- Information
- AI Chat
Toaz - Bradley Aerodynamics Problem Solutions.
Civil Engineering (BSCE 01)
Ateneo de Davao University
Preview text
BERNOULLI’S EQUATION
- A pipe is gradually tapering in size, diminishing by 0 square foot per foot run. What is change
in velocity per foot run where pipe is 4 square foot in cross section if the velocity there is 90 feet
per seconds. Is the velocity increasing or decreasing?
Given:
dA/ds = -0 sq. foot per foot run
A = 4 sq. ft.
V = 90 ft/s
Required:
dV/ds
Solution:
dA
ds
A
=
−
dV
ds
V
−0.
4
=
−
dV
ds
90
dV
ds
=+2.
ft
s per foot run , increasing
- A circular pipe, 100 ft. long, gradually tapers from 3 ft. in diameter at one end to 2 ft. at other.
Fluid is flowing from bigger toward smaller end. What is the rate of increase in velocity at
entrance if the velocity there is 80 ft. per sec.?
Given:
V = 80 ft/s
Da = 3 ft.
Db = 2 ft.
L = 100 ft
Required:
dV/ds
Solution:
dA
ds
A
=
−
dV
ds
V
A 1
= ¼ π (3 ft)
2
= 9/4 π ft
2
A 2
= ¼ π (2 ft)
2
= π ft
2
dA
ds
=
π ft
2
−
9
4
π ft
2
100 ft
=−0 ft
−0 ft
9
4
π ft
2
=
−
dV
ds
80 ft / s
dV
ds
=0.
ft
s per ft. run
- A water pipe 8 inches in diameter gradually tapers down to 4 inches in diameter. The rate of
flow is 1000 gallon/minute. If the pressure is 20 lb. per sq. in where the diameter is 8 inches.
What is the pressure where diameter is 4 inches? Water weighs 62 lb. per cu. ft.)
Given:
Q = 1000
gallon
minute
(
231000 ft
3
1728 gallon
)(
1 minute
60 seconds
)
=2.
ft
3
seconds
V 1
=
Q
A
1
=
ft
3
s
π (
4
12
ft)
2
=6.
ft
s
V
2
=
Q
A
2
=
ft
3
s
π (
2
12
ft)
2
=25.
ft
s
P 1
=
lb
ft
2
( atm)+
[
20
lb
¿
2
∗ 144
]
( gage )=4996.
lb
ft
2
¿
Required:
P 2
Solution:
P
1
ρV
1
2
2
=P
2
ρV
2
2
2
P
2
=P
1
ρ
2
( V
1
2
−V
2
2
)
Given:
V 1
=
50
ft
s
V
2
= 35
ft
s
ρ =
lb
ft
3
Required:
P 1
- P 2
Solution:
P
1
ρV
1
2
2
=P
2
ρV
2
2
2
P
1
−P
2
=
ρ
2
(V
2
2
−V
1
2
)
P
1
−P
2
=
lb
ft
3
2
(
ft
s
2
)
(( 50
ft
s
)
2
−( 35
ft
s
)
2
)
P
1
−P
2
=1235.
lb
ft
2
=8 psi
- The diameter of a horizontal pipe is 4 in. in which tetrabromosthane (spec. grav. = 3) is flowing
at rate of 0 cu. ft. per sec. The pressure is 30 psi (gage). If the tube gradually decreases to 3
in. What is the pressure?
Given:
Q = 0.
ft
3
seconds
V 1
=
Q
A
1
=
ft
3
s
1
4
π (
4
12
ft )
2
=5.
ft
s
V
2
=
Q
A
2
=
ft
3
s
1
4
π (
3
12
ft )
2
=10.
ft
s
P 1
=
lb
ft
2
( atm)+
[
30
lb
¿
2
∗ 144
]
( gage )=6436.
lb
ft
2
¿
ρ
substance
= (
specific gravity∗ρ
water
)
=
(
3 ∗62.
lb
ft
3
)
=187.
lb
ft
3
Required:
P 2
Solution:
P
1
ρV
1
2
2
=P
2
ρV
2
2
2
P
2
=6436.
lb
ft
2
lb
ft
3
2
(
ft
s
2
)
((5.
ft
s
)
2
−(10.
ft
s
)
2
)
P
2
=6230.
lb
ft
2
¿
(gage)
P
2
=28 psi
- Alcohol (spec. grav. = 0) is flowing through a horizontal pipe, which is 10 in. in diameter, with
velocity of 40 ft. per sec. At a smaller section of pipe, there is 6 psi less pressure. What is the
diameter?
Given:
V 1
=
40
ft
s
A
1
=
1
4
π (
10
12
ft )
2
=0 ft
2
P 1
- P 2
=
[
6
lb
¿
2
∗ 144
]
= 864
lb
ft
2
ρ
substance
= (
specific gravity∗ρ
water
)
=
(
0∗62.
lb
ft
3
)
=49.
lb
ft
3
Required:
D 2
Solution:
A
2
=
A
1
√
1 +
2
V
1
2
ρ
subs
(P
1
−P
2
)
Given:
Q = 10
ft
3
seconds
V 1
=
Q
A
1
=
10
ft
3
s
π (
9
12
ft)
2
=5.
ft
s
V 2
=
Q
A
2
=
10
ft
3
s
π (
3
12
ft)
2
=50.
ft
s
P 1
=
lb
ft
2
( atm)+
[
20
lb
¿
2
∗ 144
]
( gage )=4996.
lb
ft
2
¿
Required:
P 2
Solution:
P
1
ρV
1
2
2
=P
2
ρV
2
2
2
P
2
=P
1
ρ
2
( V
1
2
−V
2
2
)
P
2
=4996.
lb
ft
2
lb
ft
3
2
(
ft
s
2
)
((5.
ft
s
)
2
−(50.
ft
s
)
2
)
P
2
=2513.
lb
ft
2
¿ (gage)
P
2
=2 psi
- Air flows through a horizontal pipe at the rate of 3000 cu. ft. per sec. If the pressure is 30 lb. per
sq. in. where diameter is 3 ft., what is the pressure where diameter is 2 ft?
Given:
Q = 3000
ft
3
seconds
V 1
=
Q
A
1
=
3000
ft
3
s
1
4
π ( 3 ft )
2
=424.
ft
s
V 2
=
Q
A
2
=
3000
ft
3
s
1
4
π ( 2 ft )
2
=954.
ft
s
P 1
=
lb
ft
2
( atm)+
[
30
lb
¿
2
∗ 144
]
( gage )=6436.
lb
ft
2
¿
Required:
P 2
Solution:
P
1
ρV
1
2
2
=P
2
ρV
2
2
2
P
2
=P
1
ρ
2
( V
1
2
−V
2
2
)
P
2
=6436.
lb
ft
2
slugs
ft
3
2
(( 424.
ft
s
)
2
−(954.
ft
s
)
2
)
P
2
=5566.
lb
ft
2
¿
(gage)
P
2
=23 psi
- Water flows through a horizontal pipe at the rate of 800 gallons per minute. What is the
difference in pressure between a point where diameter is 2 in. and a point where the diameter
is 1 in?
Given:
Q = 800
gallon
minute
(
231000 ft
3
1728 gallons
) (
1 minute
60 seconds
)
=1.
ft
3
seconds
V 1
=
Q
A
1
=
ft
3
s
1
4
π (
2
12
ft )
2
=81.
ft
s
V 2
=
Q
A
2
=
ft
3
s
1
4
π (
1
12
ft )
2
=326.
ft
s
ρ =
lb
ft
3
P
2
=12981.
lb
ft
2
¿
(gage)
P
2
=75 psi
VENTURI TUBE
- A Venturi Tube narrows down from 4 in. in diameter to 2 in. in diameter. What is the rate of
flow water if pressure at the throat is 2 lb. per sq. in less than larger section?
Given:
A 1
=
π
4
d
1
2
=
π
4
(
4
12
)
2
=
1
36
π ft
2
A
2
=
π
4
d
2
2
=
π
4
(
2
12
)
2
=
1
144
π ft
2
A 2
/A 1
= 1/
P 1
- P 2
=
[
2
lb
¿
2
∗ 144
]
= 288
lb
ft
2
Required:
Q
Solution:
Q= A
2
√
P
1
−P
2
ρ
2
[ 1 −
####### (
A
2
A
1
####### )
2
]
Q=
1
144
π ft
2
√
288
lb
ft
2
lb
ft
3
2 (32.
ft
s
2
)
[ 1 −
(
1
4
)
2
]
Q=0.
ft
3
s
- A Venturi tube is 6 in. in diameter at entrance, where the pressure is 10 lb. per sq. in. The throat
is in 4 in. in diameter there the pressure is 6 lb. per sq. in. What is the flow of water??
Given:
A 1
=
π
4
d
1
2
=
π
4
(
6
12
)
2
=
1
16
π ft
2
A 2
=
π
4
d
2
2
=
π
4
(
4
12
)
2
=
1
36
π ft
2
A
2
/A
1
= 4/
P
1
- P
2
=
10 psi− 6 psi=
[
4
lb
¿
2
∗ 144
]
= 576
lb
ft
2
Required:
Q
Solution:
Q= A
2
√
P
1
−P
2
ρ
2
[ 1 −
####### (
A
2
A
1
####### )
2
]
Q=
1
36
π ft
2
√
576
lb
ft
2
lb
ft
3
2 (32.
ft
s
2
)
[ 1 −
(
4
9
)
2
]
Q=2.
ft
3
s
- A 12 in. by 6 in. Venturi meter is located in a horizontal water line. If the pressure gages read 30
lb. per sq. in. and 16 lb. per sq. in, what is the flow?
Given:
A 1
=
π
4
d
1
2
=
π
4
(
12
12
)
2
=
1
4
π ft
2
A 2
=
π
4
d
2
2
=
π
4
(
6
12
)
2
=
1
16
π ft
2
A
2
/A
1
= 1/
P 1
- P 2
=
30 psi− 16 psi=
[
14
lb
¿
2
∗ 144
]
= 2016
lb
ft
2
Required:
Q
Solution:
Q= A
2
√
P
1
−P
2
ρ
2
[ 1 −
####### (
A
2
A
1
####### )
2
]
Q=
1
16
π ft
2
√
2016
lb
ft
2
lb
ft
3
2 (32.
ft
s
2
)
[ 1 −
(
1
4
)
2
]
Q=9.
ft
3
s
STAGNATION POINT
Given:
P
s
−P
o
=0 psi∗ 144 = 36
lb
ft
2
ρ
10000
=0.
slugs
ft
3
Required:
V
Solution:
V =
√
P
s
−P
o
ρ
2
V =
√
36
lb
ft
2
slugs
ft
3
2
V =202 fps∗0= 120 knots
- An airplane is flying at sea level. The difference between impact and static pressure is 0 psi.
What is the airspeed in knots?
Given:
P
s
−P
o
=0 psi∗ 144 = 36
lb
ft
2
ρ
sealevel
=0.
slugs
ft
3
Required:
V
Solution:
V =
√
P
s
−P
o
ρ
2
V =
√
36
lb
ft
2
slugs
ft
3
2
V =174 fps∗0= 103 knots
- An airplane is flying at sea level. The difference between impact and static pressure is 0 psi.
What is the airspeed in knots?
Given:
V = 160
N. Mil
Hour
(
6080 ft
1 N. mil
)(
1 hr
3600 s
)
=270.
ft
s
ρ
10000
=0.
slugs
ft
3
Required:
P s
- P o
Solution:
P
s
−P
o
=
ρ V
2
2 P
s
−P
o
=
slugs
ft
3
(270.
ft
s
)
2
2
P
s
−P
o
=63.
lb
ft
2
P
s
=0.
lb
¿
2
EQUATION OF STATE
- Find the density of air at 23 inHg pressure and 15 F.
Given:
T = 15 F + 459 = 474 R
P = 23 inHg
Required:
Density
Solution:
ρ = ρ
o
P
P
o
T
o
T
ρ = 0.
slugs
ft
3
23 inHg
29 inHg
518 R
574 R
ρ = 0.
slugs
ft
3
- Find the density of air at 23 inHg pressure and 15 F.
Given:
T = -1 F + 459 = 458 R
P = 17 inHg
Required:
Density and Temperature
Solution:
ρ = ρ
o
(
P
P
o
)
1
γ
ρ = 0.
slugs
ft
3
(
3 atm
1 atm
)
1
ρ = 0.
slugs
ft
3
T = T
o
ρ
o
ρ
P
P
o
T = ( 519 R)
slugs
ft
3
slugs
ft
3
3 atm
1 atm
T = 709 R−459.
T = 250 F
- Air at standard pressure and temperature is permitted to expand adiabatically to one-half
atmospheric pressure. What are the density and temperature?
Given:
P = ½ atm
Required:
Density and Temperature
Solution:
ρ = ρ
o
(
P
P
o
)
1
γ
ρ = 0.
slugs
ft
3
(
1
2
atm
1 atm
)
1
ρ = 0.
slugs
ft
3
T = T
o
ρ
o
ρ
P
P
o
T = ( 519 R)
slugs
ft
3
slugs
ft
3
1
2
atm
1 atm
T = 425 R−459.
T = − 34 F
- Air at standard pressure and temperature is adiabatically compressed to 50 psi. What is
temperature?
Given:
P = 50 psi
Required:
Temperature
Solution:
ρ = ρ
o
(
P
P
o
)
1
γ
ρ = 0.
slugs
ft
3
(
50 psi
14 psi
)
1
ρ = 0.
slugs
ft
3
T = T
o
ρ
o
ρ
P
P
o
T = ( 519 R)
slugs
ft
3
slugs
ft
3
50 psi
14 psi
T = 792 R−459.
T = 332 F
- Air at standard pressure and temperature is permitted to expand adiabatically to one-half
standard density. What are the pressure and temperature?
Given:
½ Density
Required:
Pressure and Temperature
Solution:
P = P
o
(
ρ
ρ
o
)
γ
P = 2116.
lb
ft
2
(
slugs
ft
3
2
slugs
ft
3
)
Solution:
a = 49 √T
a = 49. √
a = 1159.
ft
s
- Find the speed of sound at 2000 ft. altitude, where temperature is normally -12 F.
Given:
T = -12 F = 447 R
Required:
Speed of Sound, a
Solution:
a = 49. √
T
a = 49. √
a = 1037
ft
s
- Find the speed of sound on at 30000 ft. altitude, where temperature is normally -48 F.
Given:
T = -48 F = 441 R
Required:
Speed of Sound, a
Solution:
A = 49. √
T
a = 49 √441.
a = 994
ft
s
- Find the speed of sound when barometric pressure is 27 inHg and density is 0 slugs per
cu. ft.?
Given:
P = 27 inHg
(
lb
ft
2
29
¿=1923.
lb
ft
2
ρ = 0 slugs per cu. ft.
Required:
Speed of Sound, a
Solution:
a =
√
γP
ρ
a =
√
(1)¿ ¿ ¿
a = 1137.
ft
s
- Find the speed of sound when barometric pressure is 8 inHg and density is 0 slugs per
cu. ft.?
Given:
P = 8 inHg
(
lb
ft
2
29
¿=628.
lb
ft
2
ρ = 0 slugs per cu. ft.
Required:
Speed of Sound, a
Solution:
a =
√
γP
ρ
a = √
(1)¿ ¿ ¿
a = 987.
ft
s
BERNOULLI’S EQUATION FOR COMPRESSIBLE FLUIDS
- What would be the velocity, where the pressure is 13 psi if the fluid is incompressible?
Given:
P
1
=14.
lb
¿
2
*144=2116.
lb
ft
2
( gage) +2116.
lb
ft
2
( abs ) =4232.
lb
ft
2
(atm)
P
2
=13.
lb
¿
2
*144=
lb
ft
2
( gage) +2116.
lb
ft
2
( abs ) =4060.
lb
ft
2
(atm) V
1
= 500 ft
s
ρ= 0 slugs
ft
3
Required:
Toaz - Bradley Aerodynamics Problem Solutions.
Course: Civil Engineering (BSCE 01)
University: Ateneo de Davao University
This is a preview
Access to all documents
Get Unlimited Downloads
Improve your grades
