quiz 3 – 2013.11.29 survival mode question liquid water at 1 atm is being pumped at a rate of...
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Quiz 3 – 2013.11.29 SURVIVAL MODE
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 process2. No heat loss in the tanks and along the pipes3. Uniform pipe diameter4. Constant fluid properties
Q ∆z
WS
Solution
2
2cv
Sc c
d mU gm H z Q W
dt g g
0 0
3
3
kg m kg1000 0.189 189
m s sm V
18.76 mc
gz
g
2
m9.81s
2
mkgs1.00
N
1 kJ1000 N m
kJ0.184
kg
Solution
kJ kJ kJ758 2 756
s s sSQ W
Sc
gm H z Q W
g
S
c
Q W gH z
m g
kJ756 kJ kJs 0.184 4.184kg kg kg189s
H
kJ kJ
4.184 4.184kg kg K
H T 1 KT
Outline
1.Mass Balance
2.Energy Balance
3.Momentum Balance
Momentum Balance
2 21 1 1 1 2 2 2 2 1 1 1 2 2 2
s f tot
dP v Au v Au P Au P Au
dtF m g
Rate of increase of momentum Rate of change
in momentum
Change in pressure forces
Force of solid surface on fluid
Force of gravity on fluid
Momentum Balance
We can rewrite the following terms as:
Substituting into the balance equaton:
mA
v
2 21 21 1 2 2 1 1 1 2 2 2
1 2
s f tot
m mdP v u v u P Au P Au
dt v v
F m g
Momentum Balance
For turbulent flows:
2v vv
v
0.95 0.99 for turbulent
0.75 for laminar
Momentum velocity correction factor
Momentum Balance
Rewriting:
1 1 1 2 1 2 1 1 1 2 2 2
s f tot
dP mv u mv u P Au P Au
dtF m g
s f tot
dP mv PA u F m g
dt
Momentum Balance
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 y-
direction.3. This equation assumes that the flow is turbulent,
and the velocity profile is flat.
s f tot
dP mv PA u F m g
dt
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 Plane 2
Assumptions:1. Steady-state flow2. A1 = A2 (cross-sectional area)3. Incompressible fluid4. Unidirectional flow5. No gravity effects6. No Fs→f
Overall Mass Balance
v0
(1/2)v0 Plane 1 Plane 2
Assumptions:1. Steady-state flow2. A1 = A2 (cross-sectional
area)3. Incompressible fluid4. Unidirectional flow5. No gravity effects6. No Fs→f
1 1 2
1 1 1 1 1 1 2 2 2
1 1 1 1 2 2since incompressible flow:
a b
a a a b b b
a a b b
m m m
v A v A v A
v A v A v A
00 1 1 2 2
1 23 2 3
vv A A v A
1 2A A2 0
23
v v
Overall Momentum Balance
Assumptions:1. Steady-state flow2. A1 = A2 (cross-sectional
area)3. Incompressible fluid4. Unidirectional flow5. No gravity effects6. No Fs→f
2 21 21 1 2 2 1 1 1 2 2 2
1 2s f tot
m mdP v u v u P Au P Au F m g
dt v v
Since the flow is turbulent and unidirectional:
0 mv PA
1 1 1 1 2 2 1 1 2 20 a a b bm v m v mv P A P A
2 2 1 1 1 1 1 1 1 1 2 2 2a a a b b bP A P A v A v v A v v A v
1 1 2 1 2 22 2 1 1 0 1 0 0 1 0 0 2 03 2 3 2 3 3P A P A v A v v A v v A v
2 2 21 1 42 2 1 1 0 1 0 1 0 23 6 9P A P A v A v A v A
Overall Momentum Balance
Assumptions:1. Steady-state flow2. A1 = A2 (cross-sectional area)3. Incompressible fluid4. Unidirectional flow5. No gravity effects6. No Fsf
2 2 21 1 42 2 1 1 0 1 0 1 0 23 6 9P A P A v A v A v A
1 2since :A A
2 2 21 1 42 1 0 0 03 6 9P P v v v
212 1 018P P v
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
f s s fF F
Assumptions:1. Steady-state flow2. Incompressible fluid3. Only the x- and the y-
direction are involved4. Significant gravity effects5. No friction loss
Overall Mass Balance
1 2
1 1 2 2 (incompressible flow)
m m
v A v A
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
2
2 S
v Pg z F W
Assumptions:1. Steady-state flow2. Incompressible fluid3. Only the x- and the y-
direction are involved4. Significant gravity effects5. No friction loss
2 22 1 2 1
2 1 02
v v P Pg z z
2 1since is given, we need to solve for :P P
2 211 2 1 2 1 22P g z z v v P
From the energy balance, we now have pressure values.
Overall Momentum Balance
s f tot
dP mv PA u F m g
dt
Assumptions:1. Steady-state flow2. Incompressible fluid3. Only the x- and the y-
direction are involved4. Significant gravity effects5. No friction loss
f s totF mv PA u m g
Resolving the forces into its x- and y-components:
, 1 1 1 2 2 2 1 1 1 2 2 2
,y 1 1 1 2 2 2 1 1 1 2 2 2
f s x x x x x
f s y y y y tot
F mv u mv u P Au P Au
F mv u mv u P Au P Au m g
Overall Momentum Balance
Based on the figure, the unit vectors are:
1 1
2 2
1 0
cos sinx y
x y
u u
u u
Plugging in the unit vectors:
, 1 1 2 2 1 1 2 2
,y 2 2 2 2
cos cos
sin sinf s x
f s tot
F mv mv P A P A
F mv P A m g
Overall Momentum Balance
, 1 1 2 2
1 1 2 2
,y 2 2
2 2
cos
cos
sin
sin
f s x
f s
tot
F mv mv
P A P A
F mv
P A m g
The force exerted by the fluid on the bend have components:
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.