1 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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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
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Descriptions of Flow
¢ When we move into fluids in motion, we expand the characteristics that we consider and therefore the parameters that we have to consider
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Descriptions of Flow
¢ 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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Descriptions of Flow
¢ 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.
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Descriptions of Flow
¢ 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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Descriptions of Flow
¢ 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.
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Descriptions of Flow
¢ 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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Descriptions of Flow
¢ Another description of flow conditions is based on how many variables we need to describe the velocity pattern of the flow
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Descriptions of 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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Descriptions of Flow
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.
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Descriptions of Flow
Conservation of Mass 12
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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Descriptions of Flow
Conservation of Mass 13
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.
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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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Conservation of Mass 15
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.
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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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Conservation of Mass 17
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?
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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?
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 19
Conservation of Mass ¢ We are going to limit our analysis to non-
nuclear applications ¢ Mass can neither be created or destroyed
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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 21
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
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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 23
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 = −
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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 25
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
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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 27
Conservation of Mass ¢ Each differential part of the mass flow is
crossing through some area on the surface of our control volume
tub in outmass mass masst t t
Δ = −Δ Δ Δ
m
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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
tub in outmass mass masst t t
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
tub in outmass mass masst t t
Δ = −Δ Δ Δ
m =ρdVdt
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Conservation of Mass ¢ Now the mass movement across a differential
area, n, of the control volume is defined as
tub in outmass mass masst t t
Δ = −Δ Δ Δ
mn = ρVndAn
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Conservation of Mass 31
Conservation of Mass ¢ Vn is the velocity of the mass flow normal to the
control volume surface through the differential element dAn
tub in outmass mass masst t t
Δ = −Δ Δ Δ
mn = ρVndAn
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Conservation of Mass ¢ If we sum up the mass movements across the
complete surface area of the boundary we have
tub in outmass mass masst t t
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 33
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
tub in outmass mass masst t t
Δ = −Δ Δ Δ
m = ρVn dAn
A∫
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Conservation of Mass ¢ This reduces the integral to
tub in outmass mass masst t t
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 35
Conservation of Mass ¢ The VavgA term is know as the volumetric flow
rate
tub in outmass mass masst t t
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.
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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 37
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.
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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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Conservation of Mass 39
Homework Problem 9-1
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Homework Problem 9-2
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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.
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Homework Problem 8-3
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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.