Is water dripping from a burette a good example of half life? Explain?
Yes. The head (column of water over the orifice) is decreasing, thus the pressure is decreasing thus rate of flow decreases with time. The drip must be slow though for the effect to be obvious. The If the column is falling such that the column has significant momentum then the other posters are correct
An incompressible fluid is flowing through a pipe with a constriction. The pipe is on an incline with an angle?
An incompressible fluid is flowing through a pipe with a constriction. The pipe is on an incline with an angle of 30.0 degrees. The narrow section is 10.0 m from the wide section and the narrow section is lower than the wide section. The velocity of the fluid in the wide section of the pipe is 4.00 m/s. The diameter of the wide section is 12.00 cm and the diameter of the narrow section is 9.00 cm. The pressure of the fluid in the wide section is 250 kPa. What is the pressure in the narrow section of the pipe? (density of the fluid is 1,000 kg/m3) The velocity is inversely proportional to the cross-sectional area of the pipe. v1 * A1 = v2 * A2 v1 * (π * 4.5^2) = 4.00 * π * 6^2 v1 = (4.00 * π * 6^2) ÷ (π * 4.5^2) = 7.11 m/s This is the velocity in the narrow section. The narrow section is 10.0 m from the wide section and the narrow section is lower than the wide section. So, the difference in height = 10.0 * sin 30˚ The narrow section is 5 m is lower than the wide section. Now it’s Bernoulli time! P1 + ½ * 1000 * v1^2 + density * g * h1 = P2 + ½ * density * v2^2 + density * g * h2 v1 = 7.11 m/s, v2 = 4 m/s h1 = 0 m, h2 = 5 m P2 = 250 P1 + ½ * 1000 * 7.11^2 + 1000 * 9.8 * 0 = 250 + ½ * 1000 * 4^2 + 1000 * 9.8 * 8
What are the different types of losses in pump?
A pump can be defined as “a mechanical device that adds energy to a fluid to increase its flow rate and static pressure.This process can be accomplished with positive displacement pumps or kinetic-energy pumps.Losses in a Centrifugal Pump• Mechanical friction power loss due to friction between the fixed and rotating parts in the bearing and stuffing boxes.• Disc friction power loss due to friction between the rotating faces of the impeller (or disc) and the liquid.• Leakage and recirculation power loss. This is due to loss of liquid from the pump and recirculation of the liquid in the impeller. The pressure difference between impeller tip and eye can cause a recirculation of a small volume of liquid, thus reducing the flow rate at outlet of the impellerHead lossesHead losses are potential energy that has been lost because of frictional resistance of the piping system (pipe, valves, fittings, and entrance and exit losses). Unlike velocity head, friction head cannot be ignored in system calculations. Head loss values vary as the square of the flow rate. Head losses can be a significant portion of the total head.Control lossesControl losses occur on the discharge side of a centrifugal pump that has been equipped with a back pressure valve to control flow rate. As the liquid flows through the control valve, energy is lost. Next to static head, control losses are frequently the most important factor in calculating the pump’s total dynamic head. For pump applications, control losses are treated separately from head losses, even though they are included in the hf term.thanks
Why does the velocity of water at the valve increase with decreasing the area of the valve ?
All the other posters have actually answered your original question thoroughly. But, having designed a valve system as part of a design team, let me give you a practical perspective.Essentially, the goal of a valve system is to deliver a controlled flow rate in a specified flow rate range within a specified energy budget, as well as a specified materials budget. The valve has actuator elements in it that open and close that control the flow rate by constricting a flow path. You may also have a requirement to open and close the valve within a specific time. When the valve is fully open, the flow path is the widest with the minimum flow resistance. When it is closing or closed, it has the narrowest path with the maximum flow resistance, depending on the sealing. The resistance can be a "steady-state" resistance, where the flow path is fixed and the pressure drop is due to viscosity (i.e friction) or inertia, i.e. due to changes in flow direction as the fluid goes around the obstruction. The resistance is measured by the valve Cv, which is a standard number used to determine performance (see: Flow coefficient ). In addition, there is the additional path resistance that comes from the tubing and piping that is used to convey the fluid to the valve. Now, the energy for the flow itself can come from a passive source, such as an overhead tank, or a prime mover, such as a pump. With a passive source, such as a water tank, the flow rate for the system is determined by the differential height between the tank fluid level and the valve, as well as the resistance due to the valve and the piping/tubing. When the valve is fully open, the resistance is the lowest and the flow rate is the highest. When the valve is close to fully closed, the resistance is the highest and the flow rate is the lowest. The flow rate divided by the effective open area of the valve determines the average velocity of the fluid coming out. When the valve is open, the area is the largest and usually the average velocity is the lowest. When the valve is nearly closed, the area is the lowest, and the average velocity is the highest. When the valve is fully closed, the resistance is effectively infinite, the flow rate is zero and the average velocity is hence zero.
HELP ME PLEASE HISTORY?
There is so much information about the decline and fall of the Western Roman Empire that the problem you will have is too much information. Narrow your topic and look at just one or two causes. As an example you could easily do a 10 page essay on Diocletian's Edict on Maximum Prices and what can be inferred about the state of the Roman Empire. Some Examples of causes: Role of Military. Lead Pipes Birthrate Roman Arena System of Taxation Structure of Government Rise of Parthian Empire Rise of Christianity Agricultural Policies Threats of Barbarian invasions ETC
Water flows at 0.62 m/s through a 3 cm diameter hose. At the end of the hose is a 0.35 cm diameter nozzle.?
The hose area is A = Pi*r^2 = 3.14*(0.03/2m)^2 = 7.07E-4m^2 The flow rate is Q = A.v = 7.07E-4m^2 * 0.62m/s = 4.38E-4 m^3/s The nozzle area is An = Pi*r^2 = 3.14*(0.0035/2m)^2 = 9.62E-6m^2 a) The speed of water at the nozzle is Vn = Q/An = 4.38E-4m^3/s/9.62E-6m^2 = 45.5m/s b) The pressure right before the nozzle is given by ΔP = P1-P2 = 1/2ρV2² - 1/2ρV1² where ΔP, P1 and P2 are in Pa ρ is water density = 1000 kg/m^3 V1 and V2 are in m/s In your case ΔP = 1/2*1000*(45.5m/s)² - 1/2*1000*(0.62m/s)² ΔP = 1 034 933 Pa = 10.2 atm So right before the nozzle, the pressure is 11.2 atm. And because the speed is so slow, it can be assumed that the pipe friction losses are minimum. For example at this speed, a plastic pipe 100meter long will cause losses of 0.014 atm. So the pressure at the pump is 11.2 atm.
What does "habitat fragmentation" mean?
It is the breaking up of the area in which an population lives into smaller parts. You can see it occuring in areas where suburban sprawl is a problem. It can also be caused by the building of highways, pipelines, etc. It could also be due to natural occurences, such as volcanic activity, earthquakes, rivers changing course, etc.