What is Fluid Pressure?
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Fluid pressure is the pressure that a fluid exerts at any point within it. The product of the difference in height, density, and free-fall acceleration determines the difference in pressure between the two levels.
Furthermore, hydraulic systems and variations in the fluid&#;s velocity might cause the fluid pressure to be amplified. The manometer is the device used to measure fluid pressure.
Fluid Pressure Formula
Pfluid = P ρgh
where
P = reference point&#;s pressure
Pfluid = fluid pressure absorbed at the point
Ρ = fluid density
g = accelerated gravity
h = Altitude from a reference point
Divide the mass of the liquid under consideration by the volume of the liquid under consideration to determine Liquid density.
ρ = m / v
m = Liquid mass
v = Considered liquid volume
At liquid atmospheric pressure, the system&#;s total pressure is:
Liquid = Po + ρgh
Where
Po = Atmospheric pressure
Liquid pressure is measured using a manometer in relation to an external source, which is often thought of as the earth&#;s atmosphere.
&#; Here, a liquid like mercury is used to monitor pressure while the gas whose pressure has to be computed is placed at the opposite end of the U-tube.&#; The end that is filled with gas is sealed, while the other end is left open.&#; However, both the gas pressure and the atmospheric pressure have an impact on the liquid.&#; When the liquid is level in both tubes, the air pressure within is said to be equal to the air pressure outside.&#; When the liquid rises over the straight level, it is also said to be lighter than the outside air&#;s pressure.&#; When the liquid is below the straight level, it is claimed that the air in the tube has a heavier pressure than the air outside.
We analyze fluid pressure in two scenarios:
&#; open circumstances or open channel flow&#; closed conduit or in a closed state
Hydrostatic pressure, which is sometimes called static fluid pressure, is the pressure of the fluid that is shown above. Since the pressure caused by moving fluid may not seem like much, it is only taken into account here in relation to the depth of the fluid. The static pressure of a liquid has nothing to do with its volume, mass, total surface area, or the type of container it is in.
As stated by Pascal&#;s Law,
&#;A restricted liquid&#;s external static pressure is &#;spread or transmitted uniformly throughout the liquid in all directions.&#;
Any surface in contact with the fluid is at right angles to the static pressure&#;s action. Pascal discovered that for a static fluid, the pressure at a location would be the same for all planes going through the fluid. Pascal&#;s principle or the fluid-pressure transmission principle are other names for Pascal&#;s law. French mathematician Blaise Pasca introduced the concept of Pascal law in .
Two scenarios that need to be considered are considered.
Open channel or open channel flow
Closed channel or channel affected
Gravity&#;s influence on fluid pressure is that it causes the fluid body to move in the opposite direction of its net acceleration. It also depicts the fluid pressure variation as a function of depth.
Q. What are the factors affecting fluid pressure?
Fluid pressure is influenced by two things. They are the fluid&#;s depth as well as its density. At greater depths, a fluid exerts more pressure. As you get deeper into a fluid, all of the fluid above it presses down on you.
Q. What are the methods for calculating fluid pressure?
Use the formula pgh = fluid pressure to compute fluid pressure, where p is the density of the liquid, g is the acceleration of gravity, and h is the fluid&#;s height. To solve the equation, multiply the variables and obtain the product of the three.
Q. Why does fluid flow?
The density differential between liquid and solid (shrinkage flow) or the impact of gravity on density gradients within the fluid phase itself can generate fluid flow during solidification.
Q. Where can we use fluid pressure?
Want more information on hydraulic expansion of heat exchanger tubes? Feel free to contact us.
In automotive brakes, the liquid pressure is applied. When a vehicle&#;s brakes are deployed, the car comes to a halt due to liquid pressure. ii. Injection syringes are filled at atmospheric pressure.
Q. Explain how fluid pressure works?
Fluid particles move in all directions at random all of the time. The particles continue to collide with each other and anything else in their path as they advance. These impacts create pressure, which is distributed evenly in all directions.
We hope you enjoyed studying this lesson and learned something cool about Fluid Pressure! Join our Discord community to get any questions you may have answered and to engage with other students just like you! We promise, it makes studying much more fun&#;
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How are both of the above concepts and their corresponding use of definite integrals similar to problems we have encountered in the past involving formulas such as distance equals rate times time and mass equals density times volume ?
Figure
6.64
Three settings where we compute the accumulation of a varying quantity: the area under \(y = f(x)\text{,}\) the distance traveled by an object with velocity \(y = v(t)\text{,}\) and the mass of a bar with density function \(y=\rho(x)\text{.}\)We have seen several different circumstances where the definite integral enables us to measure the accumulation of a quantity that varies, provided the quantity is approximately constant over small intervals. For instance, to find the area bounded by a nonnegative curve \(y = f(x)\) and the \(x\)-axis on an interval \([a,b]\text{,}\) we take a representative slice of width \(\Delta x\) that has area \(A_{\text{slice} } = f(x) \Delta x\text{.}\) As we let the width of the representative slice tend to zero, we find that the exact area of the region is
\begin{equation*} A = \int_a^b f(x) \, dx\text{.} \end{equation*}
In a similar way, if we know the velocity \(v(t)\) of a moving object and we wish to know the distance the object travels on an interval \([a,b]\) where \(v(t)\) is nonnegative, we can use a definite integral to generalize the fact that \(d = r \cdot t\) when the rate, \(r\text{,}\) is constant. On a short time interval \(\Delta t\text{,}\) \(v(t)\) is roughly constant, so for a small slice of time, \(d_{\text{slice} } = v(t) \Delta t\text{.}\) As the width of the time interval \(\Delta t\) tends to zero, the exact distance traveled is given by the definite integral
\begin{equation*} d = \int_a^b v(t) \, dt\text{.} \end{equation*}
Finally, if we want to determine the mass of an object of non-constant density, because \(M = D \cdot V\) (mass equals density times volume, provided that density is constant), we can consider a small slice of an object on which the density is approximately constant, and a definite integral may be used to determine the exact mass of the object. For instance, if we have a thin rod whose cross sections have constant density, but whose density is distributed along the \(x\) axis according to the function \(y = \rho(x)\text{,}\) it follows that for a small slice of the rod that is \(\Delta x\) thick, \(M_{\text{slice} } = \rho(x) \Delta x\text{.}\) In the limit as \(\Delta x \to 0\text{,}\) we then find that the total mass is given by
\begin{equation*} M = \int_a^b \rho(x) \, dx\text{.} \end{equation*}
All three of these situations are similar in that we have a basic rule (\(A = l \cdot w\text{,}\) \(d = r \cdot t\text{,}\) \(M = D \cdot V\)) where one of the two quantities being multiplied is no longer constant; in each, we consider a small interval for the other variable in the formula, calculate the approximate value of the desired quantity (area, distance, or mass) over the small interval, and then use a definite integral to sum the results as the length of the small intervals is allowed to approach zero. It should be apparent that this approach will work effectively for other situations where we have a quantity that varies.
We next turn to the notion of work: from physics, a basic principle is that work is the product of force and distance. For example, if a person exerts a force of 20 pounds to lift a 20-pound weight 4 feet off the ground, the total work accomplished is
\begin{equation*} W = F \cdot d = 20 \cdot 4 = 80 \ \text{foot-pounds}\text{.} \end{equation*}
If force and distance are measured in English units (pounds and feet), then the units of work are foot-pounds. If we work in metric units, where forces are measured in Newtons (where \(1 N = 1 \ kg \cdot m/s^2\)) and distances in meters, the units of work are Newton-meters. Note that Newton-meters are also called Joules.
Note that gravity pulls on an object of mass \(m\) with a force of \(mg \text{,}\) where \(g \) is the gravitational constant \(g = 9.8 m/s^2 \text{.}\) Since pounds are a unit of force, the value \(g \) is already incorporated into the pound. However, grams and kilograms are units of mass, so the force to move them needs to incorporate \(g \text{.}\)
For example, if a 20-kg weight is lifted 4 meters off the ground, the total work accomplished is
\begin{equation*} W = F \cdot d = (20 \text{ kg})\cdot (9.8 \ m/s^2) \cdot 4 \text{ m} = 784 \ kg \cdot m^2/s^2 = 784 \text{ Newton-meters}\text{.} \end{equation*}
Of course, the formula \(W = F \cdot d\) only applies when the force is constant over the distance \(d\text{.}\) In Example6.65, we explore one way that we can use a definite integral to compute the total work accomplished when the force exerted varies.
Example
6.65
A bucket is being lifted from the bottom of a 50-foot deep well; its weight (including the water), \(B\text{,}\) in pounds at a height \(h\) feet above the water is given by the function \(B(h)\text{.}\) When the bucket leaves the water, the bucket and water together weigh \(B(0) = 20\) pounds, and when the bucket reaches the top of the well, \(B(50) = 12\) pounds. Assume that the bucket loses water at a constant rate (as a function of height, \(h\)) throughout its journey from the bottom to the top of the well.
Find a formula for \(B(h)\text{.}\)
Compute the value of the product \(B(5) \Delta h\text{,}\) where \(\Delta h = 2\) feet. Include units in your answer. Explain why this product represents the approximate work it took to move the bucket of water from \(h = 5\) to \(h = 7\text{.}\)
Is the value in (b) an over- or under-estimate of the actual amount of work it took to move the bucket from \(h = 5\) to \(h = 7\text{?}\) Why?
Compute the value of the product \(B(22) \Delta h\text{,}\) where \(\Delta h = 0.25\) feet. Include units in your answer. What is the meaning of the value you found?
More generally, what does the quantity \(W_{\text{slice} } = B(h) \Delta h\) measure for a given value of \(h\) and a small positive value of \(\Delta h\text{?}\)
Evaluate the definite integral \(\int_0^{50} B(h) \, dh\text{.}\) What is the meaning of the value you find? Why?
Solution
Since \(B(h)\) is changing at a constant rate, it is linear, so we must find a linear function with \(B(0)=20\) and \(B(50)=12\text{.}\) Therefore, \(B(h)\) has an intercept of \(b=20\text{,}\) and a slope of \(m=\frac{12-20}{50-0}=-\frac 4{25}\) so \(B(h)=-\frac{4}{25}h+20\) feet.
If \(\Delta h = 2\text{,}\) then \(B(5) \Delta h=38.4\) foot-pounds. At \(h = 5\text{,}\) the bucket weighed \(B(5)\) pounds, so if the bucket's weight stopped changing then, it would have taken \(B(5) \Delta h=38.4\) foot-pounds of work to raise it \(\Delta h = 2\) feet. Since the bucket's weight actually did not stay constant, this number is slightly off, but since the bucket's weight did not change by too much over the time period, it will not be off by much.
The answer above is an over-estimate of the work it would have taken. The answer above would correspond to the bucket's weight remaining constant after \(h = 5\text{,}\) but since the bucket continued to get lighter after that point, it would have taken less work than our calculation showed.
If \(\Delta h = .25\text{,}\) then \(B(22) \Delta h=4.12\) foot-pounds. This represents the approximate amount of work it would have taken to raise the bucket \(.25\) feet further once the bucket had been raised 22 feet.
More generally, the quantity \(W_{\text{slice} } = B(h) \Delta h\) measures the amount of work it would take to raise the bucket an additional \(\Delta h\) feet once you've raised it \(h\) feet already.
By the power rule,
\begin{equation*} \int_0^{50} B(h) \, dh =\int_0^{50} -\frac 4 {25} h +20 \, dh \\ = -\frac{2}{25} h^2 +20 h \Large |_0^{50}\\ = (-\frac{2}{25} (50)^2 +20(50))-(0)=800 \text{.} \end{equation*}
This represents the amount of work, in foot-pounds, that it would take to raise the bucket from \(h=0\) (the bottom of the well) to \(h=50\) (the top of the well). This calculation provides the right answer because, in order to find the amount of work, we'd want to take very thin slices of the form \(W_{\text{slice} } = B(h) \Delta h\text{,}\) sum them, and then take the limit as the slices get thinner (meaning as \(\Delta h \rightarrow 0 \)). Of course, this would be a Reimann sum, and in the limit as \(\Delta h \rightarrow 0 \) it would turn into an integral.
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