2.016 Hydrodynamics
Professor A.H. Techet
0.1
Derivation of unsteady Bernoulli’s Equation
Conservation of Momentum says
m�a = F�
so
�
DV
F�
=
Dt
V
This is the acceleration and forces acting on Bob the Fluid Blob. The total derivative of the velocity is
expanded like this:
ρ�a = ρ
� ∂y ∂ V
� ∂z
� (t, x, y, z)
�
� ∂x ∂ V
DV
∂V
∂V
=
+
+
+
Dt
∂t
∂x ����
∂t
∂y ����
∂t
∂z ����
∂t
u
v
w
�
�
�
�
�
∂V
∂V
∂V
∂V
DV
=
+u
+v
+w
Dt
∂t
∂x
∂y
∂z
�
�
�
∂V
∂
∂
∂ �
V
=
+ u
+v
+w
∂t
∂x
∂y
∂z
�
�
��
�
∂ ∂ ∂
∂V
�
+ (u, v, w) ·
=
, ,
V
∂x ∂y ∂z
∂t
�
�
DV
∂V
� · �)
� V
�
=
+ (V
Dt
∂t
� ×V
� = 0), so (V
� · �)
� V
� = �(
� 1V
� ·V
� ) and
For irrotational flow, (�
2
�
�
�
�
1� �
DV
∂V
�
=
+�
V ·V
Dt
∂t
2
� = �φ,
� and we have
Also for irrotational flow, we can use the velocity potential V
�
�
��
�
�
DV
∂ �φ
1
�
� · �φ
�
ρ
=ρ
+�
�φ
Dt
∂t
2
The forces acting on Bob are pressure and gravity, so the momentum equation becomes
�
�
��
�
∂ �φ
1
�
� − ρg kˆ = −�p
� − �(ρgz)
�
� · �φ
�
ρ
+�
= −�p
�φ
����
∂t
2
d
(ρgz)
dz
�
�
�
∂φ 1 � �
�
�
� ρ
+ ρ �φ · �φ + p + ρgz = 0
2
∂t
And in one last glorious step, we integrate all the spacial derivatives (i.e. knock the nabla out), and we
have the unsteady Bernoulli’s Equation;
�
∂φ 1 � �
�
ρ
+ ρ �φ · �φ
+ p + ρgz = F (t)
∂t
2
where F (t) is some function of t (is the ”constant of integration”).
1