conservation of mass - memphis of mass.pdf · conservation of mass descriptions of flow ! up until...

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Civil Engineering Hydraulics

Conservation of Mass

Descriptions of Flow

¢ 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

Conservation of Mass 2 Monday, September 17, 2012

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¢ 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.


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¢ The first categorization is based on what the flow is in contact with. l  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.



¢ The first categorization is based on what the flow is in contact with. l  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.


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¢ The first categorization is based on what the flow is in contact with. l  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.



¢ 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


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¢ Another description of flow conditions is based on how many variables we need to describe the velocity pattern of the flow



¢  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


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Conservation of Mass 11

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.

Monday, September 17, 2012



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).


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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.



Control Volume ¢ We can start by looking at a bathtub that is

bring filled while the drain plug is not in place


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Control Volume ¢  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 ¢  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


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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?



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.


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Conservation of Mass ¢ We are going to limit our analysis to non-

nuclear applications ¢ Mass can neither be created or destroyed



Conservation of Mass ¢ Back to the tub

l  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


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Conservation of Mass ¢ During some period of time, Δt, 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



Conservation of Mass ¢  It probably isn’t too hard to see that during the

time Δt, that we have a change of mass in the tub of 20 kg of water


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Conservation of Mass ¢ We can write an expression of the time Δt for

the change in the mass of water in the tub

tub in outmass mass massΔ = −



Conservation of Mass ¢ Please notice that we are talking about mass

here, that is important because it allows us to work in both compressible and incompressible flows.


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Conservation of Mass ¢  If we look at the change in the mass per unit of

time we can divide all the terms by Δt tub in outmass mass masst t t

Δ = −Δ Δ Δ



Conservation of Mass ¢ 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

tub in outmass mass masst t t

Δ = −Δ Δ Δ

m


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Conservation of Mass ¢ Each differential part of the mass flow is

crossing through some area on the surface of our control volume


Δ = −Δ Δ Δ

m



Conservation of Mass ¢  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


Δ = −Δ Δ Δ

m =ρdVdt


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Conservation of Mass ¢ The V is slashed in this case to show that it

represents volume rather than velocity


Δ = −Δ Δ Δ

m =ρdVdt



Conservation of Mass ¢ Now the mass movement across a differential

area, n, of the control volume is defined as


Δ = −Δ Δ Δ

mn = ρVndAn


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Conservation of Mass ¢ Vn is the velocity of the mass flow normal to the

control volume surface through the differential element dAn


Δ = −Δ Δ Δ

mn = ρVndAn



Conservation of Mass ¢  If we sum up the mass movements across the

complete surface area of the boundary we have


Δ = −Δ Δ Δ

m = ρVn dAn

A∫

Vn is the velocity normal to the surface through the area dAn


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Conservation of Mass ¢ Since it is usually impossible to find the velocity

at every differential element, we use the average velocity through the total cross section


Δ = −Δ Δ Δ

m = ρVn dAn

A∫



Conservation of Mass ¢ This reduces the integral to


Δ = −Δ Δ Δ

m = ρVavg A

A is the area through which the mass flow is entering (or exiting) the control volume.


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Conservation of Mass ¢ The VavgA term is know as the volumetric flow

rate


Δ = −Δ Δ Δ

m = ρVavg A

Vavg A = V = Q

A is the area through which the mass flow is entering (or exiting) the control volume.



Conservation of Mass ¢ Converting to a differential expression we have

dmasstub

dt= min − mout

m = ρVavg A

Vavg A = V = Q


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Conservation of Mass ¢ A pipeline with a 30 cm inside

diameter is carrying liquid at a flow rate of 0.025 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.



Conservation of Mass ¢ A 4-ft-diameter tank containing solvent

(acetone) is sketched in Figure 3.10. 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.


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Homework Problem 9-1





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¢ The last problem will require you to utilize some of the fundamentals from the class and it will require you to do some integration.




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 = 2ghWhere 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.


conservation of mass - memphis of mass.pdf · conservation of mass descriptions of flow ! up until...

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