# ME 3560 Fluid Mechanics Chapter V. Finite control Volume Analysis Summer 2016

```ME3560 – Fluid Mechanics
Summer 2016
ME 3560 Fluid Mechanics
Chapter V. Finite control Volume
Analysis
1
Chapter V. Finite control Volume Analysis
ME3560 – Fluid Mechanics
Summer 2016
5.1 Conservation of Mass–Continuity Equation
• Reynolds Transport Theorem states that: The rate of change of a
property in a control system is equal to the property accumulated within
the control volume plus the rate at which the property is leaving the
control volume.
r
dB
∂
=
ρ b dV + ∫ ρ b (V ⋅ nˆ ) dA
∫
dt Sys ∂ t cv
cs
•The previous equation applied to the mass of a control system (B = m
and b = 1) results in:
r
∂
ˆ
0=
ρ
d
V
+
ρ
(
V
⋅
n
) dA
∫
∫
∂ t cv
cs
2
Chapter V. Finite control Volume Analysis
Summer 2016
ME3560 – Fluid Mechanics
r
∂
ρ dV + ∫ ρ (V ⋅ nˆ ) dA = 0
∫
∂ t cv
cs
• This equation is also know as continuity equation.
•The first term represents the mass accumulated within the CV.
•The second term represents the amount of mass per unit time crossing
the boundaries of the control surface (flux of mass)
•If the process occurs under steady state conditions
r
∫ ρ (V ⋅ nˆ ) dA = 0
m& = ρVA
cs
•If the density of the fluid is constant (incompressible flow):
∫
r
(V ⋅ nˆ ) dA = 0
cs
3
Chapter V. Finite control Volume Analysis
Summer 2016
ME3560 – Fluid Mechanics
5.2 Newton's Second Law:
The Linear Momentum
r
dB
∂
=
ρ b dV + ∫ ρ b (V ⋅ nˆ ) dA
∫
dt Sys ∂ t cv
cs
• To determine the
r Linear Momentum Equation: B = mV and b = V
r
r r
d (mV )
∂
ˆ
=
ρ
V
d
V
+
ρ
V
(
V
⋅
n
) dA
∫
∫
dt Sys ∂ t cv
cs
• Additionally, we know that Newton’srSecond Law states that
r
d (mV )
= ∑F
dt Sys
Chapter V. Finite control Volume Analysis
4
ME3560 – Fluid Mechanics
Summer 2016
• Thus, the conservation of momentum equation can be presented as
r
r r
r
∂
ˆ
ρ
V
d
V
+
ρ
V
(
V
⋅
n
) dA = ∑ F
∫
∫
∂ t cv
cs
• That is: The amount of momentum accumulated per unit time in the
CV plus the net rate of momentum flowing through the CS is equal to
the forces acting on the CV.
•In general the right hand side of the previous expression can be
expressed as:
r
r
r
∑ F = ∑ FSurface + ∑ F Body
5
Chapter V. Finite control Volume Analysis
Summer 2016
ME3560 – Fluid Mechanics
r
r
r
∑ F = ∑ FSurface + ∑ F Body
• ΣFsurface are external forces acting on the CV.
• ΣFbody is the component of the weight in the corresponding direction.
In general the coordinate system will be selected such that the
components of the weight in two of the three directions are zero.
• ΣFsurface – Net force due to Normal Stress (pressure)
r
Fpressure =
ˆ
p
n
∫CS dAr
• If p = uniform =p0 Fpressure =
CV
∫ pnˆdA = 0
CS
•Gauss Theorem
∫ Φn̂ds = ∫ ∇ΦdV • If Φ=Const then∇Φ=0
S
V
Chapter V. Finite control Volume Analysis
6
ME3560 – Fluid Mechanics
Summer 2016
• In conclusion, the net force due to a uniform pressure on the CV is zero.
Therefore, it is necessary to use “gage pressures” when applying the
momentum equation.
•The momentum equation is a vector equation which can be written as:
r
∂
ρ u dV + ∫ ρ u (V ⋅ nˆ ) dA = ∑ Fx
∫
∂ t cv
cs
r
∂
ρ v dV + ∫ ρ v (V ⋅ nˆ ) dA = ∑ Fy
∫
∂ t cv
cs
r
∂
ˆ
ρ
w
d
V
+
ρ
w
(
V
⋅
n
) dA = ∑ Fz
∫
∫
∂ t cv
cs
7
Chapter V. Finite control Volume Analysis
ME3560 – Fluid Mechanics
Summer 2016
Selection of Control Volume
•A horizontal jet of water exits a nozzle with a uniform speed of 10 ft/s,
strikes a vane, and is turned through an angle θ.
• Determine the anchoring force needed to hold the vane stationary if
gravity and viscous effects are negligible
8
Chapter V. Finite control Volume Analysis
Summer 2016
ME3560 – Fluid Mechanics
•A horizontal jet of water exits a nozzle with a uniform speed of 10 ft/s,
strikes a vane, and is turned through an angle θ.
• Determine the friction between the fluid and the vane.
Fy
Fx
9
Chapter V. Finite control Volume Analysis
Summer 2016
ME3560 – Fluid Mechanics
5.3 First Law of Thermodynamics–The Energy Equation
∆Esys = Q + W
Time rate of
total energy in
the system
=
dE
dt
Net rate of heat
system
•E = Total Energy = U + P.E. + K.E.
= Q& + W&
sys
+
Net rate of
the system
P.E. = ½ mV2 K.E. = mgz
•B = E and b = e = u + ½ V2 + gz
•Units for E are J, cal, BTU
•Units of e are J/kg, BTU/slug, lb⋅ft/slug
10
Chapter V. Finite control Volume Analysis
Summer 2016
ME3560 – Fluid Mechanics
•Applying Reynolds Transport Theorem:
r
∂
ˆ
Q& + W& =
ρ
e
d
V
+
ρ
e
(
V
⋅
n
)dA
∫
∫
∂ t CV
Cs
• Heat and Work added to the control volume are positive.
•In the most general situation
W& = W& Shaft + W& pressure + W& Shear
W&
= Tω
Shaft
W& pressure
r
p r
= − ∫ pV ⋅ nˆdA = − ρ ∫ V ⋅ nˆ dA
CS
CS
ρ
11
Chapter V. Finite control Volume Analysis
ME3560 – Fluid Mechanics
W& Shear
r
= FTangential ⋅ V
Summer 2016
• The work due to shear stress can be eliminated
by selecting the CV on solid surfaces or setting
the inlets and exits perpendicular to the velocity
such that τ⋅V = 0.
•Most of the time the effect of viscous shear
stresses are grouped with other terms to
determine the “losses”.
• Thus, the energy equation can be written as
( p V
 r
∂
& + W&
ˆ


e
ρ
d
V
+
u
+
+
+
gz
ρ
V
⋅
n
dA
=
Q
Shaft
∫
∫


∂ t cv
ρ 2

cv 
2
12
Chapter V. Finite control Volume Analysis
Summer 2016
ME3560 – Fluid Mechanics
• Realizing that h = u + p/ρ
2

 r
∂
V
& + W&
ˆ


e
ρ
d
V
+
h
+
+
gz
ρ
V
⋅
n
dA
=
Q
Shaft
∫
∫


∂ t cv
2

cv 
•If u, p, V and z are all assumed to be uniformly distributed over the flow
cross-sectional areas involved:
( p V2
 r
ˆ


+
+
+
⋅
u
gz
V
n
dA =
ρ
∫cv  ρ 2


( p V2

 u + +
+ gz m&
∑
ρ 2
flow 

out
( p V2

− ∑  u + +
+ gz m&
ρ 2
flow 

in
13
Chapter V. Finite control Volume Analysis
Summer 2016
ME3560 – Fluid Mechanics
( p V

∂
&
&
eρdV = Q + WShaft + ∑  u + +
+ gz m&
∫
ρ 2
∂ t cv
flow 

2
in
( p V

− ∑  u + +
+ gz m&
ρ 2
flow 

2
out
( p V

 u + +
+ gz m&
∑
ρ 2
flow 

2
processes:
out

( p V2
− ∑  u + +
+ gz m& = Q& + W& Shaft
ρ 2
flow 

14
in
Chapter V. Finite control Volume Analysis
Summer 2016
ME3560 – Fluid Mechanics
•By dividing this equation by the mass flow rate:


( p V2
( p V2
 u + +
+ gz  − ∑  u + +
+ gz  = q& + w&
∑
ρ 2
ρ 2
flow 

 flow 
out
in
•For only one inlet and one exit:
pout − pin
ρ
+
−V
(
(
+ g ( zout − zin ) = q& + w& − uout + uin
2
2
out
2
in
V
•Applying the previous equation to a situation where there is not shaft
work done by or on the CV.
pout − pin
ρ
+
2
out
V
−V
(
(
+ g ( zout − zin ) = q& − uout + uin
2
2
in
15
Chapter V. Finite control Volume Analysis
ME3560 – Fluid Mechanics
Summer 2016
2
Vout
pin Vin2
+
+ gzout =
+
+ gzin − loss
ρ
2
ρ
2
(
(
loss = uout − uin − q&
pout
2
Vin2
pout Vout
+
+ gzin =
+
+ gzout + loss
ρ
2
ρ
2
pin
•In general, if the device analyzed has work (shaft) interactions:
2
Vin2
pout Vout
+
+ zin + hshaft =
+
+ zout + hL
γ 2g
γ
2g
pin
16
Chapter V. Finite control Volume Analysis
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