + π‘ƒπ‘Žπ‘‘π‘š β†’ 𝑃 = 850 βˆ— 9.81 βˆ— 0.55 + 96000 = 100.586 π‘˜π‘π‘Ž
Where the gauge pressure is:
π‘ƒπ‘”π‘Žπ‘’π‘”π‘’ = πœŒπ‘”β„Ž = 850 βˆ— 9.81 βˆ— 0.55 = 4.586 π‘˜π‘π‘Ž
Water Resources Engineering College – Al-Qasim Green University
Fluid Mechanic
Ammar Ali Abd
Email: [email protected]
Case study:
What if the manometer has more than one fluid as shown in the flowing figure:
Where, the manometer here has three different fluid on three layers. So, the pressure
can be calculated easily by:
π‘ƒπ‘Žπ‘‘ π‘π‘œπ‘–π‘›π‘‘ 1 = 𝜌1 π‘”β„Ž1 + 𝜌2 π‘”β„Ž2 +𝜌3 π‘”β„Ž3 + π‘ƒπ‘Žπ‘‘π‘š
Case study:
Manometers are particularly well-suited to measure pressure drops across a horizontal
flow section between two specified points due to the presence of a device such as a
valve or heat exchanger or any resistance to flow. This is done by connecting the two
legs of the manometer to these two points, as shown in Figure below. The working
fluid can be either a gas or a liquid whose density is 𝜌1 . The density of the manometer
fluid is 𝜌2 , and the differential fluid height is h.
A relation for the pressure difference P1 = P2 can be obtained by starting at point 1 with
P1, moving along the tube by adding or subtracting the πœŒπ‘”β„Ž terms until we reach point
2, and setting the result equal to P2:
𝜌1 𝑔(β„Ž + π‘Ž) + 𝑃1 = 𝜌1 π‘”π‘Ž + 𝜌2 π‘”β„Ž + 𝑃2
𝜌1 π‘”β„Ž + 𝜌1 π‘”π‘Ž + 𝑃1 = 𝜌1 π‘”π‘Ž + 𝜌2 π‘”β„Ž + 𝑃2
𝑃1 βˆ’ 𝑃2 = 𝜌2 π‘”β„Ž βˆ’ 𝜌1 π‘”β„Ž
Note that we jumped from point A horizontally to point B and ignored the part
underneath since the pressure at both points is the same.
Water Resources Engineering College – Al-Qasim Green University
Fluid Mechanic
Ammar Ali Abd
Email: [email protected]
Example 2: The water in a tank is pressurized by air, and the pressure is measured by
a multifluid manometer as shown in Figure below. The tank is located on a mountain
at an altitude of 1400 m where the atmospheric pressure is 85.6 kPa. Determine the air
pressure in the tank if h1 =0.1 m, h2 =0.2 m, and h3=0.35 m. Take the densities of
water, oil, and mercury to be 1000 kg/m3, 850 kg/m3, and 13,600 kg/m3, respectively.
Solution: now based on the figure:
𝑃1 + 𝜌1 π‘”β„Ž1 + 𝜌2 π‘”β„Ž2 = 𝑃2 + 𝜌3 π‘”β„Ž3
𝑃1 + 1000 βˆ— 9.81 βˆ— 0.1 + 850 βˆ— 9.81 βˆ— 0.2 = 85600 + 13600 βˆ— 9.81 βˆ— 0.35
𝑃1 = 131.3 π‘˜π‘π‘Ž
2- The Barometer and Atmospheric Pressure
Atmospheric pressure is measured by a device called a barometer; thus, the
atmospheric pressure is often referred to as the barometric pressure.
Water Resources Engineering College – Al-Qasim Green University
Fluid Mechanic
Ammar Ali Abd
Email: [email protected]
The pressure at point B is equal to the atmospheric pressure, and the pressure at C can
be taken to be zero since there is only mercury vapor above point C and the pressure is
very low relative to Patm and can be neglected to an excellent approximation. Writing a
force balance in the vertical direction gives.
π‘ƒπ‘Žπ‘‘π‘š = πœŒπ‘”β„Ž
where 𝜌 is the density of mercury, g is the local gravitational acceleration, and h is the
height of the mercury column above the free surface. Note that the length and the
cross-sectional area of the tube have no effect on the height of the fluid column of a
barometer.
Example 3: Determine the atmospheric pressure at a location where the barometric
reading is 740 mm Hg and the gravitational acceleration is g =9.81 m/s2.
Assume the temperature of mercury to be 10°C, at which its density is
13,570 kg/m3.
Solution: The barometric reading at a location in height of mercury column is given.
The atmospheric pressure is to be determined.
π‘ƒπ‘Žπ‘‘π‘š = πœŒπ‘”β„Ž β†’ π‘ƒπ‘Žπ‘‘π‘š = 13570 βˆ— 9.81 βˆ— 0.74 = 98.5 π‘˜π‘π‘Ž
Note that density changes with temperature, and thus this effect should be considered
in calculations.
Example 4: The piston of a vertical piston–cylinder device containing a gas has a mass
of 60 kg and a cross-sectional area of 0.04 m2, as shown in Figure below. The local
atmospheric pressure is 0.97 bar, and the gravitational acceleration is 9.81 m/s2.
Determine the pressure inside the cylinder.
Solution: A gas is contained in a vertical cylinder with a heavy piston. The pressure
inside the cylinder and the effect of volume change on pressure are to be determined.
To solve it can easily make a force balance on the piston:
π‘‘β„Žπ‘’ π‘Žπ‘‘π‘šπ‘œπ‘ π‘β„Žπ‘’π‘Ÿπ‘–π‘ π‘π‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’ + π‘‘β„Žπ‘’ π‘€π‘’π‘–π‘”β„Žπ‘‘ π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘–π‘ π‘‘π‘œπ‘› = π‘π‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ 𝑓𝑙𝑒𝑖𝑑
π‘ƒπ‘Žπ‘‘π‘š βˆ— 𝐴 + 𝑀 = 𝑃 βˆ— 𝐴
Where, 𝑀 = π‘šπ‘” β†’ 𝑀 = 60 βˆ— 9.81 = 588.6 𝑁, and now:
Water Resources Engineering College – Al-Qasim Green University
Fluid Mechanic
Ammar Ali Abd
Email: [email protected]
𝑃=
588.6
+ 97000 = 111.7 π‘˜π‘π‘Ž
0.04
If the gas behaves as an ideal gas, the absolute temperature doubles when the volume
is doubled at constant pressure.
Case Study
The pressure balance can be illustrated:
𝑃𝐴 + 𝛾1 β„Ž1 = 𝑃𝐡 + 𝛾3 β„Ž3 + 𝛾2 𝑙2 π‘ π‘–π‘›πœƒ
Where, 𝛾 = πœŒπ‘”, and π‘ π‘–π‘›πœƒ =
β„Ž2
𝑙2
Water Resources Engineering College – Al-Qasim Green University
Fluid Mechanic
Ammar Ali Abd
Email: [email protected]
Example 5: The underground storage tank used in a service station contains gasoline
filled to the level A. Determine the gage pressure at each of the five identified points.
Note that point B is located in the stem, and point C is just below it in the tank. Take
gasoline density = 730 kg/m3.
Solution: Since the tube is open-ended, point A is subjected to atmospheric pressure,
which has zero-gauge pressure.
𝑝𝐴 = 0
The pressures at points B and C are the same since they are at the same horizontal
level with h = 1 m.
𝑝𝐡 = 𝑝𝐢 = πœŒπ‘”π‘Žπ‘ π‘œπ‘™π‘–π‘›π‘’ π‘”β„Ž = 730 βˆ— 9.81 βˆ— 1 = 7161.3 π‘π‘Ž
For the same reason, pressure at points D and E is the same. Here,
h = 1 m+ 2 m = 3 m.
𝑝𝐷 = 𝑝𝐸 = πœŒπ‘”π‘Žπ‘ π‘œπ‘™π‘–π‘›π‘’ π‘”β„Ž = 730 βˆ— 9.81 βˆ— 3 = 21.48 π‘˜π‘π‘Ž
Example 6: The soaking bin contains ethyl alcohol used for cleaning automobile
parts. If h = 7 ft, determine the pressure developed at point A and at the air surface B
within the enclosure. Take gea = 49.3 lb/ft3.
Solution: The gauge pressures at points A and B are.
𝑝𝐴 = π›Ύπ‘’π‘Ž β„Žπ΄ = 49.3 βˆ— (7 βˆ’ 2) = 246.5
𝑝𝐡 = π›Ύπ‘’π‘Ž β„Žπ΅ = 49.3 βˆ— (7 βˆ’ 6) = 49.3
𝑙𝑏
𝑙𝑏
= 1.7 2 = 1.7 𝑝𝑠𝑖
2
𝑓𝑑
𝑖𝑛
𝑙𝑏
𝑙𝑏
= 0.342 2 = 1.7 𝑝𝑠𝑖
2
𝑓𝑑
𝑖𝑛
Water Resources Engineering College – Al-Qasim Green University
Fluid Mechanic
Ammar Ali Abd
Email: [email protected]
Example 7: The funnel is filled with oil and water to the levels shown. Determine the
depth of oil hβ€² that must be in the funnel so that the water remains at a depth C, and
the mercury level made h = 0.8 m. Take oil density = 900 kg/m3, water density = 1000
kg/m3, mercury density = 13550 kg/m3.
Solution: the level C to D can be calculated by:
β„ŽπΆπ· = 0.2 + β„Žβ€² + 0.4 βˆ’ 0.8 = β„Žβ€² βˆ’ 0.2
And based on the figure:
𝑝𝐴 + πœŒπ‘œπ‘–π‘™ βˆ— 𝑔 βˆ— β„Žβ€² + πœŒπ‘€π‘Žπ‘‘π‘’π‘Ÿ βˆ— 𝑔 βˆ— 0.4 = 𝑝𝐷 + πœŒπ»π‘” βˆ— 𝑔 βˆ— (β„Žβ€² βˆ’ 0.2)
β„Žβ€² = 0.2458 π‘š
Note: pressure at the end bottom with height 0.1 m, it is equal for both side so we can
eliminate the term, it will be as below:
𝑝𝐴 + πœŒπ‘œπ‘–π‘™ βˆ— 𝑔 βˆ— β„Žβ€² + πœŒπ‘€π‘Žπ‘‘π‘’π‘Ÿ βˆ— 𝑔 βˆ— 0.4 + πœŒπ»π‘” βˆ— 𝑔 βˆ— 0.1 = 𝑝𝐷 + πœŒπ»π‘” βˆ— 𝑔 βˆ— (β„Žβ€² βˆ’ 0.2) + πœŒπ»π‘” βˆ— 𝑔 βˆ— 0.1
↓
So it is same and can be deleted
↓
𝑝𝐴 + πœŒπ‘œπ‘–π‘™ βˆ— 𝑔 βˆ— β„Žβ€² + πœŒπ‘€π‘Žπ‘‘π‘’π‘Ÿ βˆ— 𝑔 βˆ— 0.4 = 𝑝𝐷 + πœŒπ»π‘” βˆ— 𝑔 βˆ— (β„Žβ€² βˆ’ 0.2)
Water Resources Engineering College – Al-Qasim Green University
Fluid Mechanic
Ammar Ali Abd
Email: [email protected]
Example 8: Butyl carbitol, used in the production of plastics, is stored in a tank having
a U-tube manometer. If the U-tube is filled with mercury to level E, determine the
pressure in the tank at point A and B. Take SHg = 13.55, and Sbc = 0.957.
Solution: based on the figure:
𝑝𝐴 + πœŒπ‘€π‘Žπ‘‘π‘’π‘Ÿ βˆ— 𝑔 βˆ— (0.3 + 0.05) = 𝑝𝐸 + πœŒπ‘šπ‘’π‘Ÿπ‘π‘’π‘Ÿπ‘¦ βˆ— 𝑔 βˆ— 0.12
𝑝𝐴 + 0.957 βˆ— 1000 βˆ— 9.81 βˆ— 0.35 = 13.55 βˆ— 1000 βˆ— 9.81 βˆ— 0.12
𝑝𝐴 = 12.7 π‘˜π‘π‘Ž
Note: the pressure at point E is open to atmosphere so, it is zero.
And the pressure at point B:
𝑝𝐡 + πœŒπ‘€π‘Žπ‘‘π‘’π‘Ÿ βˆ— 𝑔 βˆ— (0.25 βˆ’ 0.05) = 𝑝𝐸 + πœŒπ‘šπ‘’π‘Ÿπ‘π‘’π‘Ÿπ‘¦ βˆ— 𝑔 βˆ— 0.12
𝑝𝐡 + 0.957 βˆ— 1000 βˆ— 9.81 βˆ— 0.2 = 13.55 βˆ— 1000 βˆ— 9.81 βˆ— 0.12
𝑝𝐡 = 17.8 π‘˜π‘π‘Ž
Example 9: Water in the reservoir is used to control the water pressure in the pipe at
A. If h = 200 mm, determine this pressure when the mercury is at the elevation shown.
Take density of mercury = 13 550 kg/m3. Neglect the diameter of the pipe.
Solution: based on the figure:
𝑝𝐴 + πœŒπ‘€π‘Žπ‘‘π‘’π‘Ÿ βˆ— 𝑔 βˆ— 0.25 = 𝑝𝐸 + πœŒπ‘€π‘Žπ‘‘π‘’π‘Ÿ βˆ— 𝑔 βˆ— (0.2 + 0.4 + 0.15) + πœŒπ‘šπ‘’π‘Ÿπ‘π‘’π‘Ÿπ‘¦ βˆ— 𝑔 βˆ— 0.1
Water Resources Engineering College – Al-Qasim Green University
Fluid Mechanic
Ammar Ali Abd
Email: [email protected]
𝑝𝐴 = 18.2 π‘˜π‘π‘Ž
Example 10: A solvent used for plastics manufacturing consists of cyclohexanol in
pipe A and ethyl lactate in pipe B that are being transported to a mixing tank.
Determine the pressure in pipe A if the pressure in pipe B is 15 psi. The mercury in the
manometer is in the position shown, where h = 1 ft. Neglect the diameter of the pipe.
Take Sc = 0.953, SHg = 13.55, and Sel = 1.03
Solution: based on the figure:
𝑝𝐴 + π›Ύπ‘π‘¦π‘π‘™π‘œβ„Žπ‘’π‘₯π‘Žπ‘›π‘œπ‘™ βˆ— 1.5 + π›Ύπ‘šπ‘’π‘Ÿπ‘π‘’π‘Ÿπ‘¦ βˆ— 1 = 𝑝𝐡 + 𝛾ethyl βˆ— 0.5
𝑝𝐴 + 0.953 βˆ— 62.4 βˆ— 1.5 + 13.55 βˆ— 62.4 βˆ— 1 = 2160 + 1.03 βˆ— 62.4 βˆ— 0.5
𝑝𝐴 = 1257.4
𝑙𝑏
𝑓𝑑 2
𝑝𝐴 = 8.73 𝑝𝑠𝑖
Note: the units of pressure, 𝑝𝑠𝑖 =
𝑙𝑏
𝑖𝑛2
,
π‘Žπ‘›π‘‘ π‘‘β„Žπ‘’ 𝑓𝑑 = 12 𝑖𝑛.
Example 11: The two pipes contain hexylene glycol, which causes the level of mercury
in the manometer to be at h = 0.3 m. Determine the differential pressure in the pipes,
pA - pB. Take 𝜌hgl = 923 kg/m3, 𝜌Hg = 13 550 kg/m3. Neglect the diameter of the pipes.
Solution: based on the figure:
𝑝𝐡 + π›Ύβ„Žπ‘”π‘™ βˆ— 0.1 + 𝛾𝐻𝑔 βˆ— 0.3 = 𝑝𝐴 + π›Ύβ„Žπ‘”π‘™ βˆ— 0.1
Water Resources Engineering College – Al-Qasim Green University
Fluid Mechanic
Ammar Ali Abd
Email: [email protected]
𝑝𝐴 βˆ’ 𝑝𝐡 = 𝛾𝐻𝑔 βˆ— 0.3 = 39.9 π‘˜π‘π‘Ž
Test yourself:
1- The pressure in the tank at the closed valve A is 300 kPa. If the differential
elevation in the oil level in h = 2.5 m, determine the pressure in the pipe at B.
Take oil density = 900 kg/m3.
Answer(329kpa)
2- The two tanks A and B are connected using a manometer. If waste oil is poured
into tank A to a depth of h = 0.6 m, determine the pressure of the entrapped air
in tank B. Air is also trapped in line CD as shown. Take oil density = 900 kg/m3,
water density= 1000 kg/m3.
Answer (15.1 kpa)
3- Air is pumped into the water tank at A such that the pressure gage reads 20 psi.
Determine the pressure at point B at the bottom of the ammonia tank. Take
ammonia density = 1.75 slug/ft3.
Answer (22.6psi)
Water Resources Engineering College – Al-Qasim Green University
Fluid Mechanic
Ammar Ali Abd
Email: [email protected]
4- The Morgan Company manufactures a micromanometer that works on the
principles shown. Here there are two reservoirs filled with kerosene, each
having a crosssectional area of 300 mm2. The connecting tube has a
crosssectional area of 15 mm2 and contains mercury. Determine h if the
pressure difference pA - pB = 40 Pa. What would h be if water were substituted
for mercury? Mercury density= 13550 kg/m3, Ke density= 814 kg/m3.
Answer (h=18mm)
5- Determine the difference in pressure pB – pA between the centers A and B of
the pipes, which are filled with water. The mercury in the inclined-tube
manometer has the level shown SHg = 13.55.
Answer(18.5kpa)
Water Resources Engineering College – Al-Qasim Green University
Download
Random flashcards
Arab people

15 Cards

Pastoralists

20 Cards

Radioactivity

30 Cards

African nomads

18 Cards

Create flashcards