# Titanic problem: ```ES3C9
FLUID MECHANICS FOR MECHANICAL ENGINEERS
Example Class 3
(1) Reservoir pumping
A necked-down section in a pipe flow, called a Venturi, develops a low throat
pressure which can aspirate fluid upward from a reservoir, as in figure 2.
Using Bernoulli’s equation with no losses, derive an expression for the
velocity V1 which is just sufficient to bring reservoir fluid into the throat.
D2
D1
V1
V2
Water
P2=Pa
h
Pa
Water
Fig. 2:
Water will begin to aspirate into the throat when pa  p1   gh . Hence:
Volume flow:
Bernoulli 
 D12
4
V1 
 D22
V2  V1  V2  D2 / D1   V1  V2 2 with   D2 / D1
2
4
1
1


z  0 : p1  V12  patm  V22  patm  p1  V12  V22      1V22
2
2
2
2
Solve if pa  p1   gh  V2 
Similarly, V1,min   V2,min 
2 gh
  1
2 gh
1   D1 / D2 

(2) Non-dimensional analysis
The Morton number Mo, used to correlate bubble-dynamics studies, is a
dimensionless combination of acceleration of gravity g, viscosity  ,
density  , and tension coefficient Y. If Mo is proportional to g, find its form.
The relevant dimensions are the following:
 g   LT 2  ,    ML1T 1  ,    ML3  ,    MT 2 
To have g in numerator, we need the combination:
Mo   g  
a
b
c
a
b
c
 L  M  M  M 
0 0 0
  2 
  3   2  M LT
T
LT
L
T
 
    
M a  b  c  0

L 1  a  3b  0
T 
2  a  2c  0
a  b  c  0

1  2a  3c  0
2  a  2c  0

Solve for a = 4, b= - 1, c= - 3, or: Mo 
g 4

(3) The Titanic (length 269m, maximum width 28m, and height 30.5m) sank on
April 14, 1912, hitting an iceberg and sinking 160 minutes later. Recently, a
sonar study of the bow of the Titanic on the ocean floor has revealed that the
holes caused by the iceberg are much smaller than originally thought. Until
this study, it was assumed that a large, 100m long gash was ripped in the
Titanic’s side, but now the sonar reveals that the area of the hole was only
1.4m2 (the size of a typical door!). The hole of the Titanic was approximately
6.1m below sea level at the start of the sinking.
Was the hole large enough to sink the titanic in 160 minutes?
Fig. 1:
Governing Equations:
By selecting the entire ship as an open system, the conservation of mass is given by:
d
 d    V .n dAr  0
dt 
1
In this problem, however, only part of the selected volume is filled with fluid.
Therefore, at any instant t, the density  is defined only over the volume  f  t  that
 
the fluid occupies. Therefore, the unsteady term reduces to
d
d
 d    d f

dt 
dt  f
For a streamline originating at the free surface (0) and terminating at the hole (1), the
Bernoulli equation is expressed as
V02
V12
P0  
  gz0  P1  
  gz1
2
2
where P, V, and z are respectively the pressure, velocity, and elevation from an
arbitrary reference level, of the points.
Basic Assumptions: The problem can be simplified by assuming inviscid,
incompressible flow, with constant uniform velocity at the hole. We also represent the
Titanic as an empty rectangular box – a rather conservative assumption. The water
inside the ship is assumed opened to the atmosphere.
At the hole, we have the following conditions: V  u1i and n  i .
Substituting the appropriate terms in the conservation of mass, we get
d
 d f    u1dAr  0
dt f
hole
Since sinking is defined as the point when the Titanic is fully filled, the upper limit of
the integration is set to  f  t  229726 m3 , where t is the total volume of the
d f
 u1 Ah , where
dt
Ah is the area of the hole. By integration, the final result then is  f  u1 Ahtsink . The
velocity of the water at the hole can be evaluated using the Bernoulli equation by
selecting an arbitrary point at the free surface where P0  Patm , V0  0 , and z  h0 , and
a point at the hole where P1  Patm , V1  0 , z  0 and (the reference elevation is
chosen at the level of the hole). Therefore, the water velocity at the hole
is u1  2 gh0  10.94 m s-1 .
The analysis then predicts that under the assumptions and conditions stated, the
Titanic. For uniform conditions at the hole, and constant density
Titanic would have sunk in tsink 

t
 15000s  4.2h
Ah 2 gh0
This is higher than the actual sinking time. A more detail analysis of the same
problem must consider the fact that while the Titanic sunk both the depth of the hole
and the level of the water inside the ship increased. The analysis must also consider
that the Titanic was not empty but partially filled with people, luggage ...
Nevertheless, since the predicted 4.2 hours is not very far off from 2.7 hours, the
analysis suggests that the Titanic could have sunk because of a hole that was much
smaller than originally thought.
u Ah

```