ENGI2103 Thermo-­‐Fluid Engineering II •
•
•
Some definitions and nomenclature
Weight (W) = mass (m) x gravitational acceleration (g) W = mg
mass (m) = density (ρ) x volume (Vol)
•
Specific weight (γ) = weight (W) ÷ volume (Vol)
•
•
•
W = mg = ( ρVol ) g
W ( ρVol ) g
γ =
=
= ρg
Vol
Vol
By defini?on Another property! The combination of terms ( ρ g ) appears frequently in Fluid Mechanics,
and in Heat Transfer, therefore a new terms was coined “specific weight”
and defined.
Specific weight appears frequently in the textbook, however, I use ( ρ g )
One formula to commit to memory is the formula for the pressure at at a
certain depth in a fluid P = ρ gh
January 23, 2023 ENGI2103.2.W2023.pptx 1 What is the pressure in a column of fluid? 1 • A column of fluid of height h.
P2 − P1 = ∆ P = ρ gh
• Consider the Dalhousie Life Science Centre (LSC), an 8-storey
building.
• The floors are 12 feet high, therefore, the top floor is 7 x 12 = 84
feet (25.6 m) above the lowest floor.
• The LSC underwent a major renovation in 2011 – 2012. The
2 renovation included the installation of pressure-flush toilets.
• Pressure-flush toilets? Gravity-flush toilets? Read up on them!
• Pressure-flush toilets require a water pressure of at least 35 psi (240 kPa)
• The water supplied by the Halifax Water Commission is 60 psi (410 kPa).
• This is the water pressure at the lowest level of the LSC.
P2 − P1 = ∆ P = ρ gh
P1 = P2 − ρ gh
P1 = 410 kPa −1, 000 kg/m 3 × 9.807 m/s2 × 25.6 m = 410 − 251 = 159 kPa
• P1 is too low, therefore, booster pumps were installed on the ground floor
January 23, 2023 ENGI2103.1.W2023.pptx 2 2.20 lb m
m3
3
ρ = 1000 kg/m ×
×
=
62.3
lb
/ft
m
kg
3.2813 ft 3
∆ P = ρ gh
3
BG units ∆ P = 62.3 lb m /ft × 32.2 ft/s ×1 ft
3
2
2.31 H = 1 PSI lb m -ft/s 2
∆ P = 2,006
ft 2
lb m -ft/s2
∆ P = 13.9
in 2
force
∆P=
!!
area
F = mg
1 lb f ≡ 1 lb m × 32.2 ft/s 2
lb f
2
lb m - ft/s =
32.2
13.9 lb f
∆ P1ftH2O =
= 0.432 psi
2
32.2 in
January 23, 2023 ENGI2103.2.W2023.pptx 3 Grundfos booster pump set January 23, 2023 ENGI2103.2.W2023.pptx 4 Grundfos booster pump set January 23, 2023 ENGI2103.2.W2023.pptx 5 Friday February"8,"2013"
16,000"
14,000"
Dal LH2O_Flow_D"
SC Domes?c Hot Water PRV closes at 4:40 PM By 4:40 PM ≈ 15,000 L consumed Liters"of"water"
12,000"
10,000"
No water draw overnight 8,000"
6,000"
4,000"
Water draws start at 8:00 AM PRV"opens"
PRV opens at"3:08"PM"
at 3:08 PM Booster pumps caused over pressure in the tanks By 3:00 PM ≈ 3,000 L consumed 2,000"
0"
January 23, 2023 WEL_log_2013_02.xlsx
ENGI2103.2.W2023.pptx 6 2013/02/18
Friday Water"temperature"to/from"solar"tanks"and"hot"glycol"temp"(°C)."
60"
Solar_Hot_Water"
February"8,"2013"
Cold_Water_IN"
50"
40"
T solar heated water HX_glycol_IN"
PRV opens 3:08 PM Water draws start at 8:00 AM 30"
20"
T glycol Dal LSC Domes?c Hot Water PRV"opens"at"3:08"PM."
The"temperature"
increases"because"the"
water"in"tank"#4"was"
hoUer"than"in"tank"#5"
PRV closes at 4:40 PM Solar starts 10"
T cold water 0"
WEL_log_2013_02.xlsx
January 23, 2023 ENGI2103.2.W2023.pptx 2013/02/18
7 The Great Molasses Flood, or the Boston Molasses Disaster The Great Molasses Flood was a disaster that occurred on January 15, 1919, in
Boston, Massachusetts. A large storage tank filled with 2.3 million US gal (8,700
m3) of molasses weighing 12,000 tonnes burst. A 40-foot wave of molasses
rushed through the streets at 56 km/h, killing 21 and injuring 150. The wave of
molasses buckled elevated railroad tracks, crushed buildings and inundated the
neighborhood. Structural defects in the tank combined with unseasonably warm
temperature contributed to the disaster.
Molasses tank Why did it fail? January 23, 2023 ENGI2103.2.W2023.pptx 8 Municipal water storage tank Sec?on #6 One sec?on of the wall of the tank Sec?on #1 January 23, 2023 ENGI2103.2.W2023.pptx 9 Municipal water storage tank One sec?on of the wall of the tank •
•
The Fluid Mechanics specialist will determine the forces acting on the wall
of the tank.
The stress specialist will determine the specifications for the wall of the
tank, including the thickness of each section
January 23, 2023 ENGI2103.2.W2023.pptx 10 Pressure ac?ng on the wall of the tank z = 0 Water tank The pressure
distribution
D = 46 m h = 15 m z = h
Section #1 (highest forces act on this section)
•
•
Each section is 2.5 m in height. Six sections = 15 m high.
Find the pressure at the midpoint of Section #1 (z = 13.75 m).
•
The force on this section is
P = ρ gz = 1000 × 9.81× 13.75 = 135 kPa = 19.6 psi
2
• This pressure acts on an area A = π Dhsection = π × 46 × 2.5 = 361 m
F = P = 135 × 361 = 48, 700 kN = 10,960,000 pound-force (lb f )
January 23, 2023 ENGI2103.2.W2023.pptx 11 Stress in the lowest sec?on? F = 10,960,000/2 = 5,480,000 lbf t = thickness
Ac = t x 2.5 m = t x 98.4 in.
stress σ =
σ y,steel = 51,000 psi
σ y,steel
ty =
5, 480,000 lb f
= 51,000 psi =
2 t y × 98.4 in 2
(
5, 480,000 lb f
2 × 98.4 in × 51,000
January 23, 2023 lb f
)
F
5, 480, 000 lb f
=
2Ac 2 (t × 98.4 in 2 )
Thickness (ty) will produce a
yield stress, which is not good!
= 0.55 inch
in 2
ENGI2103.2.W2023.pptx What should t be?
12 F = 9,990,000/2 = 5,000,000 lbf 24,000 m3
σy =
D = 150 ft = 45.7 m
F 5, 000, 000 lb f
=
= 17.9 psi
2
2A 2t × 98.4 in
t = 1.03 in
h = 48 ft = 14.6 m
Largest “tank” in the world:
437,000 m3; a concrete reservoir
January 23, 2023 t = 0.31 in
Why 1.03?
Largest oil tank in the world:
250,000 m3; D = 100 m; H = 32 m
ENGI2103.2.W2023.pptx 13 Compressible fluids • In a compressible fluid (e.g., an ideal gas) the pressure is
P = ρ RT
• The pressure variation in a fluid is
∆P
= ρg
∆z
• Or
∫
1
2 g
dP
= −∫
dz
1
P
RT
P
RT
dP
gP
= ρg = −
dz
RT
∫
2
1
dP
g
=−
P
RT
g 2
g
ln P 1 = −
dz = −
( z2 − z1 )
∫
1
RT
RT
⎛ g
⎞
P2 = P1 exp ⎜ −
z2 − z1 )⎟
(
⎝ RT
⎠
2
January 23, 2023 “ - ” indicates that P
decreases as z increases
dP
• Or
= − ρg
dz
• The density is ρ =
2
∆ P = ρ g∆ z
∫
2
1
dz
dP
g
=−
dz
P
RT
If T ≠ f(z)
⎛ P2 ⎞
g
ln ⎜ ⎟ = −
z2 − z1 )
(
RT
⎝ P1 ⎠
ENGI2103.1.W2023.pptx 14 Pressure drop to top of the Burj Khalifa (air) • Consider the Burj Khalifa in Dubai, world’s tallest building (828 m)
⎡ g
⎤
P2 = P1 exp ⎢ −
z2 − z1 ) ⎥
(
⎣ RT
⎦
2 • In the above equation we employ the absolute pressures.
⎡ ⎛
⎤
⎞
9.81 m/s2
P2 = (101 kPa ) exp ⎢ − ⎜
(828 m )⎥
⎟
⎣ ⎝ 287 J/kg·K × 301 K ⎠
⎦
P2 = 101 kPa exp ( −0.094 )
P2 = 101 kPa × 0.91
P2 = 101 kPa × 0.91 = 91.9 kPa = 13.4 psi
1 ∆ P = P1 − P2 = 101− 91.9 kPa = 9.1 kPa = 1.3 psi
• Assuming the air is incompressible, calculate the ∆ P.
P
101
ρ=
=
= 1.17 kg/m 3 ∆ P = ρ g∆ z = 1.17 × 9.81× 828 = 9.5 kPa
RT 0.287 × 301
January 23, 2023 ENGI2103.1.W2023.pptx 15 Atmospheric pressure at an al?tude of 35,000 feet • Commercial aircraft fly at 35,000 feet, ± 5,000 feet.
⎡ g
⎤
P2 = P1 exp ⎢ −
z2 − z1 ) ⎥
(
⎣ RT
⎦
35,000 ft = 10,700 m
• Employ absolute pressures.
⎡ ⎛
⎤
⎞
9.81 m/s 2
P2 = (101 kPa ) exp ⎢ − ⎜
(10, 700 m )⎥
⎟
⎣ ⎝ 287 J/kg·K × 288 K ⎠
⎦
P2 = (101 kPa ) exp ( −1.27)
P2 = 101 kPa × 0.28
P2 = 101 kPa × 0.30 = 28.3 kPa = 4.1 psi
15°C
At sea level, Patm = 14.7 psi
•
This calculation takes into account the compressibility of air in the
atmosphere?
• However, the temperature is assumed to be constant at 288 K (15°C).
• The temperature at 35,000 feet is well below 15°C.
January 23, 2023 ENGI2103.1.W2023.pptx 16 Temperature"and"Pressure"of"Standard"Atmosphere"
Int. Civil Aviation Org. publishes the Standard Atmosphere.
30"
Temperature"
20"
⎛ 71.5°C ⎞
T = 15 − ⎜
z = 15 − 0.0065z
⎟
⎝ 11, 000 m ⎠
10"
Temperature,"°C"
Pressure"
0"
35,000 ft = 10,700 m
P = 23 kPa/
(10"
(20"
(30"
We just calculated
28.3 kPa
(40"
240"
220"
200"
180"
160"
140"
120"
100"
80"
(50"
60"
(60"
40"
(70"
20"
T at 10,700 m = -55°C
(80"
0"
2,000"
4,000"
6,000"
Pressure,"kPa"
40"The
0"
8,000" 10,000" 12,000" 14,000" 16,000" 18,000" 20,000"
Al<tude,"m"
Atmosphere.stn.xlsx
January 23, 2023 23-01-20
ENGI2103.2.W2023.pptx 17 Atmospheric pressure at an al?tude of 35,000 feet (air at Tstandard) • Account for the change in temperature
∫
2
1
2 g
dP
= −∫
dz
1
P
RT
T = 15 − 0.0065z
dz = −154dT
∫
2
1
dP
g 2 dz
=− ∫
P
R 1 T
∫
1
2
dP
g 2
dz
=− ∫
P
R 1 (15 − 0.0065z )
15 − T
z=
= 2, 308 − 154T
0.0065
T −15
z=
−0.0065
2
∫
2
dP
g 2 −154dT 154g 2 dT
=− ∫
=
1
P
R
T
R ∫1 T
⎛ P2 ⎞ 154g ⎛ T2 ⎞
ln ⎜ ⎟ =
ln ⎜ ⎟
R
⎝ P1 ⎠
⎝ T1 ⎠
⎛ P2 ⎞
⎛ T2 ⎞
⎛ T2 ⎞
ln ⎜ ⎟ = 5.26 ln ⎜ ⎟ = ln ⎜ ⎟
⎝ P1 ⎠
⎝ T1 ⎠
⎝ T1 ⎠
⎛T ⎞
P2 = P1 ⎜ 2 ⎟
⎝ T1 ⎠
January 23, 2023 5.26
5.26
P2 ⎛ T2 ⎞
=⎜ ⎟
P1 ⎝ T1 ⎠
⎛ 218 ⎞
P2 = 101 kPa ⎜
⎝ 288 ⎟⎠
5.26
ENGI2103.1.W2023.pptx 5.26
T2 = -55°C = 218 K
= 101 kPa × 0.237 = 23 kPa
T1 = 15°C = 288 K
18 Tokyo, Japan – Frankfurt, Germany Russia
82°N latitude
Current route
10,300 km, 14 hours
Frankfurt
Tokyo
January 23, 2023 ENGI2103.2.W2023.pptx 19 Tokyo, Japan – Frankfurt, Germany Tokyo
Russia
Frankfurt
January 23, 2023 Normal route, before
Russian airspace closed
9,400 km, 11.5 hours
ENGI2103.2.W2023.pptx 20 Atmospheric pressure at an al?tude of 35,000 feet (air at Tmin) • The coldest ground temperature recorded in the northern hemisphere is
-70°C (203 K) in Greenland (1991).
T = 15 − 0.0065z
T = −70 − 0.0065z
For a ground T of 15°C
For a ground T of -70°C (203 K)
T = −70 − 0.0065 × 10, 700 m = −140°C
⎛T ⎞
P2 = P1 ⎜ 2 ⎟
⎝ T1 ⎠
5.26
⎛ 133 ⎞
P2 = 101 kPa ⎜
⎝ 203 ⎟⎠
5.26
= 11 kPa
At 10,700 m and -140°C
P
11
ρ=
=
= 0.29 kg/m 3
RT 0.287 ×133
ρ=
The lower the density the
lower the drag forces
P
24
=
= 0.38 kg/m 3
RT 0.287 × 219
At 10,700 m and -55°C
January 23, 2023 T at 10,700 m (133 K)
Aircraft will not operate
in these conditions.
The lower the temperature the
higher the engine efficiency
ENGI2103.1.W2023.pptx 21