Quiz 2 – 2013.11.27
Questions
1. What is the Reynolds number?
2. Differentiate the flow patterns observed in laminar
flow from those in turbulent flow.
3. How does temperature affect the dynamic viscosity
of a fluid?
TIME IS UP!!!
Overall Balances
Outline
1.Mass Balance
2.Energy Balance
3.Momentum Balance
Mass Balance
For an overall mass balance, no mass is being
generated. Why?
 rate of mass output   rate of mass input 
 from control volume    to control volume 

 

 rate of accumulation


0

 of mass within control volume 
Mass Balance
Imagine the control volume as having infinitesimal surfaces dA.
We need to find the net outflow of mass across the control
surface.
Mass Balance
For every dA element, a streamline of velocity vector v passes
through it.
Mass Balance
For every dA element, a unit normal vector n exists.
Mass Balance
The component of velocity vector v in the direction of the
unit normal vector n is given by:
v n  v n cos
Mass Balance
The rate of mass efflux through dA:
 (v n)d A   v n cos dA
Mass Balance
What do we get when we integrate over the entire
control surface?

v
n
d
A



Mass Balance

v
n
d
A



POSITIVE: net outflow of mass
NEGATIVE: net inflow of mass
ZERO: ?
Mass Balance
Rate of mass outflow across control surface (and
control volume):

v
n
d
A



Rate of mass accumulation in control volume:


dV



t
Mass Balance
 rate of mass output   rate of mass input 



from
control
volume
to
control
volume

 

 rate of accumulation


0
 of mass within control volume 


v
n
d
A


d
V

0






t
Overall Mass Balance


v
n
d
A


d
V

0






t
dM
m 
0
dt
Overall Mass Balance
A well-stirred storage vessel contains 10000 kg of
dilute methanol solution (xMetOH = 0.05). A constant
flow of 500 kg/min of pure water is suddenly
introduced into the tank and a constant rate of
withdrawal of 500 kg/min of solution is started.
These two flows are continued and remain
constant. Assuming that the densities of the
solutions are the same and that the total contents
of the tank remain constant at 10,000 kg of
solution, calculate the time for the alcohol content
to drop to 1.0 wt.%.
Outline
1.Mass Balance
2.Energy Balance
3.Momentum Balance
Forms of Energy
• Possessed/Carried by fluid
– Internal Energy
– Potential Energy
– Kinetic Energy
– PV-work
• Transferred between system and
surroundings
– Heat
– Shaft work
Internal Energy (U)
• Intrinsic property of the fluid
• Molecules in random motion
Potential Energy (mgz)
• Position of the fluid with respect to an
arbitrary reference plane
g
z
gc
Kinetic Energy (mv2/2α)
• Due to fluid motion
• Correction factor, a
– To account for velocity distribution
– Ranges from 0.5 (laminar) to 1.0 (turbulent)
PV Work (PV)
• Work done by surroundings to push the fluid
into the system
P
S
d
WPV
V 
 Fd  PS    PV
S
Heat (Q)
• Net heat passing through the boundary of the
system
– Positive if heat is transferred to the system from
the surroundings
– Negative if system to the surroundings
• Excludes heat generated by friction
Shaft Work (Ws)
• Net work done on the system by the
surroundings
• Convention (IUPAC)
– Positive if work done on the system
– Negative if work done by the system
Total Energy Balance
Energy balance from point 1 to point 2:


v12
 gz1  P1V1 
m  U1 
2a


Q  Ws
U1, v1, P1, V1, S1
Q
z1
Ws
U2, v2, P2,
V2, S2
z2
Datum/reference plane


v22
 gz2  P2V2 
 m  U2 
2a


d(mU)

dt
Total Energy Balance
Energy balance from point 1 to point 2:



 d(mU)
v12
v22
m  U1 
 gz1  P1V1   Q  Ws  m  U2 
 gz2  P2V2  
2a
2a
dt






d(mU)
v2
 m  U 
 gz  PV   Q  Ws
dt
2a




d(mU)
v2
 m  H 
 gz   Q  Ws
dt
2a


Total Energy Balance
Water at 93.3°C is being pumped from a large
storage tank at 1 atm abs at a rate of 0.189
m3/min by a pump. The motor that drives the
pump supplies energy at the rate of 1.49 kW.
The water is pumped through a heat exchanger,
where it gives up 704 kW of heat and is then
delivered to a large open storage tank 15.24 m
above the first tank.
What is the final
temperature of the water to the second tank?
Mechanical Energy Balance
A modification of the total energy balance
- shaft work
- kinetic energy
- potential energy
- flow work (PV)
Does not include heat and internal energy.
- Why?
Energy converted to heat is lost work
- loss of mechanical energy by friction
Ideal Fluids
No shear stress; zero viscosity
For isothermal flow and Q=WS=0,
v12
v22
P1V1 
 gz1  P2V2 
 gz2
2a
2a
P1
v12
P2
v22

 z1 

 z2
1 g 2 ga
2 g 2 ga
Bernoulli Equation
Bernoulli Equation
Restrictions:
P1
v12
P2
v22

 z1 

 gz2
1 g 2 ga
 2 g 2 ga
1. Valid only for incompressible fluids
2. No devices that add/remove energy should
be between points 1 and 2
3. No heat transfer occurring in the system
4. No loss of energy due to friction
Real Fluids
• Friction losses: SF (energy dissipation)
• Total heat absorbed by the fluid Q  Q   F
• Total work done by fluid,
-W = -WS + SF
– Additional work must be done by the fluid to
overcome fluid friction
Real Fluids

 Q Ws
v2
 U 
 gz  PV   
2a

 m m
Q  Q  F
Q  Q   F
 v 2  Q  SF WS
U  (PV )    gz    


m
m
 2a 
• Note: energy per mass units
• kJ/kg or ft-lbf/lbm
• For incompressible flow:
v 2
P
 g z 
 SF  WS
2a
