Download Fluids Assignment 1 – Fluid Properties and more Quizzes Fluid Mechanics in PDF only on Docsity!
A VERTICAL CYLINDRICAL TANK WITH A DIAMETER OF 12 m AND A DEPTH OF 4 m IS FILLED
WITH WATER TO THE TOP WITH WATER AT 20°C. IF THE WATER IS HEATED TO 50°C , HOW
MUCH WATER WILL SPILL OVER? UNIT WEIGHT OF WATER AT 20°C AND 50°C IS 9.79 kN/m
3
AND 9.69 kN/m
3
, RESPECTIVELY.
SOLUTION:
GIVEN:
d = 12 m
h = 4 m
γ
@ 20℃
= 9. 79 kN/m
3
γ
@ 50℃
= 9. 69 kN/m
3
V
@ 20℃
= πr
2
h = π
6 m
2
4 m
= 144π m
3
SOLUTION:
γ =
V
γ
@ 20℃
W
@ 20℃
V
@ 20℃
- 79 kN/m
3
W
@ 20℃
144π m
3
)
W
@ 20℃
= W
@ 50℃
= 4428. 892 kN
γ
@ 50℃
W
@ 50℃
V
@ 50℃
- 69 kN/m
3
( 4428. 892 kN)
V
@ 50℃
V
@ 50℃
= 457. 058 m
3
V
spilled
= V
@ 50℃
− V
@ 20℃
V
spilled
= 457. 058 m
3
− 144π m
3
V
spilled
= 4. 669 m
3
A RIGID STEEL CONTAINER IS PARTIALLY FILLED WITH A LIQUID AT 15 atm. THE VOLUME OF
THE LIQUID IS 1.23200 L. AT A PRESSURE OF 30 atm , THE VOLUME OF THE LIQUID IS 1.
L. FIND THE AVERAGE BULK MODULUS OF ELASTICITY OF THE LIQUID OVER THE GIVEN RANGE
OF PRESSURE IF THE TEMPERATURE AFTER COMPRESSION IS ALLOWED TO RETURN TO ITS
INITIAL VALUE. WHAT IS THE COEFFICIENT OF COMPRESSIBILITY?
SOLUTION:
GIVEN:
i
= 15 atm
i
= 1. 23200 L
f
= 30 atm
f
= 1. 23100 L
SOLUTION:
ε
B
∆P
∆V V
ε
B
( 30 atm − 15 atm)( 101325 Pa/atm)
1. 23100 L − 1. 23200 L
1. 23200 L
ε
B
= 1872486000 Pa or 1. 872 GPa
β =
ε
B
- 872 GPa
β = 0. 534 GPa
− 1
AIR IS KEPT AT A PRESSURE OF 200 kPa AND A TEMPERATURE OF 30°C IN A 500 - L CONTAINER.
WHAT IS THE MASS OF THE AIR?
SOLUTION:
GIVEN:
P = 200 kPa = 200 × 10
3
Pa
T =
K
V = 500 L
R = 287 Pa − m
3
/kg − K
SOLUTION:
ρ =
P
RT
ρ =
200 × 10
3
Pa
287 J
kg − K
) ( 30 + 273 K)
ρ = 2. 300 kg/m
3
ρ =
m
V
; m = ρV
m =
- 3 kg/m
3
500 L
- 001 m
3
1 L
m = 1. 150 kg
IF 12 m
3
OF NITROGEN AT 30°C AND 125kPa abs IS PERMITTED TO EXPAND ISOTHERMALLY TO
30 m
3
, WHAT IS THE RESULTING PRESSURE? WHAT WOULD THE PRESSURE AND TEMPERATURE
HAVE BEEN IF THE PROCESS HAD BEEN ISENTROPIC?
SOLUTION:
GIVEN:
P
1
= 125 kPa
V
1
= 12 m
3
T
1
K
V
2
= 30 m
3
SOLUTION:
ISOTHERMAL
P
1
V
1
= P
2
V
2
125 kPa
12 m
3
= P
2
( 30 m
3
P
2
= 50 kPa
ISENTROPIC
P
1
V
1
k
= P
2
V
2
𝑘
( 125 kPa)( 12 m
3
- 4
= P
2
( 30 m
3
- 4
P
2
= 34. 657 kPa
T
2
T
1
P
2
P
1
k− 1
k
T
2
( 30 + 273 K)
- 657 kPa
125 kPa
4 − 1
4
T
2
= 210. 023 K
BENZINE AT 20°C HAS A VISCOSITY OF 0.000651 Pa-s. WHAT SHEAR STRESS IS REQUIRED TO
DEFORM THIS FLUID AT A STRAIN RATE 4900 s
- 1
SOLUTION:
GIVEN:
μ = 0. 000651 Pa − s
dU
dy
= 4900 s
− 1
SOLUTION:
τ = μ
dU
dy
τ = ( 0. 000651 Pa − s)( 4900 s
− 1
τ = 3. 190 Pa
A SHAFT 70 mm IN DIAMETER IS BEING PUSHED AT A SPEED OF 400 mm/s THROUGH A BEARING
SLEEVE 70.2 mm IN DIAMETER AND 250 mm long. THE CLEARANCE, ASSUMED UNIFORM, IS
FILLED WITH OIL AT 20°C WITH ν = 0.005 m
2
/s AND SP. GR. = 0 .9. FIND THE FORCE EXERTED BY
THE OIL IN THE SHAFT.
SOLUTION:
GIVEN:
s
oil
v = 0. 005 m
2
/s
dU = 400 mm/s = 0. 4 m/s
d
inner
= 70 mm = 0. 07 m
d
outer
= 70. 2 mm = 0. 0702 m
SOLUTION:
μ = ρv
μ = ( 0. 9 )( 998 kg/m
3
)( 0. 005 m
2
/s)
μ = 4. 491 kg/m − s
dγ =
- 0702 m − 0. 07 m
= 0. 0001 m
τ = ( 4. 491 kg/m − s)
( 0. 4 m/s)
( 0. 0001 m)
τ = 17964 Pa
F = τA
F = ( 17964 Pa) (π × 0. 07 m × 0. 25 m)
F = 987. 622 N
FIND THE ANGLE OF THE SURFACE TENSION FILM LEAVES THE GLASS FOR A VERTICAL TUBE
IMMERSED IN WATER IF THE DIAMETER IS 0.25 in AND THE CAPILLARY RISE IS 0.08 in. USE σ =
0.005 lb/ft?
SOLUTION:
GIVEN:
σ = 0. 005 lb/ft
h = 0. 08 in
d = 0. 25 in
SOLUTION:
h =
4σcosθ
γd
; γ = 62. 468 lb/ft
3
- 08 in (
1 ft
12 in
4 ( 0. 005 lb/ft)cosθ
lb
ft
3
- 25 in
1 ft
12 in
θ = 64 .291°
WHAT FORCE IS REQUIRED TO LIFT A THIN WIRE RING 6 cm IN DIAMETER FROM A WATER
SURFACE AT 20°C? (σ OF WATER AT 20°C = 0.0728 N/m ) NEGLECT THE WEIGHT OF THE RING
SOLUTION:
GIVEN:
d = 6 cm = 0. 06 m
𝑊𝐴𝑇𝐸𝑅 @ 20℃
= 0. 0728 N/m
SOLUTION:
F = 2σL
F = 2σ(πd)
F = 2 ( 0. 0728 N/m)(π × 0. 06 m)
F = 0. 027 N
