New answers tagged

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This is really three separate questions. Static and dynamic pressure: It is the static pressure that really matters in practical situations. The dynamic pressure is related to the kinetic energy of the fluid which, when it changes, causes a corresponding change in the static pressure. Condenser/evaporator application: The basic Bernoulli equation ...


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Actually. the velocity can be determined as follows. V = sq. root of 2gh where: V = velocity in ft./sec. g = acceleration const. 32.2 ft per sec per sec. And h = head in feet of liquid. h = P * 2.31/SG where: P = pressure in psi. SG = liquid specific gravity. With the above information you can now calculate F.


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You need to get $p_4-p_3$. Taking the datum of elevation z as that of points 3 and 4, we have $$p_{atm}+(10)\rho g=p_2+\frac{1}{2}\rho v^2$$ $$p_3+\frac{1}{2}\rho v^2=p_2+\frac{1}{2}\rho v^2$$ $$p_{atm}+(120)\rho g=p_5+\frac{1}{2}\rho v^2+(120-h)\rho g$$where h is the depth of point 5 below the surface of the tank on the right. $$p_4+\frac{1}{2}\rho v^2=...


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Using Bernoulli's equation and the momentum conservation equation, we can show that water flowing out of a pipe with cross-section $A$ at speed $v$ exerts a force $F$ on a wall (at 90 degrees), acc.: $$F=\rho Av^2$$ With $\rho$ the density of the water. But your specification of "8" pipe with 500psi stream of water exiting it and hitting a wall at 90 ...


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The pressure drop shown in the pipe in situation 2 is based on an assumed flow rate, but that flow rate can't be achieved when the pressure is only 50 Pa. You cannot deliver a fixed flow rate and a fixed static pressure: when the flow rate is kept constant, like produced by volumetric pumps, the pressure will vary with resistance: if you block the exit the ...


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When the suction cup presses down it squeezes all the air out, which creates an area of low pressure and sucks the cup in. When you lift the cup the area inside the cup stretches out, and the area of low pressure turns slightly towards high pressure which lets some air out. The air keeps going until the low pressure turns completely to high pressure and cup ...


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Your premise "but higher the velocity, greater is the temperature and pressure must be high" is wrong. To see why, you must recognize that the velocity field $\mathbf v $ that you calculate in fluid dynamics refers to a "fluid particle", which is composed of maybe some $10^{10\pm 5}$ molecules. On the other hand, the (local) temperature, pressure, density ...


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Assuming that your preliminary calculations are correct, you now have everything that you need: n, R and T. And unless you have been given a modulus of elasticity for the tyre, you must assume that it is rigid, so you have V as well (as a commentator hinted above). Just plug them into the ideal gas equation. If you know the modulus of elasticity of the tyre,...


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This is a correct way to solve exercises involving viscosity (within certain constraints, e.g., constant viscosity). Eqn. 1 is a version of the Bernoulli equation, modified to include a frictional head loss, and is definitely valid, provided the velocities used are the average velocities. Eqn. 1 without the $h_L$ is valid along a streamline, even for a ...


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A few clarifications on this thread in case anyone is reading in the future and is getting confused. I know however, that the temperature did in fact change, hence it's not an adiabatic process. That isn't really how an adiabatic process is defined. The temperature can change within a system (and often does) and it still be adiabatic. An adiabatic ...


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Shouldn't water pressure change, when water goes from liquid to vapor (boiler) and vice versa (condenser)? No, it shouldn't. Because water is on a constant pressure line during the processes (from point $1$ to $2$ in the boiler and from point $3$ to $4$ in the condenser) for both cases. In addition, phase of a matter doesn't depend on the pressure only. ...


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A reversible process is characterized by a continuous sequence of thermodynamic equilibrium states for whatever system you are considering. So, for your system to experience a reversible process, its pressure and temperature must differ only slightly from that of its surroundings throughout the entire process. And there can be no spatial temperature or ...


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It depends on what you consider to be the system. If the system is the entire container, then there are no thermodynamic operations, quasistatic or not, on the system by the external environment. And as you said, the system is not in thermal equilibrium. If you talk about a thermodynamic operations you need to define a system and an environment, in this ...


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In a complex number $z=a+i b$,the real part is $a$ and abs is $\sqrt {a^2+b^2}$,. I am not an expert of acoustics but in general, when pressure is given in complex quantity and you want to measure its magnitude then take absolute. If you are comparing two complex numbers then equate real with real and imaginary with imaginary.


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Difference between real and absolute value in general: Look at count_to_10 's answer. For acoustics and preasure measurement: Absolute pressure - pressure against perfect vacuum. Real pressure: Usually defined as the pressure against a reference-environment. Also called differential pressure. For example the pressure of the air inside a football against the ...


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Your assumption is correct. At point (a), assuming the liquid is static, the pressure is: P = (H x SG)/2.31 Where: P = pressure in PSI: SG = Liquid specific gravity However, at point (B), there is a dynamic component and some of the pressure energy P is converted to velocity energy (hv). hv = (V^2/2g) where: hv = Velocity head in feet of liquid. ...


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Your guess is correct. I think you are expected to assume that $a$ is sufficiently far from the hole that $v_a=0$. If the question does not ask you to make this assumption you should state it explicitly yourself. It does not follow that there is a discontinuity in velocity at the hole just because there is a discontinuous change in cross-section. ...


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In your approximation there is a velocity discontinuity where the pipe joins the container. Assuming that is a good approximation then you are correct and $P_a=P_b+\rho v^2/2$. In real life though, the pressure will change smoothly because the closer you get to the pipe, the larger the effect of the motion of the water inside the tank.


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Plungers are more effective when pushing water instead of air because water doesn't compress. If you let the bell of the plunger fill up with water before thrusting, you have a "solid" column of water from the plunger to the clog. Thrusting the plunger moves column of water like it was a solid piston and applies the entire force of the thrust against the ...


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The movement of the "dookey" is caused by pressure created from a snug, well fitted toilet plunger. When you press and and lift the plunger, it pushes and pulls the dookey. Rocking it back and forth until its free. So there is pressure when you push, and suction when you pull.


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The force straining a grain (which may soon be squashed) is equal and opposite to the weight (gravity force) supplied by the grain pile above. That's Newton's third law. If ALL the weight of the pile were held up by one single kernel, F_grain ~= Mass_of_pile * g and it would be easy for the ton of grain to smash the bottom kernel. It doesn't work ...


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The physical processes that control the structure of conical piles are fascinating and imperfectly understood even today. However we can approach your question in approximate way. The angle that the surface of the pile makes with the ground is called the angle of repose. Predicting this theoretically is hard because it is is sensitive to the exact nature of ...


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The second one is correct. Work is defined by $$w=F \cdot dx$$ where x is travel distance, which can be calculated by $$dx \cdot S = dV$$ Here you can see dV is not $dV_B$. It should be total volume change.


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It depends what work you refer to. If it is the work made by gas B on the piston at the right, then this work is $-P\Delta V$ because the length the piston moves is given by the change in the total volume (regardless of the motion of the other piston). This will be also the net work done on the system A+B. To compute the net work by gas B you need to add ...


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You can check your answer by figure below: $$\Delta V_B=S(x_{B,2}-x_{B,1})-S(x_{A,2}-x_{A,1})= \Delta V-\Delta V_A$$


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The stress induced in a pipe by internal pressure is $$P_r \over t$$ where $P$ is the applied pressure, $r$ the pipe radius and $t$ the pipe wall thickness. The strain induced in the pipe wall is $$\Delta r \over r$$ where $\Delta r$ is the change in radius. The basic stress strain relationship is $$\sigma = E \epsilon$$ where $\sigma$ is the stress, $...


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Pascal's law: Pascal's law or the principle of transmission of fluid-pressure (also Pascal's Principle) is a principle in fluid mechanics that states that pressure exerted anywhere in a confined incompressible fluid is transmitted equally in all directions throughout the fluid such that the pressure variations (initial differences) remain the same. Due ...


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Your analysis and intuition are correct. The force needed in the second setup is larger, even though the weight of the water is the same. To understand why, consider the horizontal part of the container, $0.5\ \text{m}$ off the ground. This wall is above the water, so the water's pressure pushes up on it. Then in reaction, the wall pushes down on the water, ...


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The quick answer is that a long as the surface area of the piston, and the volume/mass of the remain unchanged, the pressure stayed the same. So the erreanous step is to assume that $ P_2 = \rho g h $. This invalid as the cross section area changes with height.


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The Bernoulli equation inherently takes into account the fact that the flow approaching the exit hole is converging toward the exit hole and thereby accelerating. So the pressure in close proximity to the exit hole (within just a few exit hole diameters away) is decreasing while the flow velocity is increasing. This is how the pressure decreases from ...


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Your professor is correct. The hole $S_2$ is exposed to the atmosphere and the pressure at that point is indeed $p_{atm}$. The $\rho gh$ term you worry about has in fact already been 'accounted' for in the Bernoulli equation (third term on the LHS).


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the water is no longer constant at the same level because it moves. It is likely very close to $p_{atm}+\rho g h $ an the far left side, but not on left side where the hole is, it should be less because it is moving. Luckily the details are not important because you know that the hole is open to the atmosphere, so the pressure there must be $1 atm$, why ...


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1. Why the isobaric surface chosen is the one passing through $X$ and $X′$ and not the one passing through $Y$ and $Y′$? If you look at this passage from your link Equating the pressure at the level XX'(pressure at the same level in a continuous body of static fluid is equal) In your example you would not regard a static fluid since there is a flow ...


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I presume that you wish to avoid grinding the lumps of iron into dust. The difficulty with your proposal is finding some way of grinding the shavings to dust without also grinding the lumps of iron. If you can find some way to separate them first that would work - but then the problem would be solved without any need for grinding. To separate them you ...


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If your question is about how to reduce small particles of iron into smaller particles, I would try using some form of grinder designed for very small particle output. Possibly some form of conical burr grinder similar to the hand cranked ceramic burr ones used for grinding coffee beans. A possible problem is that too much heat caused by grinding will result ...


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It's because the centre of buoyancy and centre of gravity don't necessarily lie on the same point. This creates two types of mechanical equilibrium: stable and unstable. It turns out that when a human body is floating with its face inside the water, the body is in stable equilibrium. That is because in that position, the centre of gravity lies below centre ...


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Turns out it's more of a bio question. https://www.google.com/search?q=why+do+dead+people+float&oq=why+do+dead+people+float&aqs=chrome..69i57.6523j0j1&sourceid=chrome&ie=UTF-8 As for the head part, your head (mainly your brain) is more dense than water.


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Vacuum or pressure is quantified in terms of bar or torr. 1 bar or 760 torr is roughly one atmosphere and in terms of vacuum measurement 1 mili-bar is taken as almost equal to 1 torr. at 1 bar pressure and at room temperature the number of atoms per cc is ~$2\times 10^{19}$. The number of particles per cc in vacuum can range from $10^{-4}$ to $10^6$. The ...


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We want to find relation between pressure and depth in a stationary liquid. If the pressure $P$ is a function of depth $y$, i.e. $P=P(y)$; then we say if the depth changes from $y$ to $y+\mathrm dy$, the pressure will change from $P$ to $P+\mathrm dP$. $$P=P(y)$$ $$\Longrightarrow\;\large{\frac{\mathrm dp}{\mathrm dy}}=\large{\frac{P(y+\mathrm dy)-P(y)}{\...


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It's the volume that determines the pressure (for a given amount of gas at a given temperature), so if you consider a rectangular container, increasing the surface area of two opposite sides and/or increasing the distance between those sides will increase the volume and therefor decrease the pressure. But increasing the total area while keeping the volume ...


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The argument in the video seems flawed: it is true that the surface area has increased, but that is irrelevant, because when you calculate pressure you look at the number of molecules hitting a unit of area, which depends on the gas density, and here on the volume, but not on the total area. The derivation is made here: https://en.wikipedia.org/wiki/...


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As you increase the surface area you must bring parts of the surfaces closer together and this will increase the rate at which the surfaces are hit by the molecules which will compensate for the increase in area.


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The density of water (and other fluids) depends both on the pressure and the temperature. A graph for water is here: You may see that at 1 bar (1 atmosphere), the density is highest around 4 °C. That's the conditions where the density reaches the nice 1,000 kilograms per cubic meter. Water contracts when it gets warmer than that, but also when it gets ...


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As mentioned above density of liquid nitrogen at atmospheric pressure is c. 0.8g/cm3 or 800Kg/M3. Thus 1 litre of Liquid N2 is about 800g Warm up to room temperature and I know I should have 22.4 litres of gas at STP (Standard temperature and pressure) for each mole. Nitrgoen molecules (N2) have an atomic mass of 28, so I have 28.6 moles of gas at STP. 28....


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Other answers have handled the oxygen issue quite well. In regards to the temperature, space itself has no temperature because it's a vacuum. Objects in space, however, do have a temperature. If a human is exposed, unprotected, to space near the sun (or any other star), the temperature change in their body could very well be terminal. Even near our ...


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Assuming the deck doesn't bulge to much at at the sides (as the video suggests), the press exerts work $W$ on the deck acc.: $$W=\int_0^yFdy$$ With $F$ the force exerted by the press and $y$ the displacement. This work is converted to potential energy and stored in the deck. As suggested in the answer to this question, the high pressure is likely to ...


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Why do we not include the surface area of the container in the formula? Because it is not needed. Pressure $p$ is force $F$ per unit of surface area $A$: $$p=\frac{F}{A}$$ The pressure a gas exerts on the walls of a container is the collective force collisions of the gas molecules exert on the container walls, per unit of surface area. If we look at ...


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Pressure is a measurement of force/unit area. It doesn't measure the total force exerted over the entirety of the surface, but the force exerted on one "unit area" of the surface. One unit of area depends on the measurement system you're using, but by doing that surface area can be disregarded.



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