This is not the document you are looking for? Use the search form below to find more!

0.00 (0 votes)

drain

- Added:
**February, 01st 2011** - Reads:
**294** - Downloads:
**2** - File size:
**117.78kb** - Pages:
**7** - content preview

- Name:
**sdfsdf**

Content Preview

IOP PUBLISHING

EUROPEAN JOURNAL OF PHYSICS

Eur. J. Phys.**30 **(2009) 277–283

doi:10.1088/0143-0807/30/2/006

**An exact solution to the draining**

reservoir problem of the

incompressible and non-viscous liquid

**Seok-In Hong**

Department of Science Education, Gyeongin National University of Education, Anyang 430-739,

Korea

E-mail: sihongtao@hanmail.net

Received 21 October 2008, in ﬁnal form 6 December 2008

Published 13 January 2009

Online at stacks.iop.org/EJP/30/277

**Abstract**

The exact expressions for the drain time and the height, velocity and

acceleration of the free surface are found for the draining reservoir problem

of the incompressible and non-viscous liquid. Contrary to the conventional

approximate results, they correctly describe the initial time dependence of the

liquid velocity and acceleration. Torricelli’s law does not hold in the initial

transient region, which imposes restrictions on the validity of the analogy

between the drain system and the electric circuit (Ohm’s law).

Recently, the classic draining reservoir experiments in ﬂuid dynamics have been performed

in relation to physics education. de Oliveira*et al *[1] measured the speed of water squirting

through a pin-hole tube located at the lateral bottom of a cylindrical reservoir and the height of

the free surface of water by making use of rulers. A simple empirical correction was made to

Torricelli’s law which is derived from the steady Bernoulli equation (BE), taking energy losses

into account. Guera*et al *[2] derived the drain time from the steady BE and the assumption of

constant acceleration for the free surface of water, and compared it with experimental results

obtained by a motion sensor. For a cylindrical reservoir draining through a tube, Sianoudis

and Drakaki [3] measured the liquid height using a motion sensor. This allowed conﬁrmation

of Poiseuille’s law (the relationship between the ﬂow rate and pressure difference between the

ends of the tube) for viscous liquids. In addition, Aguilar [4] used a level sensor to measure

the water level in a draining reservoir.

Draining is obviously an unsteady process. In this paper, the draining problem [5, 6]

of the incompressible and non-viscous liquid in a reservoir is revisited in search of the exact

solution of the unsteady BE. It will be compared with solutions of the approximate unsteady

BE or the steady BE. In particular, we emphasize that the well-known Torricelli law must

be modiﬁed for such an ideal liquid (see (21)) and its limitations need to be addressed

conceptually with undergraduate students in introductory physics courses. This work will

0143-0807/09/020277+07$30.00

c 2009 IOP Publishing Ltd

Printed in the UK

277

278

S-I Hong

A1

Usurface

H(t)

A2

U2

**Figure 1. **Draining of a cylindrical reservoir. Plane 1 is the cross-sectional plane of the cylinder

with area A1 ﬁxed over the initial free surface and plane 2 is the outlet plane with area A2. Usurface

and U2 are the velocity of the falling free surface with height H (t) from the plane 2 and that of the

liquid at plane 2, respectively.

be beneﬁcial to undergraduate and graduate students as well as instructors who have some

knowledge of mathematical physics and ﬂuid dynamics because they can gain some insight

and better understanding of the draining problem. For instructors, it can also be useful for

forming the basis for homework problems about draining.

We assume frictionless and adiabatic (no heat transfer across the boundaries of the system)

ﬂow. In this situation the derivation is independent of the shape of the reservoir, but for

simplicity we take the reservoir to be cylindrical (see ﬁgure 1). The BE for the unsteady ﬂow

[5] is given by

d

ρE dV +

ρU AE + pU A = 0,

(1)

dt

where ρ, U and*p *are the density, velocity and absolute pressure of the liquid respectively,

(≡

2 −

1) denotes the difference between average quantities leaving (plane 2) and

entering (plane 1) the system, and V , A are the volume of the system and the area of the plane

respectively.*E *is the total energy per unit liquid mass at any point within the boundaries of

the system:

E = U 2 + g(−z),

(2)

2

where only kinetic and gravitational potential energies are retained with the constant internal

energy neglected. Let us take planes 1 and 2 as in ﬁgure 1. For convenience, we chose

the downward vertical direction as the z-axis with the origin at plane 2. Plane 1 with the

cross-sectional area A1 of the cylinder is ﬁxed over the initial free surface of the liquid so that

there is no ﬂow across plane 1. Plane 2 is the outlet with area A2 (

A1). H (t) is the height

of the falling free surface from plane 2 at time*t*.

Suppose the liquid velocity*U *is independent of the position within the liquid volume and

depends only on time:

U (t ) = Usurface(t) = − dH (t) (> 0).

(3)

dt

At plane 1, there is no ﬂow (U1 = 0) of liquid and so ρU AE + pU A 1 = 0. Substituting (2)

and (3) into (1) and using the conservation of mass (i.e., A1Usurface = A2U2) and z2 = 0, we

have the equation for H (t)

d2H

dH

2

H

− 1(J + 1)

+ gH = 0,

(4)

dt2

2

dt

An exact solution to the draining reservoir problem

279

with J ≡ c2 − 2 and c ≡ A1/A2 (

1). This equation was studied*numerically *in [7] under

the supervision of the author in the context of the water ﬂow analogy for electric circuits.

Introducing new variables to ﬁnd the exact solution to H (t),

dH

2

K ≡ U 2 = 1

,

λ ≡ ln H,

(5)

2

2

dt

(4) is rewritten as

dK − (J + 1)K + geλ = 0.

(6)

dλ

With the initial condition (H (0) = H0, U (0) = 0) for the draining problem, (6) can be solved

exactly:

H

J +1

1/2

U (H ) =

2gH0

− H

.

(7)

J

H0

H0

(To avoid confusion, we will present the argument*H *explicitly when the velocity *U *or the

acceleration*a *is a function of *H*.) This can be integrated to give

√π 1

1

1

1

J

t =

2J H0

2J

−

H

2F1

,

, 1 +

; H

.

(8)

g

2J

1 + 1

H

2J 2

2J

H

2

2J

0

0

Here

and 2F1 are the gamma function and the hypergeometric function, respectively. The

ﬁrst term in the brackets of (8) is 2F1(1/(2J ), 1/2, 1 + 1/(2J ); 1).

The liquid velocity*U *has the maximum value

Umax =

2gH0(J + 1)− J+1

2J

(9)

at the height of the free surface

H∗ = H0(J + 1)− 1J ,

(10)

or equivalently at time

1

1

1

t∗ = T −

2J H0 (J + 1)− 12J 2F1

,

, 1 +

; 1

.

(11)

g

2J 2

2J J + 1

Here the exact drain time*T *is obtained by setting H = 0 in (8):

1

T =

π H0

2J

.

(12)

2gJ

1 + 1

2

2J

Since J

1, the exact drain time*T *can be expanded in power series of 1/J as

γ + ψ 1

1

T =

2H0J

1 −

2

+ O

,

(13)

g

2J

J 2

where γ (≈ 0.5772) is the Euler–Mascheroni constant and ψ(1/2) ≈ −1.9635. Here ψ is the

digamma function deﬁned as ψ(σ ) ≡ d ln (σ )/dσ . Note that the drain time is about*c *(or

√

√

J ) times as large as the free-fall time

2H0/g.

Since 0

H /H0

1, the hypergeometric series

1

1

1

J

1

H

J

3

H

2J

2F1

,

, 1 +

; H

= 1 +

+

+ · · ·

(14)

2J 2

2J

H0

2(1 + 2J )

H0

8(1 + 4J )

H0

280

S-I Hong

is convergent [8]. In the leading-order approximation for J large, the exact solution (8) reduces

to the approximate one

2

H (t ) = H0 1 − t

(15)

tD

√

with the (approximate) drain time tD =

2H0J /g. The same approximate solution (15) with

a slightly different value of drain time [2, 6]

tD =

2H0(J + 1)

(16)

g

is obtained by neglecting the acceleration of the liquid in (4) under the quasi-steady (i.e.,

unsteady but slowly varying) ﬂow assumption. On the other hand, the neglect of the kinetic

energy in the ﬁrst term of (1) replaces J + 1 in (16) by J + 2 [5]. The solution (15) satisﬁes one

initial condition H (0) = H0, but does not satisfy the other initial condition U (0) = 0 by the

property of the ﬁrst-order differential equation. All these tDs are the same as the exact drain

time*T *(12) up to the leading order of 1/J . Hence we set tD = T in (15) for consistency with

the exact solution (8) for*H *from now on. Then the velocity and the acceleration of the liquid

are given by

H

U = 2H0

,

(17)

T

H0

= 2H0 1 − t ,

(18)

T

T

a = − 2H0 = constant,

(19)

T 2

√

respectively. Compare (17) with the exact solution (7) using 2H0/T ≈

2gH0/J .

In ﬁgures 2, 3 and 4, the full line represents the graph of the exact result while the dotted

line denotes that of the approximate one stemmed from (15), for H0 = 1 m and c = 10 (i.e.,

J = 98). Figure 2 is the graph of*H *as a function of *t *where the dotted line (15) is in good

agreement with the full line (8). Figures 3(a) and (b) are the graphs of*U *as a function of *t *and

*H*, respectively. Note that the approximate results (17) and (18) (dotted lines) derived from

(15) are completely different from the exact ones (full lines) at the initial stage (in time or

height). The liquid velocity U (t) increases rapidly from the rest to the maximum value Umax

as soon as the liquid starts to drain and then decreases at nearly constant rate (acceleration)

to reach 0 at the end of draining. In this case, the exact drain time (12) is 4.50 s while the

approximate one (16) is 4.49 s.

In the real-water experiment with H0 = 20 cm, an inverted soda bottle (reservoir) with

a diameter of 10.5 cm and outlets with four different diameters (7.2 mm, 8.8 mm, 9.8 mm

and 12.1 mm) were used [2]. For these four cases, the values of*c *(or J ) are 15, 12, 11 and

8.68 (210, 140, 110 and 73.3) respectively. The per cent difference between the calculated

(by (16)) and measured drain times is less than 6% for all cases [2]. The assumption of the

incompressible and non-viscous liquid is approximately valid for this experiment.

At this point it is appropriate to discuss the limitation of Torricelli’s law,

U Torricelli =

2gH ,

(20)

2

An exact solution to the draining reservoir problem

281

1

0.8

0.6

m

H 0.4

0.2

0 0

1

2

3

4

t s

**Figure 2. **Graph of *H *as a function of *t *for H0 = 1 m and c = 10 (i.e., J = 98). The full

and dotted lines are the exact and approximate results (i.e., (8) and (15) with the exact drain time

*T *(≈ 4.50 s)), respectively.

a

0.4

0.3

s

m 0.2

U

0.1

0 0

1

2

3

4

t s

b

0.4

0.3

s

m 0.2

U

0.1

0 0

0.2

0.4

0.6

0.8

1

H m

**Figure 3. **Graph of *U *as a function of (a) *t *and (b) *H *for H0 = 1 m and c = 10 (i.e., J = 98).

In (a), the full and dotted lines represent the exact result (7) (with*H *substituted by H (t) which

is obtained from (8) numerically) and approximate one (18) respectively. In (b), the full and

dotted lines denote the exact (7) and approximate (17) results respectively. Liquid velocity has the

maximum value Umax (≈ 0.43 m s−1) at time t∗ (≈ 0.14 s) or height H∗ (≈ 0.95 m).

which is derived from the steady BE. In order to correctly describe the entire process of

draining, this approximate result must be replaced by the exact one derived from (7):

2

J

1/2

U2 = cU =

1 +

1 −

H

U Torricelli.

(21)

J

H

2

0

282

S-I Hong

a 10

8

2

6

s

m

4

a

2

0

0

1

2

3

4

t s

b 10

8

2

6

s

m

4

a

2

0

0

0.2

0.4

0.6

0.8

1

H m

**Figure 4. **Graph of the acceleration *a *as a function of (a) *t *and (b) *H *for H0 = 1 m and c = 10

(i.e., J = 98). In (a), the full and dotted lines represent the exact result (22) (with*H *substituted

by H (t) which is obtained from (8) numerically) and approximate one (19) respectively. In (b),

the full and dotted lines denote the exact (22) and approximate (19) results respectively. Note that

from (22), a = −g/J = −0.10 m s−2 at t = T or H = 0.

For J

1 and (H /H0)J

1, (21) reduces to Torricelli’s result (20). This condition is well

satisﬁed except at the initial stage of draining (see ﬁgure 3(b)). For example, (H /H0)J ≈ 0.051

for c = 10 (i.e., J = 98) and H/H0 = 0.97.

It is well known that the drain system serves as an analogical model for Ohm’s law. But

this analogy is valid as long as the liquid velocity U2 at the outlet is given by Torricelli’s

law (20): the electric current, the electric resistance and the electric potential difference (i.e.,

voltage) correspond to the volume ﬂow rate Q2 at the outlet, the inverse of the outlet cross-

√

sectional area A2 and the liquid velocity U Torricelli (∝

H ) at the outlet in the drain system,

2

respectively. That is, Q2 = A2U Torricelli is an analogue of Ohm’s law. Note that the voltage

2

should be associated with a strictly increasing function of*H*. Therefore, the analogy is broken

in the initial transient region where Torricelli’s law does not hold.

Finally, let us calculate the acceleration*a *of the liquid from (7):

H

J

a(H ) = g (J + 1)

− 1 .

(22)

J

H0

We have a(H = H0) = g and a(H = 0) = −g/J . Figures 4(a) and (b) are the graphs of

acceleration*a *as a function of *t *and *H*, respectively. The initial behaviour of the approximate

acceleration (19) (the dotted line) is quite different from that of the exact one (22) (the full

line), as expected from liquid velocity. The acceleration approaches the value at H = 0 (or

t = T ) rapidly as soon as the liquid starts to drain.

An exact solution to the draining reservoir problem

283

In conclusion, we have solved exactly the draining reservoir problem of the incompressible

and non-viscous liquid to obtain the drain time and the height, velocity and acceleration of the

free surface. It is found that the approximate expressions ((18) and (19)) derived from (15)

fail to explain the initial time dependence of the liquid velocity and acceleration. In particular,

the exact solution to liquid velocity implies that Torricelli’s law is not valid and in turn the

pedagogical analogy between draining and Ohm’s law is broken, in the initial transient region.

After the initial rise of liquid velocity, the subsequent motion with a small and nearly constant

deceleration (a ≈ −g/J from (22)) is consistent with Torricelli’s law.

**References**

[1] de Oliveira P M C, Delﬁno A, Costa E V and Leite C A F 2000 Pin-hole water ﬂow from cylindrical bottles*Phys.*

*Educ. ***35 **110–9

[2] Guerra D, Plaisted A and Smith M 2005 A Bernoulli’s law lab in a bottle*Phys. Teach. ***43 **456–9

[3] Sianoudis I A and Drakaki E 2008 An approach to Poiseuille’s law in an undergraduate laboratory experiment

*Eur. J. Phys. ***29 **489–95

[4] Aguilar H M 2008 Level sensor for hydrodynamics experiments*Phys. Educ. ***43 **46–50

[5] Middleman S 1998*An Introduction to Fluid Dynamics *(New York: Wiley) pp 464–7

[6] Alexandrou A N 2001*Principles of Fluid Mechanics *(Upper Saddle River, NJ: Prentice-Hall) pp 122–3

[7] Kim K R 2007 Hydrodynamic analysis of water ﬂow analogy for electric circuits*MEd Thesis *Gyeongin National

University of Education, Anyang, Korea

[8] Arfken G B and Weber H J 1995*Mathematical Methods for Physicists *4th edn (San Diego: Academic) pp 796–9

# Document Outline

EUROPEAN JOURNAL OF PHYSICS

Eur. J. Phys.

doi:10.1088/0143-0807/30/2/006

reservoir problem of the

incompressible and non-viscous liquid

Department of Science Education, Gyeongin National University of Education, Anyang 430-739,

Korea

E-mail: sihongtao@hanmail.net

Received 21 October 2008, in ﬁnal form 6 December 2008

Published 13 January 2009

Online at stacks.iop.org/EJP/30/277

acceleration of the free surface are found for the draining reservoir problem

of the incompressible and non-viscous liquid. Contrary to the conventional

approximate results, they correctly describe the initial time dependence of the

liquid velocity and acceleration. Torricelli’s law does not hold in the initial

transient region, which imposes restrictions on the validity of the analogy

between the drain system and the electric circuit (Ohm’s law).

Recently, the classic draining reservoir experiments in ﬂuid dynamics have been performed

in relation to physics education. de Oliveira

through a pin-hole tube located at the lateral bottom of a cylindrical reservoir and the height of

the free surface of water by making use of rulers. A simple empirical correction was made to

Torricelli’s law which is derived from the steady Bernoulli equation (BE), taking energy losses

into account. Guera

constant acceleration for the free surface of water, and compared it with experimental results

obtained by a motion sensor. For a cylindrical reservoir draining through a tube, Sianoudis

and Drakaki [3] measured the liquid height using a motion sensor. This allowed conﬁrmation

of Poiseuille’s law (the relationship between the ﬂow rate and pressure difference between the

ends of the tube) for viscous liquids. In addition, Aguilar [4] used a level sensor to measure

the water level in a draining reservoir.

Draining is obviously an unsteady process. In this paper, the draining problem [5, 6]

of the incompressible and non-viscous liquid in a reservoir is revisited in search of the exact

solution of the unsteady BE. It will be compared with solutions of the approximate unsteady

BE or the steady BE. In particular, we emphasize that the well-known Torricelli law must

be modiﬁed for such an ideal liquid (see (21)) and its limitations need to be addressed

conceptually with undergraduate students in introductory physics courses. This work will

0143-0807/09/020277+07$30.00

c 2009 IOP Publishing Ltd

Printed in the UK

277

278

S-I Hong

A1

Usurface

H(t)

A2

U2

with area A1 ﬁxed over the initial free surface and plane 2 is the outlet plane with area A2. Usurface

and U2 are the velocity of the falling free surface with height H (t) from the plane 2 and that of the

liquid at plane 2, respectively.

be beneﬁcial to undergraduate and graduate students as well as instructors who have some

knowledge of mathematical physics and ﬂuid dynamics because they can gain some insight

and better understanding of the draining problem. For instructors, it can also be useful for

forming the basis for homework problems about draining.

We assume frictionless and adiabatic (no heat transfer across the boundaries of the system)

ﬂow. In this situation the derivation is independent of the shape of the reservoir, but for

simplicity we take the reservoir to be cylindrical (see ﬁgure 1). The BE for the unsteady ﬂow

[5] is given by

d

ρE dV +

ρU AE + pU A = 0,

(1)

dt

where ρ, U and

(≡

2 −

1) denotes the difference between average quantities leaving (plane 2) and

entering (plane 1) the system, and V , A are the volume of the system and the area of the plane

respectively.

the system:

E = U 2 + g(−z),

(2)

2

where only kinetic and gravitational potential energies are retained with the constant internal

energy neglected. Let us take planes 1 and 2 as in ﬁgure 1. For convenience, we chose

the downward vertical direction as the z-axis with the origin at plane 2. Plane 1 with the

cross-sectional area A1 of the cylinder is ﬁxed over the initial free surface of the liquid so that

there is no ﬂow across plane 1. Plane 2 is the outlet with area A2 (

A1). H (t) is the height

of the falling free surface from plane 2 at time

Suppose the liquid velocity

depends only on time:

U (t ) = Usurface(t) = − dH (t) (> 0).

(3)

dt

At plane 1, there is no ﬂow (U1 = 0) of liquid and so ρU AE + pU A 1 = 0. Substituting (2)

and (3) into (1) and using the conservation of mass (i.e., A1Usurface = A2U2) and z2 = 0, we

have the equation for H (t)

d2H

dH

2

H

− 1(J + 1)

+ gH = 0,

(4)

dt2

2

dt

An exact solution to the draining reservoir problem

279

with J ≡ c2 − 2 and c ≡ A1/A2 (

1). This equation was studied

the supervision of the author in the context of the water ﬂow analogy for electric circuits.

Introducing new variables to ﬁnd the exact solution to H (t),

dH

2

K ≡ U 2 = 1

,

λ ≡ ln H,

(5)

2

2

dt

(4) is rewritten as

dK − (J + 1)K + geλ = 0.

(6)

dλ

With the initial condition (H (0) = H0, U (0) = 0) for the draining problem, (6) can be solved

exactly:

H

J +1

1/2

U (H ) =

2gH0

− H

.

(7)

J

H0

H0

(To avoid confusion, we will present the argument

acceleration

√π 1

1

1

1

J

t =

2J H0

2J

−

H

2F1

,

, 1 +

; H

.

(8)

g

2J

1 + 1

H

2J 2

2J

H

2

2J

0

0

Here

and 2F1 are the gamma function and the hypergeometric function, respectively. The

ﬁrst term in the brackets of (8) is 2F1(1/(2J ), 1/2, 1 + 1/(2J ); 1).

The liquid velocity

Umax =

2gH0(J + 1)− J+1

2J

(9)

at the height of the free surface

H∗ = H0(J + 1)− 1J ,

(10)

or equivalently at time

1

1

1

t∗ = T −

2J H0 (J + 1)− 12J 2F1

,

, 1 +

; 1

.

(11)

g

2J 2

2J J + 1

Here the exact drain time

1

T =

π H0

2J

.

(12)

2gJ

1 + 1

2

2J

Since J

1, the exact drain time

γ + ψ 1

1

T =

2H0J

1 −

2

+ O

,

(13)

g

2J

J 2

where γ (≈ 0.5772) is the Euler–Mascheroni constant and ψ(1/2) ≈ −1.9635. Here ψ is the

digamma function deﬁned as ψ(σ ) ≡ d ln (σ )/dσ . Note that the drain time is about

√

√

J ) times as large as the free-fall time

2H0/g.

Since 0

H /H0

1, the hypergeometric series

1

1

1

J

1

H

J

3

H

2J

2F1

,

, 1 +

; H

= 1 +

+

+ · · ·

(14)

2J 2

2J

H0

2(1 + 2J )

H0

8(1 + 4J )

H0

280

S-I Hong

is convergent [8]. In the leading-order approximation for J large, the exact solution (8) reduces

to the approximate one

2

H (t ) = H0 1 − t

(15)

tD

√

with the (approximate) drain time tD =

2H0J /g. The same approximate solution (15) with

a slightly different value of drain time [2, 6]

tD =

2H0(J + 1)

(16)

g

is obtained by neglecting the acceleration of the liquid in (4) under the quasi-steady (i.e.,

unsteady but slowly varying) ﬂow assumption. On the other hand, the neglect of the kinetic

energy in the ﬁrst term of (1) replaces J + 1 in (16) by J + 2 [5]. The solution (15) satisﬁes one

initial condition H (0) = H0, but does not satisfy the other initial condition U (0) = 0 by the

property of the ﬁrst-order differential equation. All these tDs are the same as the exact drain

time

the exact solution (8) for

are given by

H

U = 2H0

,

(17)

T

H0

= 2H0 1 − t ,

(18)

T

T

a = − 2H0 = constant,

(19)

T 2

√

respectively. Compare (17) with the exact solution (7) using 2H0/T ≈

2gH0/J .

In ﬁgures 2, 3 and 4, the full line represents the graph of the exact result while the dotted

line denotes that of the approximate one stemmed from (15), for H0 = 1 m and c = 10 (i.e.,

J = 98). Figure 2 is the graph of

agreement with the full line (8). Figures 3(a) and (b) are the graphs of

(15) are completely different from the exact ones (full lines) at the initial stage (in time or

height). The liquid velocity U (t) increases rapidly from the rest to the maximum value Umax

as soon as the liquid starts to drain and then decreases at nearly constant rate (acceleration)

to reach 0 at the end of draining. In this case, the exact drain time (12) is 4.50 s while the

approximate one (16) is 4.49 s.

In the real-water experiment with H0 = 20 cm, an inverted soda bottle (reservoir) with

a diameter of 10.5 cm and outlets with four different diameters (7.2 mm, 8.8 mm, 9.8 mm

and 12.1 mm) were used [2]. For these four cases, the values of

8.68 (210, 140, 110 and 73.3) respectively. The per cent difference between the calculated

(by (16)) and measured drain times is less than 6% for all cases [2]. The assumption of the

incompressible and non-viscous liquid is approximately valid for this experiment.

At this point it is appropriate to discuss the limitation of Torricelli’s law,

U Torricelli =

2gH ,

(20)

2

An exact solution to the draining reservoir problem

281

1

0.8

0.6

m

H 0.4

0.2

0 0

1

2

3

4

t s

and dotted lines are the exact and approximate results (i.e., (8) and (15) with the exact drain time

a

0.4

0.3

s

m 0.2

U

0.1

0 0

1

2

3

4

t s

b

0.4

0.3

s

m 0.2

U

0.1

0 0

0.2

0.4

0.6

0.8

1

H m

In (a), the full and dotted lines represent the exact result (7) (with

is obtained from (8) numerically) and approximate one (18) respectively. In (b), the full and

dotted lines denote the exact (7) and approximate (17) results respectively. Liquid velocity has the

maximum value Umax (≈ 0.43 m s−1) at time t∗ (≈ 0.14 s) or height H∗ (≈ 0.95 m).

which is derived from the steady BE. In order to correctly describe the entire process of

draining, this approximate result must be replaced by the exact one derived from (7):

2

J

1/2

U2 = cU =

1 +

1 −

H

U Torricelli.

(21)

J

H

2

0

282

S-I Hong

a 10

8

2

6

s

m

4

a

2

0

0

1

2

3

4

t s

b 10

8

2

6

s

m

4

a

2

0

0

0.2

0.4

0.6

0.8

1

H m

(i.e., J = 98). In (a), the full and dotted lines represent the exact result (22) (with

by H (t) which is obtained from (8) numerically) and approximate one (19) respectively. In (b),

the full and dotted lines denote the exact (22) and approximate (19) results respectively. Note that

from (22), a = −g/J = −0.10 m s−2 at t = T or H = 0.

For J

1 and (H /H0)J

1, (21) reduces to Torricelli’s result (20). This condition is well

satisﬁed except at the initial stage of draining (see ﬁgure 3(b)). For example, (H /H0)J ≈ 0.051

for c = 10 (i.e., J = 98) and H/H0 = 0.97.

It is well known that the drain system serves as an analogical model for Ohm’s law. But

this analogy is valid as long as the liquid velocity U2 at the outlet is given by Torricelli’s

law (20): the electric current, the electric resistance and the electric potential difference (i.e.,

voltage) correspond to the volume ﬂow rate Q2 at the outlet, the inverse of the outlet cross-

√

sectional area A2 and the liquid velocity U Torricelli (∝

H ) at the outlet in the drain system,

2

respectively. That is, Q2 = A2U Torricelli is an analogue of Ohm’s law. Note that the voltage

2

should be associated with a strictly increasing function of

in the initial transient region where Torricelli’s law does not hold.

Finally, let us calculate the acceleration

H

J

a(H ) = g (J + 1)

− 1 .

(22)

J

H0

We have a(H = H0) = g and a(H = 0) = −g/J . Figures 4(a) and (b) are the graphs of

acceleration

acceleration (19) (the dotted line) is quite different from that of the exact one (22) (the full

line), as expected from liquid velocity. The acceleration approaches the value at H = 0 (or

t = T ) rapidly as soon as the liquid starts to drain.

An exact solution to the draining reservoir problem

283

In conclusion, we have solved exactly the draining reservoir problem of the incompressible

and non-viscous liquid to obtain the drain time and the height, velocity and acceleration of the

free surface. It is found that the approximate expressions ((18) and (19)) derived from (15)

fail to explain the initial time dependence of the liquid velocity and acceleration. In particular,

the exact solution to liquid velocity implies that Torricelli’s law is not valid and in turn the

pedagogical analogy between draining and Ohm’s law is broken, in the initial transient region.

After the initial rise of liquid velocity, the subsequent motion with a small and nearly constant

deceleration (a ≈ −g/J from (22)) is consistent with Torricelli’s law.

[1] de Oliveira P M C, Delﬁno A, Costa E V and Leite C A F 2000 Pin-hole water ﬂow from cylindrical bottles

[2] Guerra D, Plaisted A and Smith M 2005 A Bernoulli’s law lab in a bottle

[3] Sianoudis I A and Drakaki E 2008 An approach to Poiseuille’s law in an undergraduate laboratory experiment

[4] Aguilar H M 2008 Level sensor for hydrodynamics experiments

[5] Middleman S 1998

[6] Alexandrou A N 2001

[7] Kim K R 2007 Hydrodynamic analysis of water ﬂow analogy for electric circuits

University of Education, Anyang, Korea

[8] Arfken G B and Weber H J 1995

- References

## Add New Comment