Quiz 3 – 2013.11.29
Question
Liquid water at 1 atm is being pumped at a rate of
0.189 m3/s from a large storage tank by a pump with a
rating of 2 kW. The water is pumped through a heat
exchanger, where it gives up 758 kW of heat and is then
delivered to a storage tank at 18.76 m above the first
tank. What is the total change in temperature for the
water? For water:  = 1 g/cm3, cP = 4.184 J/g∙K.
TIME IS UP!!!
Solution
Assumptions:
1. Steady state process
2. No heat loss in the tanks and along the pipes
3. Uniform pipe diameter
4. Constant fluid properties

Q

WS
∆z
Solution
0
0
d  mU cv
 
2
g 
  m  H 
 z    Q  WS
dt
2gc
gc  
 
3


kg
m
kg


m  V   1000 3   0.189   189
m 
s 
s


m
9.81 2

g
s
z  18.76 m  
m
gc

kg 2

s
 1.00 N



1 kJ
kJ

 0.184


kg
  1000 N  m 



Solution
kJ kJ
kJ
Q  WS  758  2  756
s
s
s

g
m  H  z   Q  WS
gc 

Q  WS
g
 H 
 z
m
gc
kJ
756
kJ
kJ
s
H 
 0.184  4.184
kg
kg
kg
189
s
kJ 
kJ 
H  4.184   4.184
T

kg 
kg  K 
T  1 K
Outline
1.Mass Balance
2.Energy Balance
3.Momentum Balance
Momentum Balance
Change in pressure forces
d
2
2
P  1 v1 A1u1  2 v2 A2u2  P1 A1u1  P2 A2u2
dt
Rate of increase
Fsf  mtot g
of momentum
Rate of change
in momentum
Force of
solid
surface on
fluid
Force of
gravity on
fluid
Momentum Balance
m
We can rewrite the following terms as:  A 
v
Substituting into the balance equaton:
m1 2
m2 2
d
P
v1 u1 
v2 u2  P1 A1u1  P2 A2u2
dt
v1
v2
Fsf  mtot g
Momentum Balance
For turbulent flows:
v
2
v

v

v
Momentum velocity correction factor
0.95  0.99 for turbulent
 
0.75 for laminar
Momentum Balance
Rewriting:
d
P  m1v1u1  m2v1u2  P1 A1u1  P2 A2u2
dt
Fsf  mtot g
d
P    mv  PA  u  Fsf  mtot g
dt
Momentum Balance
d
P    mv  PA  u  Fsf  mtot g
dt
Important Notes:
1. All terms are considered vectors, so the direction
must be specified (x, y, or z).
2. The force due to gravity only acts along the ydirection.
3. This equation assumes that the flow is turbulent,
and the velocity profile is flat.
Exercise
A diagram of a liquid-liquid ejector is shown in the figure below. It is
desired to analyze the steady-state mixing of two streams, both of the
same fluid, by means of overall balances. At plane 1 the two fluids merge.
Stream 1a has a velocity v0 and a cross-sectional area (1/3)A1, and Stream
1b has a velocity (1/2)v0 and a cross-sectional area (2/3)A1. Plane 2 is
chosen far enough downstream so that the two streams have mixed and
the velocity is almost uniform at v2. The flow is turbulent and the velocity
profiles at planes 1 and 2 are assumed to be flat. Neglect Fs→f and gravity
effects.
Liquid-Liquid Ejector
We can rewrite the entire system like this:
v0
(1/2)v0
Plane 1
Assumptions:
1. Steady-state flow
2. A1 = A2 (cross-sectional area)
3. Incompressible fluid
4. Unidirectional flow
5. No gravity effects
6. No Fs→f
Plane 2
Overall Mass Balance
v0
(1/2)v0
Plane 1
Plane 2
Assumptions:
1. Steady-state flow
2. A1 = A2 (cross-sectional
area)
3. Incompressible fluid
4. Unidirectional flow
5. No gravity effects
6. No Fs→f
m1 a  m1 b  m2
1 av1 a A1 a  1 bv1 b A1 b  2v2 A2
since incompressible flow: v1 a A1 a  v1 b A1 b  v2 A2
 1  v0  2 
v0  A1    A1   v2 A2
3  2 3 
A1  A2
2
v2  v0
3
Overall Momentum Balance
m
m
d
P  1 v12 u1  2 v22 u2  P1 A1u1  P2 A2u2  Fsf  mtot g
dt
v1
v2
Since the flow is turbulent and unidirectional:
0    mv  PA
0  m1 av1 a  m1 bv1 b  m2v2  P1 A1  P2 A2
Assumptions:
1. Steady-state flow
2. A1 = A2 (cross-sectional
area)
3. Incompressible fluid
4. Unidirectional flow
5. No gravity effects
6. No Fs→f
P2 A2  P1 A1   v1 a A1 a  v1 a   v1 b A1 b  v1 b   v2 A2  v2
P2 A2  P1 A1    v0   13 A1  v0      12 v0   32 A1   12 v0      32 v0   A2   32 v0 
P2 A2  P1 A1  13 v02 A1  16 v02 A1  94 v02 A2
Overall Momentum Balance
P2 A2  P1 A1  13 v02 A1  16 v02 A1  94 v02 A2
since A1  A2 :
P2  P1  13 v02  16 v02  49 v02
P2  P1 
1
18
v
2
0
Assumptions:
1. Steady-state flow
2. A1 = A2 (cross-sectional area)
3. Incompressible fluid
4. Unidirectional flow
5. No gravity effects
6. No Fsf
Questions:
1. What conclusion can be made from the above result?
2. If we are to carry out an MEB on the system (ΣF is significant),
what result should we expect? What is ΣF is negligible?
Exercise
Fluid is flowing at steady state through a reducing pipe bend,
as shown in the figure below. Turbulent flow will be assumed
with frictional forces negligible. The volumetric flow rate of
the liquid and the pressure p2 at point 2 are known, as are the
pipe diameters at both ends. Derive the equations to calculate
the forces on the bend. Assume that the fluid density is
constant.
Exercise
Required Quantity:
Force of the fluid on the
surface
Ff s  Fs f
Assumptions:
1. Steady-state flow
2. Incompressible fluid
3. Only the x- and the ydirection are involved
4. Significant gravity effects
5. No friction loss
Overall Mass Balance
m1  m2
v1 A1  v2 A2 (incompressible flow)
From the mass balance, given the volumetric
flowrate and areas of the bend, we can obtain
the velocities at the two points.
Mechanical Energy Balance
v 2
P
 g z 
 F  WS
2

v22  v12
P2  P1
 g  z2  z1  
0
2

since P2 is given, we need to solve for P1 :
P1   g  z2  z1   12  v22  v12   P2
From the energy balance, we now
have pressure values.
Assumptions:
1. Steady-state flow
2. Incompressible fluid
3. Only the x- and the ydirection are involved
4. Significant gravity effects
5. No friction loss
Overall Momentum Balance
d
P    mv  PA  u  Fsf  mtot g
dt
Ff s    mv  PA u  mtot g
Assumptions:
1. Steady-state flow
2. Incompressible fluid
3. Only the x- and the ydirection are involved
4. Significant gravity effects
5. No friction loss
Resolving the forces into its x- and y-components:
Ff s, x  m1v1u1 x  m2v2u2 x  P1 A1u1 x  P2 A2u2 x
Ff s,y  m1v1u1 y  m2v2u2 y  P1 A1u1 y  P2 A2u2 y  mtot g
Overall Momentum Balance
Based on the figure, the unit
vectors are:
u1 x  1
u1 y  0
u2 x  cos
u2 y  sin
Plugging in the unit vectors:
Ff s, x  m1 v1  m2v2 cos   P1 A1  P2 A2 cos 
Ff s,y  m2v2 sin  P2 A2 sin  mtot g
Overall Momentum Balance
The force exerted by the fluid
on the bend have components:
Ff  s, x  m1 v1  m2v2 cos  
P1 A1  P2 A2 cos 
Ff  s,y  m2v2 sin 
P2 A2 sin  mtot g
Questions:
1. What would be the magnitude and direction of this force?
2. What will be the force exerted by the bend on the fluid?
Exercise
Water at 95°C is flowing at a rate
of 2.0 ft3/s through a 60° bend, in
which there is a contraction from
4 to 3 inches internal diameter.
Compute the force exerted on
the bend if the pressure at the
downstream end is 1.1 atm. The
density and viscosity of water at
the conditions of the system are
0.962 g/cm3 and 0.299 cp,
respectively.