3. Pulling Gasoline Up (Work on Fluids) A cylindrical tank with a length of 10 m and a radius of 5 m is on its side and half-full of diesel (see figure below). A vacuum pump is lifting the "handfuls" of diesel to the top of the tank, so that the diesel level changes from 0 to -5. length=10 D(y) y- width width of the gasoline layer y-0 area of the gasoline layer How much work is required to empty the tank through an outlet pipe at the top of the tank? The equation of work for this type of problems is: W = LpgA(y)D(y)dy Where: p = 737 kg/m³ is the density of gasoline. g=9.8 m/s² is the acceleration of gravity. A(y) is the cross-sectional area of the gasoline layer. D(y) is the distance that the slices of gasoline must be lifted. Useful integrals [hint: I recommend you to use MATLAB to avoid mistakes]: √ √ a² = x² dx = sin−1 +√a²x²+c Sx√(a²x²)dx = −(4) Where u = a²x², and thus the new limits become a' = a² - a² and ẞ' = a² - ẞ²
3. Pulling Gasoline Up (Work on Fluids) A cylindrical tank with a length of 10 m and a radius of 5 m is on its side and half-full of diesel (see figure below). A vacuum pump is lifting the "handfuls" of diesel to the top of the tank, so that the diesel level changes from 0 to -5. length=10 D(y) y- width width of the gasoline layer y-0 area of the gasoline layer How much work is required to empty the tank through an outlet pipe at the top of the tank? The equation of work for this type of problems is: W = LpgA(y)D(y)dy Where: p = 737 kg/m³ is the density of gasoline. g=9.8 m/s² is the acceleration of gravity. A(y) is the cross-sectional area of the gasoline layer. D(y) is the distance that the slices of gasoline must be lifted. Useful integrals [hint: I recommend you to use MATLAB to avoid mistakes]: √ √ a² = x² dx = sin−1 +√a²x²+c Sx√(a²x²)dx = −(4) Where u = a²x², and thus the new limits become a' = a² - a² and ẞ' = a² - ẞ²
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 94E
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![3. Pulling Gasoline Up (Work on Fluids)
A cylindrical tank with a length of 10 m and a radius of 5 m is on its side and half-full of diesel
(see figure below). A vacuum pump is lifting the "handfuls" of diesel to the top of the tank, so
that the diesel level changes from 0 to -5.
length=10
D(y)
y-
width
width of the gasoline layer
y-0
area of the gasoline layer
How much work is required to empty the tank through an outlet pipe at the top of the tank?
The equation of work for this type of problems is:
W =
LpgA(y)D(y)dy
Where:
p = 737 kg/m³ is the density of gasoline.
g=9.8 m/s² is the acceleration of gravity.
A(y) is the cross-sectional area of the gasoline layer.
D(y) is the distance that the slices of gasoline must be lifted.
Useful integrals [hint: I recommend you to use MATLAB to avoid mistakes]:
√ √ a² = x² dx =
sin−1 +√a²x²+c
Sx√(a²x²)dx = −(4)
Where u = a²x², and thus the new limits become a' = a² - a² and ẞ' = a² - ẞ²](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff165ec61-47d8-4d0c-a4e4-a7168a9b1fe8%2F494869aa-20d6-408b-a3a9-a0f778a560ff%2Fz09hdrb_processed.png&w=3840&q=75)
Transcribed Image Text:3. Pulling Gasoline Up (Work on Fluids)
A cylindrical tank with a length of 10 m and a radius of 5 m is on its side and half-full of diesel
(see figure below). A vacuum pump is lifting the "handfuls" of diesel to the top of the tank, so
that the diesel level changes from 0 to -5.
length=10
D(y)
y-
width
width of the gasoline layer
y-0
area of the gasoline layer
How much work is required to empty the tank through an outlet pipe at the top of the tank?
The equation of work for this type of problems is:
W =
LpgA(y)D(y)dy
Where:
p = 737 kg/m³ is the density of gasoline.
g=9.8 m/s² is the acceleration of gravity.
A(y) is the cross-sectional area of the gasoline layer.
D(y) is the distance that the slices of gasoline must be lifted.
Useful integrals [hint: I recommend you to use MATLAB to avoid mistakes]:
√ √ a² = x² dx =
sin−1 +√a²x²+c
Sx√(a²x²)dx = −(4)
Where u = a²x², and thus the new limits become a' = a² - a² and ẞ' = a² - ẞ²
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