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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)

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

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