fluid friction definition
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Friction loss refers to that portion of pressure lost by fluids while moving through a pipe, hose, or other limited space. In mechanical systems such as internal combustion engines, it refers to the power lost overcoming the friction between two moving surfaces.
Friction loss has several causes, including:
- Frictional losses depend on the conditions of flow and the physical properties of the system.
- Movement of fluid molecules against each other
- Movement of fluid molecules against the inside surface of a pipe or the like, particularly if the inside surface is rough, textured, or otherwise not smooth
- Bends, kinks, and other sharp turns in hose or piping
In pipe flows the losses due to friction is of two kind first the skin-friction and the other is form-friction, the former one is due to the roughness in the inner part of the pipe where the fluid comes in the contact of the pipe material and the latter one is due to the obstructions present in the line of flow, it may be due to a bend or a control valve or anything which changes the course of motion of the flowing fluid.
While friction loss has multiple applications, one of the most common is in the realm of firefighting. With the advent of modern power-takeoff (PTO) fire pumps, pressures created can sometimes overwhelm the ability of water to flow through a hose of a given diameter. As the velocity of water inside a hose increases, so does the friction loss. This resulting increase occurs as an exponential rate, thus an increase in the flow by a factor of X will result in an increase in friction loss by a factor of X2. For example, doubling the flow through a hose will quadruple the friction loss. Ultimately, as the pressure created by a fire pump goes higher and higher the amount of water actually flowing through a hose to a given point lessens, threatening firefighting operations. Conversely, friction loss can restrict the distance which water can be lifted during fire department drafting operations.
The formula used most often in firefighting to express the amount of friction loss is:
FL = CQ2L
Where FL = friction loss (expressed in psi) C = coefficient of friction (based on the inside diameter of the hose and the inside jacket material) Q = flow rate in hundreds of gallons (gpm/100) L = Length of hose in hundreds of feet (L/100)
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Answers:Bullet shot into water. Skier skidding on water. Globules in a lava lamp being slowed down as they move in the liquid. The heart having to pump harder than otherwise to push blood through your vessels.
Answers:Dry friction (also known as Coulomb friction) is the kind of friction to which we refer with the formula "F = mu*N". This involves two interacting surfaces, which are basically "dry". I.e. not involving any significant lubricating fluid. Dry friction can be both a force of constraint (static friction or traction), or a force which dissipates heat due to the sliding (sliding friction or kinetic friction). Dry friction is only dependent on if the surfaces are static or sliding. The force doesn't depend on how fast the sliding occurs. This is one big difference between dry friction types and friction involving fluids. To discuss further: 1. Static friction. The coefficient is mu_s. F=mu_s*N gives you the maximum value of the static friction force. The force of static friction will oppose what ever motion is induced by the other forces, until it reaches its maximum value. Static friction NEVER dissipates energy as heat. 2. Kinetic friction. The coefficient is mu_k. Typically mu_k is smaller than mu_s. Both coefficients only depend on the roughness nature of the surfaces in contact, not on velocity or normal force. F=mu_k*N tells you the actual value for what friction opposes the sliding. Sliding friction is the type of dry friction which dissipates energy as heat. ---------------------- Now to discuss fluid drag. Fluid drag is the general term. Air drag, air resistance, air friction are all phrases which mean the same, with one condition that the fluid involved is air (as opposed to water or oil and other options). Fluid drag does depend on velocity. It is proportional to either the square of velocity or directly proportional to velocity, depending on flow conditions. For laminar flow conditions, it is proportional to velocity relative to the fluid, and is named "Stokes Drag". For turbulent flow conditions, it is proportional to the square of velocity relative to the fluid and is named "Newton Drag". To discuss further: 1. Stokes drag occurs at slow speeds, where viscosity effects dominate (typically fluids such as oil). Stokes drag can be given by the formula F = -b*v, where b is a constant depending on geometry and fluid viscosity. 2. Newton drag occurs at fast speeds or in less viscous fluids (such as water or air), where impacting pressure effects dominate. Newton drag can be given by the formula F=1/2*Cd*rho*A*v^2, where Cd is the drag coefficient (depending on shape), rho is the fluid density, and A is the effective cross sectional area. BOTH regimes of drag are forces which dissipate energy into heat. Never does the drag force act as a force of constraint on stationary objects.
Answers:You need to change your pump. Assuming your current setup is producing 1 gpm at X psi....you would need to find a pump (i.e. pump curve) where x+50psi will efficiently pump 1 gpm This assumes you don't want to change your pipe diameter. However, changing your pump diameter will not help much either...You may increase frictional losses with a smaller pipe but this will just move you on your pump curve and throw off/change the flow-rate... I would say you need to size your pump
Answers:Cavitation will appear in the pump when the pressure at the inlet approaches the vapor pressure of the water. There is literally enough suction to boil the water. The vapor pressure of water at 20c is 17.54mmHg or 2.34 kPa. The pressure of the water starts at 101.3 kPa. At the pump the pressure is 101.3kPa + 10m head - elevation loss of x sin(30 ) - head loss due to friction. Assuming ideal flow, friction is zero. 10m head =98.1kPa So we look for the elevation that gives 2.34kPa 101.3+98.1-9.81y=2.34 y=20.08m y=x sin(30) x=40.16m pump power req. = work/time = force x dist/time dist=s-h=20m time=1sec F=mg mass=120kg (one liter is one kg) g=9.81m/s^2 120kg*9.81m/s^2*20m/ 1sec = 23,544 kg*m^2*s^-3 Which happens to be the unit of watts. 23.5kW assuming 100% efficiency.