2141-365 Fluid Mechanics for International Engineers
HW#3: Fluid Static, Reynolds Transport Theorem, and Conservation of Mass
Due: Fri., Sep 28th, 2007.
Solution HW #3: Fluid Static, Reynolds Transport Theorem, and Conservation of Mass
Problem 1. Differential Governing Equation of Motion for Static Fluid and Fluid Moving as A Rigid Body.
1.1. Write down the governing differential equation of motion (in vector form) for static fluid or fluid moving as a rigid
body.
1.2. What is the dimension of this equation?
1.3. State the physical meaning of each term in this equation, and state the underlying physical principle of this
equation.
1.4. How is this equation compared to the Newton’s second law of motion,
v
v
∑ F = ma ? What are the similarities,
and what are the differences?
1.5.
v v
V,a
Water
a.
Can we use this equation to describe the motion of a body of water that completely fills the tank of the
water truck?
b.
If the truck is moving at a constant velocity V , what will be the direction of maximum spatial rate of
change of pressure? State the principle and sketch a vector diagram to illustrate your solution.
v
If the truck is moving at a constant linear and forward acceleration a , what will be the direction of
maximum spatial rate of change of pressure? State the principle and sketch a vector diagram to illustrate
your solution. Which part has higher pressure, front or back, and top or bottom?
c.
v
Solution
v
v
1.1. If gravitation is the only body force, we have −∇p + ρg = ρa .
ANS
1.2. Force / Volume
ANS
= Net pressure force per unit volume.
1.3.
−∇p
v
ρg
= Gravitational force per unit volume.
v
v
= ma per unit volume.
ρa
The underlying physical principle of this equation is the Newton’s second law of motion.
ANS
1.4.
v
v
F = ma , i.e., Newton’s second law.
• The underlying physical principle of this equation is the same as that of
∑
•
•
However, in this case the Newton’s second law is applied to a fluid body, which is considered a continuum.
In addition, the forces that act on a fluid body are considered distributive. Hence, forces in this equation are
described as fields.
ANS
a.
Yes, under the assumption that the fluid body is moving as a rigid body. If there is a flow, however, some
modifications to take into account the viscous effect may be required.
ANS
v
v v
v v
v
v
From the Newton’s second law, −∇p + ρg = ρa , if V is constant, a = 0 , and we have − ∇p + ρg = 0 . Hence,
v
v
∇p = ρg and the direction of maximum spatial rate of change of pressure corresponds to the direction of g .
1.5.
b.
2141-365 Fluid Mechanics for International Engineers
HW#3: Fluid Static, Reynolds Transport Theorem, and Conservation of Mass
Due: Fri., Sep 28th, 2007.
−∇p
Water
v v
V,a
v
ρg , ∇ p
c.
v
v
From the Newton’s second law, we have −∇p + ρg = ρa . Hence, the direction of the maximum spatial rate of
v v
change of pressure corresponds to the direction of ∇p = ρ ( g − a ) . This is illustrated in the figure below. Hence,
the back-bottom part has higher pressure.
−∇p
v
ρa
∇p
v v
V,a
v
ρg
ANS
2141-365 Fluid Mechanics for International Engineers
HW#3: Fluid Static, Reynolds Transport Theorem, and Conservation of Mass
Due: Fri., Sep 28th, 2007.
Problem 2. Pressure Distribution and Manometer [Çengel and Cimbala, 2006, Problem 3-9, p. 104.]
The water in a tank is pressurized by air, and the pressure is measured by a multifluid manometer as shown in the
figure below. Determine the gage pressure of air in the tank if h1 = 0.2 m, h2 = 0.3 m, and h3 = 0.46 m. Take the densities
of water, oil, and mercury to be 1,000 kg/m3, 850 kg/m3, and 13,600 kg/m3, respectively.
oil
oil
Air
1
d
Air
1
a
mercury
mercury
h1
h1
c
h3
h3
Water
Water
h2
h2
b
Solution
Assumptions
1. Static fluid
2. ρ is constant within any one fluid.
3. Air pressure in the tank is uniform. (Neglect the effect of elevation change, or hydrostatic pressure, in tank air.)
Basic Equation
v v
− ∇p + ρg = 0
→
p a = pb − ρ M gh3
pb = pc + ρ O gh2
p( h) = po + ρgh .
po is the pressure at the interface.
h is positive downward.
→
p d − p a = ρ M gh3 − ρ O gh2 − ρW gh1 .
pc = p d + ρW gh1
Thus,
p d − p a = g (ρ M h3 − ρ O h2 − ρW h1 )
= 9.81
m
2
× (13,600 × 0.46 − 850 × 0.3 − 1,000 × 0.2
s
= 56,907 Pa
That is, the air in the tank has the pressure 56.9 kPa gage.
) kg3 m
m
ANS
2141-365 Fluid Mechanics for International Engineers
HW#3: Fluid Static, Reynolds Transport Theorem, and Conservation of Mass
Due: Fri., Sep 28th, 2007.
Problem 3. Resultant Pressure Force
For each of the surfaces (a-f) shown below, there is a uniform and equal pressure p A acting on the left surface (side
A).
3.1. Which surfaces have equal x -component of the net pressure force on side A?
3.2. Which surfaces have equal y -component of the net pressure force on side A?
3.3. In addition, for which surfaces you can tell directly and exactly the direction of the net pressure force on side A?
Draw a force vector on the diagram to indicate such direction.
In doing so, briefly prove your answers from relevant vector equations.
y
x
pA
A B
a
b
c
d
e
f
Solution
v
v
Consider an infinitesimal area element on side A. We have the infinitesimal force dF due to pressure on dA as
v
v
dF = − pdA .
Its components in the x and y directions are given by
v
v
dF = − pdA
dFx = − pdAx
v
,
dA
dFy = − pdAy
v
where dAx and dA y are recognized as the components of the vector dA (thus, have sign). Therefore, the net forces due to
the uniform pressure p A on side A along the x and y directions are given by
∫
Fx = − pdAx = − p A Ax ,
A
∫
Ax = dAx
A
∫
F y = − pdAy = − p A Ay ,
A
∫
Ay = dAy
A
3.1. Because all surfaces have equal uniform pressure p A and equal projected area in the x direction Ax , the net forces
due to the uniform pressure p A on side A along the x direction are equal.
ANS
3.2. Because the projected areas in the y direction Ay of (b) to (e) are the same, they have equal net force along the y
direction. In fact, since Ay = 0, F y = 0 . This is not the case with (a) and (f).
ANS
3.3. From 3.1 and 3.2 ( F y = 0 ), we can see that the net forces due to pressure on side A of (b) to (e) are equal and in the x
ANS
direction.
v v
F = Fx
2141-365 Fluid Mechanics for International Engineers
HW#3: Fluid Static, Reynolds Transport Theorem, and Conservation of Mass
Due: Fri., Sep 28th, 2007.
Problem 4. Hydrostatic Force on A Curved Surface [Adapted from Fox et al., 2004, Problem 3.68, p. 93.]
A Tainter gate (in a circular arc shape) used to control water flow from the Uniontown Dam on the Ohio River is
shown; the gate width is w = 35 m.
4.1. Determine the algebraic expressions for the magnitude and direction (the force vector), and the line of action of the
net force due to water and air pressures acting on the gate.
4.2. Determine the corresponding numerical values of the force vector and the line of action.
4.3. By inspecting the governing equations of the resultant force and the resultant moment, can you tell anything about
the resultants (force and/or moment) without actually having to solve anything yet? If so, what is it?, and how?
x
R = 20 m
D = 10 m
y
water
Solution
Assumptions
1. Static fluid
2. ρ = constant = 1,000 kg/m3
x
α
θ
R
v
v
dF = − pdA
êr
y
D
v
dA
êθ
v
4.1. Consider the infinitesimal area element dA as shown above. The net pressure force due to pressures on both sides of
the gate is given by
v
v
v
v
v
v
dA = wRdθ eˆr
dF = − pdA
+ (− p o (− dA)) air side = −( p − p o )dA = − ρghdA ,
water side
where, due to the cancellation, there is no contribution from the part of the gate above the water level. Hence, the net
pressure force is given by
α
v
v
F = − ρghdA = − ρghwRdθeˆr
∫
∫
A
0
α
∫
= − ρgwR 2 sin θ (cosθ eˆ x + sin θ eˆ y )dθ ,
eˆr = cos θ eˆ x + sin θ eˆ y , h = R sin θ
0
α
α
α
⎤
⎡ 1 α
⎤
⎡
1 − cos 2θ
dθ ⎥
= ρgwR 2 ⎢− eˆ x sin θ cos θ dθ − eˆ y sin θ sin θ dθ ⎥ = ρgwR 2 ⎢− eˆ x sin 2θ dθ − eˆ y
2
⎥⎦
⎢⎣ 2 0
⎥⎦
⎢⎣
0
0
0
α⎤
⎡ 1
⎡1
θ 1
α
⎛α 1
⎞⎤
= ρgwR 2 ⎢− eˆ x − cos 2θ 0 − eˆ y − sin 2θ ⎥ = ρgwR 2 ⎢ eˆ x (cos 2α − 1) − eˆ y ⎜ − sin 2α ⎟⎥
2 4
⎝2 4
⎠⎦
⎢⎣ 4
⎣4
0 ⎥
⎦
∫
That is,
[
∫
∫
]
∫
v 1
ANS
F = ρgwR 2 (cos 2α − 1)eˆ x − (2α − sin 2α )eˆ y .
4
v
v
Due to symmetry, the resultant force F acts in the mid-span
v plane. The line of action of F can be found from the
v
′
principle of moment. Let r denote the point of application of F in the mid-span plane. We thus have
2141-365 Fluid Mechanics for International Engineers
HW#3: Fluid Static, Reynolds Transport Theorem, and Conservation of Mass
Due: Fri., Sep 28th, 2007.
v
v
v
v v
v
r ′ × F = ∫ r × dF = − ∫ ( R eˆr ) × p dA eˆr = 0 .
A
v v
That is, r ′ // F , or
A
iˆ
ˆj
kˆ
v v
r ′ × F = x′
y′
v
0 = ( x ′F y − y ′Fx )kˆ = 0 .
Fx
Fy
0
Therefore, the line of action is given by
y ′ Fy
2α − sin 2α
=
=−
→
x ′ Fx
cos 2α − 1
y′ =
sin 2α − 2α
x′ .
cos 2α − 1
ANS
4.2. Since α = 30o, we have
v 1
⎡ 1
⎤
π
3
F = ρgwR 2 ⎢− eˆ x − ( −
= −17.2 eˆ x − 6.2 eˆ y
)eˆ y ⎥
4
3
2
⎢⎣ 2
⎥⎦
and
y ′ = 0.362 x ′ ,
or
the resultant force acts through the pivot at an angle θ = 19.9o.
MN ,
ANS
ANS
v
v
4.3. Since the gate is a circular arc with the center at the pivot and all the infinitesimal forces dF = − pdA pass through the
v v
v
v
v v
v
v v
v
v
pivot, r × dF = 0 , ∫ r × dF = 0 , and from the moment equation r ′ × F = ∫ r × dF we know that the resultant force F
A
v
must be parallel to r ′ and must also pass through the pivot.
A
ANS
2141-365 Fluid Mechanics for International Engineers
HW#3: Fluid Static, Reynolds Transport Theorem, and Conservation of Mass
Due: Fri., Sep 28th, 2007.
Problem 5. Reynolds Transport Theorem (and Streamlines and Flux)
5.1. Write down the mathematical statement of the Reynolds transport theorem for a property N whose intensive
property is η (N per unit mass).
Write down the corresponding word statement of the Reynolds transport theorem and state the physical
interpretation of each term in the equation.
Why do we often use, or often need to use, the Reynolds transport theorem in fluid flow applications? A brief
description is fine.
5.2. Consider an incompressible flow through a two-dimensional converging nozzle of length l , span width w , inlet
width 2 y o , and contraction ratio α (ratio of exit area to inlet area), as shown below. Provided that the viscous
effect and the boundary layer near the wall are neglected (assume inviscid flow), a simple approximation of the
velocity field for this flow is given by
v
V = (u , v) = (U o + bx)iˆ − byˆj ,
where the top and bottom nozzle profiles correspond to certain upper and lower streamlines. [No flow can cross a
streamline, hence a streamline can be used to represent a solid boundary.] U o = u ( x = 0) is a given velocity,
and b is a yet-to-be-determined parameter.
Consider the following quantities given:
Geometrical properties: l , w , y o , Ao = A( x = 0) = 2 wy o , and α .
Kinematic properties:
Uo
Fluid properties:
ρ
a.
b.
c.
d.
e.
Is this flow steady? Is this flow, one-, two-, or three-dimensional?
Determine the unknown parameter b in terms of U o , l , and α . Does b depend on the velocity U o ? On
what quantity does the normalized/non-dimensionalized b depend?
Find the normalized kinetic energy fluxes at the inlet and exit in terms of α and the normalized halfwidth at inlet yo / l . Let the kinetic energy flux be normalized by ρAoU o3 , where Ao is the cross section
area at the inlet.
Finally, find the rate of change of kinetic energy of a control mass that instantaneously resides within the
nozzle.
Do you think you can design a simple two-dimensional nozzle with a contraction ratio α , length l , and
inlet velocity U o ?
y
Span width w
y = yo
y = α yo
x=l
y = −α yo
y = − yo
x
2141-365 Fluid Mechanics for International Engineers
HW#3: Fluid Static, Reynolds Transport Theorem, and Conservation of Mass
Due: Fri., Sep 28th, 2007.
Solution
5.1. See Worked-Out Problem Set on The Web.
5.2.
a. It is a steady, two-dimensional flow.
ANS
b. Note that the velocity fields for y > 0 and for y < 0 are mirror images: v(- y ) = −b(− y ) = by = −( −by ) = − v( y ) .
The unknown parameter b must be such that two of the resulting streamlines correspond to the upper and lower
walls, respectively. That is, it must be such that the upper streamline passes through the points (0, y o ) and (l , α y o )
[hence, the lower streamline through the mirror image points (0,− y o ) and (l ,−α y o ) ]. Therefore, we proceed by
finding the streamlines that passes through the point (0, y o ) .
Streamline:
dy dx
=
u
v
y
−
∫
yo
x
dy
dx
=
by 0 U o + bx
∫
y
1
1 U + bx
− ln
= ln o
b yo b
Uo
Hence, the streamline that passes through the point (0, y o ) is given by
y ( x) =
U o yo
,
U o + bx
and b is given by
⎛ yo
⎞
⎜⎜
− 1⎟⎟ .
⎝ y (x) ⎠
In order for the streamline to also pass through the point (l , α y o ) , we must have
b=
Uo
x
Uo ⎛ 1
⎞
ANS
⎜ − 1⎟ .
l ⎝α
⎠
Similar conclusion is arrived if we consider the lower streamline. Hence, b depends on the velocity U o . However, the
ANS
normalized bl / U o = 1 / α − 1 depends only on α and does not depend on U o .
b=
c.
In addition, the streamline that passes through the points (0, y o ) and (l , α y o ) is then given by
y ( x)
1
=
.
x⎛1
yo
⎞
1 + ⎜ − 1⎟
l ⎝α
⎠
As a result of (b), we have the velocity field given by
v
⎛ ⎛1
⎞
⎞x⎞ U ⎛ 1
V = (u , v) = U o ⎜⎜1 + ⎜ − 1⎟ ⎟⎟iˆ − o ⎜ − 1⎟ yˆj
l ⎝α
⎠
⎠l⎠
⎝ ⎝α
⎡⎛ ⎛ 1
⎞x⎞ ⎛ 1
⎞y ⎤
= U o ⎢⎜⎜1 + ⎜ − 1⎟ ⎟⎟iˆ − ⎜ − 1⎟ ˆj ⎥
⎠ l ⎠ ⎝α
⎠l ⎦
⎣⎝ ⎝ α
The kinetic energy flux at any cross section at x can be found as follows.
2141-365 Fluid Mechanics for International Engineers
HW#3: Fluid Static, Reynolds Transport Theorem, and Conservation of Mass
Due: Fri., Sep 28th, 2007.
FKE ( x) =
v v
1
V 2 ρV ⋅ dA,
2A
v
dA = ( wdy ) iˆ
∫
2
2
⎡⎛ ⎛ 1
1
⎞ x ⎞ ⎛⎛ 1
⎞ y⎞ ⎤ v v
= U o2 ⎢⎜⎜1 + ⎜ − 1⎟ ⎟⎟ + ⎜⎜ ⎜ − 1⎟ ⎟⎟ ⎥ ρV ⋅ dA
2
⎢
⎝α
⎠ l ⎠ ⎝⎝ α
⎠ l ⎠ ⎥⎦
A ⎣⎝
2
2
y⎡
⎛ ⎛1
1
⎞ x ⎞ ⎛⎛ 1
⎞ y⎞ ⎤
wU o2 ⎢⎜⎜1 + ⎜ − 1⎟ ⎟⎟ + ⎜⎜ ⎜ − 1⎟ ⎟⎟ ⎥ ρ (U o + bx)dy
2
⎢
⎝α
⎠ l ⎠ ⎝⎝ α
⎠ l ⎠ ⎥⎦
− y⎣⎝
2
2
2
2
y⎡
y
⎡⎛
⎤
⎛⎛ 1
⎛ ⎛1
1
1
⎞ y⎞ ⎤
⎛1
⎞ x⎞
⎞ y⎞
⎞ x ⎞ ⎛⎛ 1
= wU o2 ρ (U o + bx) ⎢⎜⎜1 + ⎜ − 1⎟ ⎟⎟ + ⎜⎜ ⎜ − 1⎟ ⎟⎟ ⎥ dy = wU o2 ρ (U o + bx) ⎢⎜⎜1 + ⎜ − 1⎟ ⎟⎟ 2 y + ⎜⎜ ⎜ − 1⎟ ⎟⎟ dy ⎥
2
2
α ⎠l⎠ ⎥
⎢⎝ ⎝ α
⎢
⎠ l ⎠ ⎥⎦
⎠l⎠
⎝α
⎠ l ⎠ ⎝⎝ α
− y⎣⎝
− y⎝ ⎝
⎣
⎦
2
2
2
2
2
⎛
⎡
⎤
⎛
⎛
⎛ ⎛1
1
1
1
1
1
1 y2
x⎞
x⎞
⎞ x⎞
⎛1
⎞ 1 y
⎥ = wU o3 ρ ⎜⎜1 + ⎛⎜ − 1⎞⎟ ⎟⎟ 2 y ⎜ ⎜⎜1 + ⎛⎜ − 1⎞⎟ ⎟⎟ + ⎛⎜ − 1⎞⎟
= wU o2 ρ (U o + bx) ⎢⎜⎜1 + ⎜ − 1⎟ ⎟⎟ 2 y + ⎜ − 1⎟
2
y
2
⎥ 2
⎢⎝ ⎝ α
⎠ l ⎠ ⎜⎝ ⎝ ⎝ α
⎠ l ⎠ ⎝α
⎠l⎠
⎝α
⎠ 3 l2
⎠ 3 l2
⎝ ⎝α
⎦
⎣
2
2
2
1
y ⎛ ⎛1
⎞ x ⎞ ⎛⎛ ⎛ 1
⎞ x⎞ ⎛ 1
⎞ 1 y ⎞⎟
⎜⎜1 + ⎜ − 1⎟ ⎟⎟ ⎜ ⎜⎜1 + ⎜ − 1⎟ ⎟⎟ + ⎜ − 1⎟
= ρAoU o3
2
2
yo ⎝ ⎝ α
⎠ l ⎠ ⎜⎝ ⎝ ⎝ α
⎠ l ⎠ ⎝α
⎠ 3 l ⎟⎠
Hence, the normalized kinetic energy flux at any cross section x is given by
2
2
2
FKE ( x) 1 y ⎛ ⎛ 1
⎞ x ⎞ ⎛⎜ ⎛ ⎛ 1
⎞ x⎞ ⎛ 1
⎞ 1 y ⎞⎟
⎜
⎟
⎜
⎟
,
1
1
1
1
1
=
+
−
+
−
+
−
⎜
⎟
⎜
⎟
⎜
⎟
ρAoU o3 2 y o ⎜⎝ ⎝ α ⎠ l ⎟⎠ ⎜⎝ ⎜⎝ ⎝ α ⎠ l ⎟⎠ ⎝ α ⎠ 3 l 2 ⎟⎠
and at inlet by
2
2
FKE ( x = 0) 1 ⎛⎜ ⎛ 1
⎞ 1 y o ⎞⎟
,
ANS
=
+
−
1
1
⎜
⎟
2 ⎜⎝ ⎝ α
ρAoU o3
⎠ 3 l 2 ⎟⎠
and at exit by
2
2
2 2
FKE ( x = l ) 1 ⎛⎜ ⎛ 1 ⎞ ⎛ 1
⎞ 1 α y o ⎞⎟
.
ANS
+
−
=
1
⎜
⎟
⎜
⎟
2 ⎜⎝ ⎝ α ⎠ ⎝ α
ρAoU o3
⎠ 3 l 2 ⎟⎠
∫
∫
=
∫
d.
∫
Using the RTT for the kinetic energy, we have
v v
dKE MV (t ) dKECV (t )
1
=
+ ∫ V 2 ( ρV ⋅ dA)
dt
dt
2
CS (t )
1
2
η = V 2:
⎤
v v
d ⎡
1 2
1
= ⎢ ∫
V ( ρdV )⎥ + ∫ V 2 ( ρV ⋅ dA)
dt ⎢CV (t ) 2
⎥ CS (t ) 2
⎣
⎦
.
Since the flow is incompressible and steady, the unsteady term vanishes. Thus, the time rate of change of the kinetic
energy of the material volume that instantaneously coincides with the control volume is equal to the net efflux through
the control surface according to
dKE MV (t )
1 2 v v
1 2 v v
1 2 v v
=
V ( ρV ⋅ dA) =
V ( ρV ⋅ dA) +
V ( ρV ⋅ dA)
dt
2
2
2
.
CS (t )
Ainlet
AExit
∫
∫
= FKE ( x = l ) − FKE ( x = 0)
Thus,
∫
⎞
⎟
⎟
⎠
2141-365 Fluid Mechanics for International Engineers
HW#3: Fluid Static, Reynolds Transport Theorem, and Conservation of Mass
Due: Fri., Sep 28th, 2007.
dKE MV (t )
= FKE ( x = l ) − FKE ( x = 0)
dt
2
2
2 2
2
⎧⎪ ⎛ ⎛ 1 ⎞ 2 ⎛ 1
1
⎞ 1 yo
⎞ 1 α y o ⎞⎟ ⎛⎜ ⎛ 1
−
+
−
= ρAoU o3 ⎨ ⎜ ⎜ ⎟ + ⎜ − 1⎟
1
1
⎜
⎟
2
2
⎟ ⎜ ⎝α
2
⎠ 3l
⎠ 3 l
⎪⎩ ⎜⎝ ⎝ α ⎠ ⎝ α
⎠ ⎝
2
⎧⎪ ⎛ 1 ⎞ 2 ⎛ 1
⎫⎪
1
1 y o2
⎞
= ρAoU o3 ⎨ ⎜ ⎟ + ⎜ − 1⎟ α 2 − 1
−
1
⎬
2
3 l2
⎠
⎪⎩ ⎝ α ⎠ ⎝ α
⎪⎭
(
⎞
⎟
⎟
⎠
⎫⎪
⎬
⎪⎭
)
ANS
2141-365 Fluid Mechanics for International Engineers
HW#3: Fluid Static, Reynolds Transport Theorem, and Conservation of Mass
Due: Fri., Sep 28th, 2007.
Problem 6. Conservation of Mass and Time Scale of A Problem in Unsteady Flow
6.1. Find the explicit algebraic expression for the dimensionless time that is needed to drain a liquid from the initial
height h to the final fractional height α h from a cylindrical tank. The fluid has the density ρ . The average jet
exit velocity (shown below) is V = 2 gh , where h is the instantaneous height of fluid in the tank measured from
the center of the hole. All other geometric parameters are given in the figure.
6.2. What is the appropriate time scale for this problem (i.e., the standard unit of measure of time that is inherently in
the problem itself)?
Note:
This problem is not directly about dimensional analysis. However, we shall use the basics of dimensional analysis
in two ways:
1. In 6.1, we shall arrange the explicit solution in terms of dimensionless groups.
2. In 6.2, from the explicit solution, we shall find the appropriate unit of measure of time that is inherently
in the problem itself, i.e., the time scale of the problem.
6.3. Say, we want to drain water to half its original height from a given tank. Will the times required be different for
two different original heights if we measure the times in the unit of seconds? How about if we measure the times
in the unit of the time scale of the problem?
6.4. A 200-Litre tank has the diameter of approximately 60 cm and the height of 90 cm. Assume that there is a drain
hole of 2 cm in diameter at the bottom.
a. How much time is required to drain water from the full height of 90 cm to half? Answer in both units of
second and time scale of the problem.
b. How much time is required to drain it further by another half (from 45 cm to 22.5 cm)? Again, answer in
both units of seconds and time scale of the problem.
Make any assumptions you deem necessary.
Think-about-it questions (no need to turn in, and no score will be given):
• Does the density ρ play any role in the current problem/solution? Why? Note that from the current
information, all other quantities are either geometric or kinematic.
•
Is the jet exit velocity V = 2 gh a good approximation for water? How about if it is a tank of real viscous
oil? Will it take the same amount of time as water? Do fluid properties play any role in this problem, from
both perspectives of the current idealized one and the more realistic one?
Air
Water
ho
α ho
Dj
V = 2 gh
Dt
Solution
6.1.
Assumption 1:
Assumption 2:
Incompressible flow. Densities of water and air are constant.
Uniform velocity at exit cross section.
2141-365 Fluid Mechanics for International Engineers
HW#3: Fluid Static, Reynolds Transport Theorem, and Conservation of Mass
Due: Fri., Sep 28th, 2007.
Air
ho
h
Water
α ho
Dj
V = 2 gh
Dt
0=
⎤
v v
dM MV (t ) d ⎡
= ⎢ ρdV ⎥ + ( ρV ⋅ dA)
dt
dt ⎢⎣CV
⎦⎥ CS
Control Volume 1
∫
∫
(1)
Stationary and non-deforming CV. The CV is the original volume of water at height ho .
Equation 1 becomes
⎤
v v
⎛ dM MV (t ) ⎞ d ⎡
0 ⎜=
⎟ = ⎢ ρdV ⎥ + ( ρV ⋅ dA)
dt
⎝
⎠ dt ⎣⎢CV
⎦⎥ CS
∫
∫
⎤
⎤
v v
v v
d ⎡⎢
d ⎡
ρ a dV ⎥ +
( ρ aV ⋅ dA) + ⎢ ρ w dV ⎥ +
( ρ wV ⋅ dA)
dt ⎢V
dt ⎢V
⎥ A: Air in
⎥ A: Water out
⎣ air
⎦
⎣w
⎦
Since each species is individually conserved, we have for air
⎤
v v
d ⎡⎢
ρ a dV ⎥ +
( ρ aV ⋅ dA) = 0 ,
dt ⎢V
⎥ A: Air in
⎣ air
⎦
and for water, or from (2) and (3),
⎤
v v
d ⎡⎢
⎥+
ρ
dV
( ρ wV ⋅ dA) = 0 .
w
∫
∫
dt ⎢V
⎥ A: Water out
⎣w
⎦
Thus,
=
∫
∫
∫
∫
∫
(2)
∫
d ( ρ w Ah)
+ ρ w A j 2 gh = 0
dt
Aj
dh
=−
dt
A
αho
Aj
dh
∫ h =− A
h
(
αho
=−
ho
)
2 ho 1 − α =
t
2 g ∫ dt
0
Aj
Aj
A
(
2gt
A
2gt
t = 1− α
Hence, the dimensionless time is given by
(4)
2 gh
o
2 h
(3)
)
(
A
= 1− α
2ho / g
Aj
)
⎛D
2ho / g ⎜ t
⎜ Dj
⎝
⎞
⎟
⎟
⎠
2
2141-365 Fluid Mechanics for International Engineers
HW#3: Fluid Static, Reynolds Transport Theorem, and Conservation of Mass
Due: Fri., Sep 28th, 2007.
(
)
(
2
)
⎛D ⎞
A
= 1− α
= 1− α ⎜ t ⎟ .
⎜ Dj ⎟
Aj
2ho / g
⎝
⎠
t
t =
*
ANS
ANS
2ho / g .
6.2. From above, we can see that the time scale for this problem is
6.3. If we measure the time required to drain water in the unit of seconds, it is given by
(
t = 1− α
)
⎛D
2ho / g ⎜ t
⎜ Dj
⎝
2
⎞
⎟ ,
⎟
⎠
which depends on the original height. However, if we measure the time in the unit of the time scale of the problem, it is
given by
⎛
t
*⎜
⎜
⎝
=
⎞
⎟ =
2ho / g ⎟⎠
t
⎛D
1− α ⎜ t
⎜ Dj
⎝
(
)
2
⎞
⎟ ,
⎟
⎠
ANS
which does not depend on the original height.
6.4.
a. 90 cm to 45 cm:
⎛
⎞
⎟ =
t =
⎜
⎟
2
/
h
g
o
⎝
⎠
2ho / g = 0.428 s,
t
*⎜
(
)
2
2
⎞
⎟ = 1 − 1 / 2 ⎛⎜ 60 ⎞⎟ = 263.6 ,
⎟
⎝ 2 ⎠
⎠
(
)
t = t * × 2ho / g = 112.9 s.
and
b.
⎛D
1− α ⎜ t
⎜ Dj
⎝
ANS
45 cm to 22.5 cm:
t * does not change and we still have
⎛
while
⎞
⎟ =
t =
⎜
2ho / g ⎟⎠
⎝
2ho / g = 0.303 s,
and
t = t * × 2ho / g = 79.8 s.
t
*⎜
⎛D
1− α ⎜ t
⎜ Dj
⎝
(
)
2
⎞
⎟ = 1 − 1/ 2
⎟
⎠
(
)⎛⎜⎝ 602 ⎞⎟⎠
Air
ho
α ho
= 263.6 ,
ANS
Note
On the other hand, if we choose to use a deforming CV, we have the following.
h
2
Water
Dj
V = 2 gh
Dt
2141-365 Fluid Mechanics for International Engineers
HW#3: Fluid Static, Reynolds Transport Theorem, and Conservation of Mass
Due: Fri., Sep 28th, 2007.
Control Volume 2: Deforming CV. CV always contains water in the tank only.
Equation 1 becomes
⎤
v v
⎛ dM MV (t ) ⎞ d ⎡
0 ⎜=
⎟ = ⎢ ∫ ρdV ⎥ + ∫ ( ρV ⋅ dA)
dt
⎝
⎠ dt ⎢⎣CV
⎥⎦ CS
⎤
v v
d ⎡
⎢ ∫ ρ w dV ⎥ +
( ρ wV ⋅ dA)
∫
dt ⎢V
⎥ A: Water out
⎣w
⎦
d ( ρ w Ah)
=
+ ρ w A j 2 gh
dt
.....
=
This is the same as Eq. (4) above.
2141-365 Fluid Mechanics for International Engineers
HW#3: Fluid Static, Reynolds Transport Theorem, and Conservation of Mass
Due: Fri., Sep 28th, 2007.
7.
Mass Flux and Conservation of Mass in Compressible Flow
A steady, compressible flow of air is passed through a circular, converging nozzle as shown in the figure below. Find
the velocity at the inlet ( V1 ) and the Mach numbers at the inlet ( M 1 ) and the exit ( M 2 ). Assume that the azimuthal and the
radial components of the velocity are negligible.
Note that while Reynolds number (~ Inertial force/Viscous force) plays important role in incompressible viscous flow,
Mach number (~ local speed / speed of sound) is another dimensionless number that plays important role in compressible
flow. Mach number is defined as the ratio of the local speed of fluid V to the local speed of sound c . For a perfect gas, the
speed of sound is related to the local temperature T through c = γRT , where R is the gas constant and γ is the specific
heat ratio. For air R = 287 J/kg-K and γ = 1.4 (again dimensionless).
Make any other assumptions that you deem necessary.
A1 = 0.001 m2
p1 = 650 kPa (abs)
T1 = 335 K
m& 1 = 0.744 kg/s
A2 = 5.1× 10 −4 m2
p 2 = 454 kPa (abs)
T2 = 302 K
Solution
Assumptions:
1. Steady flow
2. Compressible flow
3. Uniform velocity and uniform properties at each cross section.
4. Perfect gas
Mass Flux:
m& 1 =
v
v
∫ ( ρV ⋅ dA)
→
m& 1 = ρ1V1 A1
(1)
A1
Thus,
V1 =
m& 1
m& RT 0.744 kg / s 287 J / kg − K × 335 K
= 1 1=
×
= 110 m / s ANS
650 k Pa
ρ1 A1 A1 p1
0.001 m 2
0 (1)
0=
C-Mass
Thus,
⎤
v v
dM MV (t ) d ⎡
= ⎢ ρdV ⎥ + ( ρV ⋅ dA)
dt
dt ⎢⎣CV
⎥⎦ CS
∫
∫
(2)
v v
v v
0 = ∫ ( ρV ⋅ dA) + ∫ ( ρV ⋅ dA)
A1
A2
= −m& 1 + ρ 2V2 A2
m& 1
m&
RT
0.744 kg / s 287 J / kg − K × 302 K
V2 =
= 1× 2 =
×
= 278.5 m / s
454 kPa
ρ 2 A2 A2 p2 5.1× 10 −4 m 2
The speeds of sound and the Mach numbers at 1 and 2 are given by
c1 = γRT1 = 366.9 m/s
M 1 = V1 / c1 = 0.3.
ANS
c2 = γRT2 = 348.3 m/s
M 2 = V2 / c2 = 0.8.
ANS