# Tag Info

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As long as you can (1) set the free water surface of your destination at a lower level than the free water surface of your source, (2) use a completely sealed, air tight siphon, and (3) initially prime the system, it should work. Your source being your datum, for example, setting the depth of your destination to the same level (in this case 10 ft) will not ...

5

If you don't care it being a little flimsy and just want quick, and the plastic is not too brittle: Use a barbed fitting with a straight thread on the outside, e.g. #10-32 of M5x.8 (these sizes happen to be interchangable), and a sealant. SUMMARY: drill 4.2mm (work up size in small increments to avoid cracking) tap hole M5x.8 (or #10-32) (with great care, ...

4

Many commercial hot water recirculation pumps will do that. Just grabbed one at random - Grundfoss Alpha1 15 40 130. I couldn't fetch the page after I ran their calculator. I used temp of 99C and head of 2m. 100C doesn't work in their calculator. Click link then click "view in product center", and fill in form. ALPHA1 circulator pump - heating, cooling, hot ...

4

P2 is given as atmospheric and P1 is given as a gauge pressure of 80kN/m^2. The gauge is assumed to measure the pressure against atmospheric ie it could be a Bourdon tube type gauge. Which means if atmospheric is taken as 1 then the inlet pressure would be 80 + 1 kN/M^2 absolute. Then, one would subtract 1 from the 81 to get the pressure difference...

4

This is actually something not easily answered because it is part of the definition of pressure. I will instead point you to other answers which hopefully make sense. The following is an excerpt from Lumen Physics. *The force exerted on the end of the tank is perpendicular to its inside surface. This direction is because the force is exerted by a static or ...

4

If you don't want to drill a hole on the box and mess around with nozzles and sealants, a solution would be to siphon water like The only difference is that that valve is going to be at the bottom side. The only "problem" is that you need to fill the hose. A common solution to that is

3

If there are no losses and no storing between the beginning $A$ and the end $B$, you can calculate the velocities based on the principle of continuity. $$Q_A= Q_B$$ whereas in a homogeneous flow $Q$ is equal to the cross-sectional area times the flow velocity: $$Q=A\cdot v$$ With the proportions of the two cross-sectional areas you can now determine, ...

3

As mensioned above it works. You would need to make sure the level of discharge at your end is lower than the lowest seasonal level of the lake and also make sure it never gets submrged at your end to interrupt the flow. Both ends of the siphon need to be terminating at vertical drops. I would use shut off valves at both ends for temporary priming and ...

3

So, I attempted to solve a simplified problem using Bernoulli, it is an approximation though, since it doesn't account for the change in flow direction and subsequent turbulence caused after the hole. I rearranged equations for volumetric flow rate and head loss. Please let me know if you spot any errors. Note: the last two lines rearrange from an earlier ...

3

Yes your can will still have buoyancy when it is submerged to the bottom. Regardless of the submersion depth, any object will lose weight equal to the weight of water it has displaced, even when held at the bottom. You confuse the hydrostatic pressure with buoyancy. Hydrostatic pressure will increase with depth, to a point that it may even crush the can in....

3

The pressure drop between the left and right sumps can be calculated by measurement. For fresh water 10 m = 1 bar (approx.) so 10 mm = 1 mbar (or just work in mm). There are a several useful calculators online. Figure 1. Resuts from Copely's flow rate calculator. Click for 100% magnification.) I've guessed that the pipe length is about 300 mm and that while ...

3

The equation $F_{\textrm{A}} = C_{\textrm{D}}A\rho\dot{x}^2/2$ doesn't tell you that drag is proportional to the square of velocity. In itself, that equation doesn't tell you anything at all about how drag depends on velocity, because that equation is just a definition of the drag coefficient $C_{\textrm{D}}$. What tells you how drag depends on velocity is ...

2

To estimate the frictional losses you can use the Darcy–Weisbach equation: $$h_f = f_D \frac{L}{D} \frac{V^2}{2\,g}$$ This head-loss ($h_f$) can be added to Bernoulli's equation: $$h - h_f = \frac{V^2}{2\,g}$$ In this case the length ($L$) is the same as your height ($h$) so these equations combine like so: $$h - f_D \frac{h}{D} \frac{V^2}{2\,g} = \frac{... 2 If a water pipe has a large diameter at the start and a small diameter at the end what part of the pipe would the water flow faster? If there are no losses of the fluid between the wide point and the narrow point, then the flow will be faster at the narrow point. This is because if you take the flow rates:$$\dot{Q}_{start} =A_{start} \times V_{start} = \...

2

I'm not sure why SolarMike deleted his response. The only thing holding the can to the ground ("proud" in naval terms) is the vacuum force, i.e. the same pressure that keeps you from lifting the can off a table if there's a perfect seal to the table. So long as the can's density is less than that of the surrounding fluid it'll experience a buoyant force. ...

2

This question is a theoretical/academic edge-case. A body in the water will experience two forces: Pressure acting on all surfaces in contact with water Gravity acting on the mass of the body The articel on buoyancy over at Wikipedia explains very good how the following equations are set-up. This article also gives the defintion of buoyancy as: In ...

2

Yes it will - the space the object occupies is lighter than the fluid around it so it wants to rise. Same as pushing a ball to the bottom of the bath - does it stay there? Edit: for those who say the shape of the ball makes all the difference: try it with a plastic hollow cube (filled with air) so the cube can lay flat on the surface...

2

The text book answer is correct. I think you missed to calculate the density of the downstream of the shock. the equation you entered is valid only in the region 1 or in region 2. means the isentropic relations are not valid across the shock. Across the shock you have to use normal shock relations. Then you would get the right answer

2

I'm going to suggest that use of a siphon-based system is just plain unworkable here. What you need is a pipe permanently installed in or below the lake bottom, with a good filter at the inlet end, and an electric pump. Personally I'd go with a submersible pump at the inlet end (and thus a proper power cable run alongside the pipe) so that priming the ...

2

punch a small hole in the bottom of a plastic container. measure the time it takes for the container to empty through the hole. experiment with different hole diameters to "scale" the effect appropriately for your expected viscosity values, and calibrate it against a standard.

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A long glass tube and a steel ball which matches the internal diameter closely - then measure the different times compared to some known liquids for the ball to drop a given distance.

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There must be pressure ( water level) difference to provide force to push water through the system. If you stop input and output for a few minutes the levels will even out , except where a baffle may prevent it .With no flow ,no pressure differential is need to move water . It is not much pressure , height of 4 " equals about 0.17 psi. Unrelated , have ...

2

TL; DR: this is actually complicated and I don't know if your question can really be answered. The movement of (practically) spheres in a fluid, you can use Stoke's law for very low Reynolds numbers, as we would expect in a thick resin (without doing the actual math, that part is up to you). Just find the equilibrium between bouyuancy and stokes' friction ...

2

Why won't we simply add work calculated in each process of a cycle,like in Carnot: W1 + W2 + W3 + W4 We can (But you need to be careful in calling energy of each process as work, $W_1$ & $W_3$ as you call them are heats added and taken from the system not mechanical work). You already answered your question, the net work of the Carnot cycle is the area ...

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The surface of C where the glass sat is smooth, then the glass may have a slightly concave bottom and the water, with capillary action, managed to form a seal between the two surfaces. Even if the glass bottom and surface C are both flat, capillary action can still form a seal around the edge leaving a void in the centre, causing the same effect of lower v. ...

1

Let's assume that you will work out the right combinations of dimensionless terms on your own. Let's consider only the question of what to plot once you have those terms. By example, in heat transfer, the Nusselt number $Nu$ is for forced convection is a function of the Reynolds number $Re$ and the Prandtl number $Pr$. In this case we can make one of two ...

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Yes. Conservation of mass dictates that the velocity at an exit with double the area of the upstream pipe is 1/2 the velocity in the upstream pipe. For an ideal fluid in an ideal (horizontal) expansion transition, there is no change in entropy. So Bernoulli's equation gives you the relationship for calculating the pipe's static pressure given that the exit ...

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They are called haptic greases. Used on knobs and levers to provide stiction but very smooth movement when actuated. Here's an explanation from Nye Technologies - https://www.nyelubricants.com/motion-control

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There is no practical way to do this. It is possible to estimate specific heats from first principles using computer simulation and quantum mechanics. There are also some semi-empirical "laws" which apply to particular situations, such as materials whose elements have large atomic numbers, (bigger than iron, and therefore irrelevant for lubricating oil) or ...

1

The divergence theorem only applies to a closed surface that encloses the volume $V$. If you construct a closed surface by "shrink wrapping" a section of pipe and closing off both ends, enclosing a volume $V$, the volumetric flow out the complete surface is zero, because the flow into one end is equal and opposite to the flow into of the other end - i.e. ...

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