Descriptions of Flow

Transcription

1 Civil Engineering Hydraulics Up until this point, we have only considered fluids that were not in motion therefore we were limited to fluids that did not have any velocity. This limited our analytical parameters to force, pressure, and the lines of action of the force 2 1

2 3 When we move into fluids in motion, we expand the characteristics that we consider and therefore the parameters that we have to consider Often, we start with a verbal description of the type of flow that we are considering These are somewhat fuzzy terms as there are flows at the boundaries which could be classified two ways but in general they are sufficient to describe fluids flows for engineering purposes. 4 2

3 The first categorization is based on what the flow is in contact with. If the flow is partially enclosed by what it is flowing through, i.e. It has at least some part of the surface in contact with the atmosphere, then the flow is considered as a Open-Channel flow l This is typical of a open ditch or a river. l 5 The first categorization is based on what the flow is in contact with. If the flow is totally enclosed by what it is flowing through, i.e. Every part of the surface in contact enclosing material, then the flow is considered as a closed-conduit flow l An example would be flow through a water hose at its capacity. l 6 3

4 The first categorization is based on what the flow is in contact with. If the flow is not in contact with any confining surface, then the flow is considered as a Unbounded flow l Flow out of a fountain would be an example here. l 7 8 While the same laws of physics apply to all of these flow conditions, we often use different methods of analysis to work on problems involving each of the flow conditions 4

5 9 Another description of flow conditions is based on how many variables we need to describe the velocity pattern of the flow 10 If we assume that we have a perfectly smooth pipe and that the velocity profile across a cross section of the pipe has the same magnitude at any distance from the center of the pipe, we have a one-dimensional flow 5

6 If we assume a circular cross section in the pipe the velocity at any section will be determined by the cross sectional area looking in the direction of flow. The flow velocity in the left hand section will be determined by the area of the yellow circle while the flow velocity in the right hand section will be determined by the area of the red circle. By making the assumption that the pipe is frictionless, the velocity will be the same at all points in the cross section. 11 In this case, the only variable the would determine the flow velocity would be the cross sectional area the flow was passing through (assuming a constant flow volume). 12 6

7 A more realistic case would be to consider friction in the walls of the pipe such that the flow at the walls was equal to 0. This would cause the flow velocity to develop a non-uniform profile as you move radially away from the center of the pipe. Here we have two variables which would determine the velocity profile or distribution, the first being the cross sectional area of the flow and the second being the distance from the center line of the flow. This would be a two-dimensional flow. 13 Control Volume 14 We can start by looking at a bathtub that is bring filled while the drain plug is not in place 7

8 Control Volume 15 If we choose the entire bathroom as our control volume, we would also have to consider water flowing into and out of the sink, commode, shower, and in through the window. Control Volume 16 If however, we choose the tub alone, then we only have to look at the fluid in the tub, fluid entering the tub, and fluid exiting the tub 8

9 Control Volume Here is where common sense comes in. l 17 If more water is coming into the tub than is going out the drain, what is happening in the tub itself? Control Volume Here is where common sense comes in. l If more water is coming into the tub than is going out the drain, what is happening in the tub itself? Since there aren t any openings in the tub (other than the top), the water will be getting deeper. The tub will be filling up. 18 9

10 We are going to limit our analysis to nonnuclear applications Mass can neither be created or destroyed 19 Back to the tub l 20 If we take a stopwatch and a way to measure the mass of water that enters and leaves the tub we can set up an experiment 10

11 21 During some period of time,, we measure a mass of 50 kg of water entering the tub and during the same time period, we measure a mass of 30 kg of water leaving the tub 22 It probably isn t too hard to see that during the time, that we have a change of mass in the tub of 20 kg of water 11

12 We can write an expression of the time for the change in the mass of water in the tub Δmasstub massin massout Please notice that we are talking about mass here, that is important because it allows us to work in both compressible and incompressible flows. 12

13 If we look at the change in the mass per unit of time we can divide all the terms by 25 We are moving to a differential expression The amount of mass flowing through a cross section per unit of time is known as the mass flow rate and given the symbol m 26 13

14 Each differential part of the mass flow is crossing through some area on the surface of our control volume m 27 If the total mass flow is the sum of n elements of flow then we can say that a differential element of the mass flow (assuming constant density) is equal to ρdv m dt 28 14

15 The V is slashed in this case to show that it represents volume rather than velocity ρdv m dt 29 Now the mass movement across a differential area, n, of the control volume is defined as m n ρvn dan 30 15

16 Vn is the velocity of the mass flow normal to the control volume surface through the differential element dan m n ρvn dan 31 If we sum up the mass movements across the complete surface area of the boundary we have m ρvn dan A Vn is the velocity normal to the surface through the area dan 32 16

17 Since it is usually impossible to find the velocity at every differential element, we use the average velocity through the total cross section m ρvn dan A 33 This reduces the integral to m ρvavg A A is the area through which the mass flow is entering (or exiting) the control volume

18 The VavgA term is know as the volumetric flow rate m ρvavg A Vavg A V Q A is the area through which the mass flow is entering (or exiting) the control volume. 35 Converting to a differential expression we have dmasstub m in m out dt m ρvavg A Vavg A V Q 36 18

19 37 A pipeline with a 30 cm inside diameter is carrying liquid at a flow rate of m3/s. A reducer is placed in the line, and the outlet diameter is 15 cm. Determine the velocity at the beginning and end of the reducer. 38 A 4-ft-diameter tank containing solvent (acetone) is sketched in Figure The solvent is drained from the bottom of the tank by a pump so that the velocity of flow in the outlet pipe is constant at 3 ft/s. If the outlet pipe has an inside diameter of 1 in., determine the time required to drain the tank from a depth of 3 ft to a depth of 6 in. 19

20 Homework Problem Homework Problem

21 Homework Problem 8-3 The last problem will require you to utilize some of the fundamentals from the class and it will require you to do some integration. 41 Homework Problem 8-3 A 4-ft-high, 3-ft diameter cylinderical water tank whose top is open to the atmosphere is initally filled with water. A discharge plug near the bottom of the tank is pulled out and a water just whose diameter is 0.5-in streams out. The average velocity of the jet is given by V 2gh Where h is the height of the water in the tank measured from the elevation of the center of the hole. Determine how long it will take for the water level to drop from 2-ft to the bottom