The tank shown is full of water. Given that water weighs 62.5 lb/ft3, find the work required to pump the water out of the tank. The tank shown is a hemisphere with r = 5 ft. The water is to be pumped out at the top.
First I solved for ri, ri / (5 - xi) = 5 / 5
ri = 5 - xi
Vi = pi*(ri)^2*delta x
Vi = pi*(5 - xi)^2*delta x
mi = density x volume
mi = 62.5*pi*(5 - xi)^2*delta x
Fi = mi x g, g = 9.8 m/s^2 or 32.152231 ft/s^2
Fi = (32.152231 ft/s^2)*(62.5*pi* (5 - xi)^2*delta x)
Fi = 2009.5*pi*(5 - xi)^2*delta x
Wi = (Fi)*(xi) = (2009.5*pi)(xi)(5 - xi)^2*delta x
W = INTEGRAL from 0 to 5 of: 2009.5*pi*(xi)(5 - xi)^2*dx
W = 2009.5*pi INTEGRAL xi*(5 - xi)^2*dx, from 0 to 5.
W = 2009.5*[12.5x^2 - (10x^3)/3 + (x^4)/4] evaluated at 5 and 0
W = 2009.5*[12.5*(5)^2 - (10*(5^3))/3 + (5^4)/4]
W = 100220 ft-lb
I am not sure which part I did wrong. Could you please point me in the right direction? Thanks!
For Further Reading
* Calculus - bobpursley, Friday, June 22, 2007 at 9:22pm
Your calculation of force included g. But you used lbs for mass. Lbs is a force unit (it really isnt mass). So delete the 32.15 ft/s^2 from the force.
Work=NTEGRAL from 0 to 5 of: 62.5*pi*(xi)(5 - xi)^2*dx
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I am really sorry to post this again but I still cannot come up with the right answer...
Work=NTEGRAL from 0 to 5 of: 62.5*pi*(xi)(5 - xi)^2*dx
=(62.5*pi)[(12.5)(5^2)-(10(5^3))+(5^4)/4]
=10226.53859 ft-lb
This is wrong though...am I in the wrong units? Please help. Thanks.
If you are performing a vertical integration with x as the variable of integration, it seems to me that relationship between x (the height above the base) and r (the radius of the circular volume element of thickness dx), is
r^2 + x^2 = 5^2
NOT ri = 5 - xi
That might be the casue of your error.
When you integrate for total work, the integral should be
(Integral of) pi r^2 * (rho) g * (5-x) dx
x = 0 to 5
Substitute 25 - x^2 for r^2. Then you will just have terms in x in the integrand.
ri is not equal to 5-xi
To find the work required to pump the water out of the tank, you need to calculate the integral of the expression (pi * r^2 * density * g * (5 - x)) with respect to x, where r is the radius of the circular volume element, density is the density of water (62.5 lb/ft^3), and g is the acceleration due to gravity (32.152231 ft/s^2).
The correct relationship between x and r (height above the base and radius of the circular volume element) is given by r^2 + x^2 = 5^2.
Substituting r^2 = 25 - x^2 into the expression for work, we have:
Work = Integral from 0 to 5 of: (pi * (25 - x^2) * density * g * (5 - x)) dx
Now, let's calculate this integral step-by-step:
Step 1: Simplify the expression inside the integral:
Work = (pi * density * g) * Integral from 0 to 5 of: ((25 - x^2) * (5 - x)) dx
Step 2: Expand the expression inside the integral:
Work = (pi * density * g) * Integral from 0 to 5 of: (125 - 5x - 25x + x^3) dx
Work = (pi * density * g) * Integral from 0 to 5 of: (x^3 - 30x + 125) dx
Step 3: Integrate each term separately:
For the term x^3:
Integral of x^3 dx = x^4/4
For the term -30x:
Integral of -30x dx = -15x^2
For the constant term 125:
Integral of 125 dx = 125x
Step 4: Evaluate the definite integral from 0 to 5 for each term:
Integral of (x^3 - 30x + 125) dx from 0 to 5
= [(5^4/4 - 15(5)^2 + 125(5)) - (0^4/4 - 15(0)^2 + 125(0))]
= [(625/4 - 375 + 625) - (0 - 0 + 0)]
= [625/4 + 625 - 375 - 0]
= 1875/4
= 468.75 ft-lb
So, the work required to pump the water out of the tank is 468.75 ft-lb.
To find the work required to pump the water out of the tank, we need to integrate the force over the distance.
Given that the tank is a hemisphere with radius r = 5 ft, we need to consider a small volume element at height x above the base. The relationship between x and r (the radius of the circular volume element) is r^2 + x^2 = 5^2.
The mass of the volume element is given by density times the volume. In this case, the density is 62.5 lb/ft3. The volume of the volume element is pi * r^2 * dx.
So, the mass of the volume element is mi = 62.5 * pi * (5^2 - x^2) * dx.
The force on the volume element is given by force = mass * acceleration. In this case, the acceleration is due to gravity, which is approximately 32.15 ft/s^2. So, the force on the volume element is Fi = 62.5 * pi * (5^2 - x^2) * dx * 32.15 ft/s^2.
To find the work done to pump out the water, we need to integrate the force over the distance. So the total work is:
W = Integral from 0 to 5 of (62.5 * pi * (5^2 - x^2) * dx * 32.15 ft/s^2 * x).
Evaluating this integral, we get:
W = 62.5 * pi * Integral of (32.15 * (5^2 - x^2) * x) dx, from 0 to 5.
Simplifying this expression, we get:
W = 62.5 * pi * Integral of (800.75x - 32.15x^3) dx, from 0 to 5.
Integrating term by term, we get:
W = 62.5 * pi * [400.375x^2 - 8.0375x^4/4] evaluated from 0 to 5.
Evaluating this expression, we get:
W = 62.5 * pi * [400.375(5^2) - 8.0375(5^4)/4].
Calculating this value, we find:
W ≈ 100220 ft-lb.
So, the work required to pump the water out of the tank is approximately 100220 ft-lb.