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Dr. Nikos J. Mourtos – AE 160 / ME 111

CONTINUITY (Conservation of Mass) Case 5: Water Reservoir - Rising / Falling Level Imagine the same water reservoir as before with two entry points (1 and 3) and one exit point (point 2). Now the water level in the reservoir may rise or fall, depending on whether there is more water going in than coming out or vice versa.

The following assumptions are based on the description of the problem:  Incompressible flow  1-D flow at all inlet / outlet pipes Since the water level is changing, it is easier to consider as our control volume the entire tank plus the water in the short cylindrical pipe sections at points 1, 2, and 3. Now the mass inside the control volume ( mcv ) is changing, hence:

m in ≠ m out If m in  m out , the water level will rise. If m in  m out , the water level will fall. The difference m in − m out will be equal to the time rate of change of the mass of the water in the reservoir. Hence, we can write Continuity as: 1

Dr. Nikos J. Mourtos – AE 160 / ME 111

m in − m out =

d ( mcv ) dt

We can eliminate the density because the flow is incompressible:

Qin − Qout =

d(∀ H 2 O,cv ) dt

where ∀ H 2 O,cv is the volume of the water in the reservoir (control volume). Writing the left-hand side of Continuity as in Case 4: V1 A1 − V2 A2 + V3 A3 =  

∑V ⋅ A = − C.S.

d(∀ H 2 O,cv )

(

d ∀ H 2 O,cv

)

dt

dt

This is Continuity for unsteady, incompressible, one-dimensional flow for a control volume with multiple entry / exit points. The minus sign on the right-hand-side is necessary because of the convention discussed in Case 4. The summation on the left-hand-side gives a positive result if the outflow is greater than the inflow, in which case the mass inside the control volume is decreasing, so the derivative on the right-hand side is negative. We can capitalize on the simple geometry of our problem to gain further insights about Continuity by simplifying the unsteady term.

∀ H 2 O,cv = Ares h

Note that

where Ares is the cross-sectional area of the reservoir and h is the depth of the water in the reservoir (see figure below). The cross sectional area of the tank does not change with time, so the unsteady term in Continuity becomes:

d(∀ H 2 O,cv ) dt

=

d dh ( Ares h ) = Ares = AresVsurface dt dt

As shown in the figure below, Vsurface is the velocity with which the surface is rising or falling.

2

Dr. Nikos J. Mourtos – AE 160 / ME 111

It is interesting to note that in this case the unsteady term can be made to look like the inflow or outflow terms. If we move this term to the left-hand-side we have: −V1 A1 + V2 A2 − V3 A3 + AresVsurface = 0

This equation represents a mass balance (or more precisely a volume balance since the flow is incompressible) for the new control volume indicated with the dotted line in the figure above. This control volume includes a fixed depth of water hcv , as shown. The mass of the water inside this control volume does not change with time, even though the water level may be rising or falling, as long as the free surface of the water stays above the top control surface or as long as:

h  hcv Hence, the flow for this control volume is steady (compare with Continuity in Case 4). Assume for a moment that the water level is rising, as shown in the figure above. The mass outflux through the top control surface is given by:

m 4 = AresVsurface

3

Dr. Nikos J. Mourtos – AE 160 / ME 111 This mass outflux together with the mass outflux from exit 2, exactly balances the influx at points 1 and 3, so for our new control volume we have:

m in = m out The last equation shows that our problem has now been transformed into a steady flow problem. We see that the choice of the control volume is critical because it can make a problem appear steady or unsteady. Usually, it is easier to solve a steady flow problem than an unsteady one, so if a control volume exists that allows such a formulation, it may be preferable. Example Consider the water reservoir shown below with four inlets on the sides, two outlets on the bottom, and a cross-sectional area of Ares = 100 m 2 .

The cross-sectional area and the corresponding velocity for each inlet / outlet are given below:

4

Dr. Nikos J. Mourtos – AE 160 / ME 111 A1 = 0.1 m 2

V1 = 1.5 m / s

A2 = 0.4 m

2

V2 = 2.5 m / s

A3 = 0.2 m

2

V3 = 0.5 m / s

A4 = 0.3 m 2

V4 = 2.75 m / s

A5 = 0.6 m 2

V5 = 1.5 m / s

A6 = 1.375 m 2 V6 = 0.75 m / s

Is the reservoir filling or emptying? At what rate is the water level rising or falling? Solution The problem implies that the reservoir may be filling or emptying, hence we must use the unsteady, incompressible form of Continuity for a control volume with multiple entry / exit points:  

∑V ⋅ A = − C.S.

(

d ∀ H 2 O,cv dt

)⇒

−A1V1 − A2V2 − A3V3 − A4V4 + A5V5 + A6V6 = 0 ⇒ −0.1(1.5 ) − 0.4 ( 2.5 ) − 0.2 ( 0.5 ) − 0.3( 2.75 ) + 0.6 (1.5 ) + 1.375 ( 0.75 ) = −

(

d ∀ H 2 O,cv dt

) = +0.14375 m

3

(

d ∀ H 2 O,cv dt

)⇒

/s

The water volume in the reservoir is increasing. As shown earlier, this rate of increase can be written as:

(

d ∀ H 2 O,cv dt Vsurface =

) = +0.14375 m

3

/ s = AresVsurface ⇒

0.14375 ⇒ Vsurface = 0.00144 m / s 100

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