Prob 2.1
3/3/2005
PROBLEM 2.1
PROBLEM STATEMENT:
Liquid water flows at a velocity of 2 m/s from the pipe inlet (point A) through a
horizontal pipe of 1 cm inside diameter (ID). The horizontal part of the pipe is 2 m
above the floor. The pipe then turns rises
through a vertical distance of 20 m, and
again turns to the horizontal at the new height, where the pipe ID then changes to 3
cm (point B). If the velocity of water is inversely proportional to the crosssectional
area of the pipe,
(a) What is the change in specific potential energy (kJ/kg) of the water between
points A and B?
(b) What is the change in the specific kinetic energy (kJ/kg) between points A and
B?
(c) If the water temperature entering the pipe is 27
°
C, where its specific internal
energy is 113.12 kJ/kg, what is the specific energy e
in
(kJ/kg) of the water at point
A, and what percentage of the specific energy is composed of potential, kinetic
and internal energy?
DIAGRAM DEFINING SYSTEM AND PROCESS:
D
B
=3 cm
v
A
=2m/s
D
A
=1 cm
20m
2m
GIVEN:
Point A:
v
A
=2 m/s, D
A
=1 cm, z
A
=2m, u
A
=113.12 kJ/kg, T
A
=27
°
C
Point B: D
B
=3 cm, z
B
= z
A
+20m
v
=Const/A
pipe
=Const/D
2
FIND:
(a) pe
2
pe
1
(b) ke
2
ke
1
(c)
e
A
and % pe and ke
ASSUMPTIONS:
Constant gravitational acceleration, g
=
9.81 m/s
2
GOVERNING RELATION:
2
c
c
gz
=
and
ke =
g
2
v
pe
g
QUANTITATIVE SOLUTION:
Using the governing equation, the change in potential energy is:
(
)
(
)(
)
2
2
2
1
2
1
c
g(z  z )
pe pe =
= 9.81 m/s
22
2
m
1(kg
m s
N)
g
=196(N
m/kg) = 0.196kJ/kg
×
−
−
−
−
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Prob 2.1
3/3/2005
The velocity at B can be calculated from the given velocityarea relationship
=
=
=
=
=
=
A
A
B
B
2
2
A
B
A
A
B
A
2
B
K / A
and
K / A
D
1cm
(A
/ A )
2m/s
0.22m/s
3cm
D
v
v
v
v
v
And the change in kinetic energy is
2
2
2
2
2
2
2
1
c
3
1
m
ke ke =
=
(0.22
2 )
1(kg
m s
N)
2
s
2g
= 1.976
J/kg = 1.976
10
kJ /kg
−
−
−
−
×
2
1
v
v
Specific energy e
A
=u
A
+ke
A
+pe
A
2
A
A
A
A
A
A
A
c
c
2
2
2
2
2
v
gz
e
u
ke
pe
u
2g
g
4m /s
9.81m/s
2m
113.12kJ/kg
2 1kg
m/s
N
1kg
m/s
N
(113.12
0.002
0.0196)kJ/kg
113.14kJ/kg
=
+
+
=
+
+
⋅
=
+
+
⋅
−
−
−
−
=
+
+
=
u represents 113.12/113.14 = 99.9% of the total specific energy in this case,
and pe and ke together account for less than 0.1%
DISCUSSION OF RESULTS:
For many of the problems
we deal with in thermodynamics, changes in potential
and kinetic energy will be very small relative to changes in internal energy, and
hence these terms will often be assumed negligible. In this problem, the
velocity
and the height
would have to be much larger for ke
and pe to be of significance.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '07
 Schmidt
 Thermodynamics, Energy, Kinetic Energy, in2, ft lbf, PDV

Click to edit the document details