Determine the work to empty a parabolic tank filled with water which can be seen

Question

Determine the work to empty a parabolic tank filled with water which can be seen as the rotation about the y-axis of the function y = 1/8 x2 on the interval 0 x 6. Note: Water weighs 62.4lb/ft3. Work is ft-lbs. You do not need to do calculations. For example 72.6856090638206 can be given as: pi (7^3/3+ (7/2)^2- 14exp(2)). Incorrect. Tries 1/8 Determine the force due to hydrostatic pressure on the flat vertical side of a tank which has the shape in feet of the boundaries y = 0, y = 7 ln x, y = 7 ln (-x) and the line y = 7 ln 19. Note that water has density 62.4lb/ft3. Force on the vertical side is lbs. You do not need to do calculations. For example 72.6856090638206 can be given as pi (7^3/3+(7/2)^2- 14exp(2)). Tries 0/8 Determine the force due to hydrostatic pressure on the flat vertical side of a tank which has the shape in feet of the boundaries y = 0, y = 4x - 7, y = -4x - 7 and the line y = 21. Note that water has density 62.4lb/ft3. Force on the vertical side is lbs. You do not need to do calculations. For example 72.6856090638206 can be given as pi (7^3/3+(7/2)^2- 14exp(2)). Tries 0/8

Explanation / Answer

1) Lets take a strip of width dy at a height y.

Volume of the strip = pi*x^2*dy

Energy required to take this volume out of tank = pi*x^2*dy*62.4*g*(4.5 - y)

x^2 = 8y

hence energy = pi*62.4*g*(4.5 - y)*8y*dy = 3.142*62.4*32.17*(4.5 - y)*8y*dy = 6307.27*(4.5 - y)*8y*dy

Total energy required = int{0 to 4.5}6307.27*(4.5 - y)*8y*dy

= int{0 to 4.5} 227061y*dy - int{0 to 4.5} 50458.2*y^2*dy

= 227061*4.5^2/2 - 50458.2*4.5^3/3 = 766324.8 lbft^2/s^2

2) Pressure on a strip at height y of width dy = 62.4*32.17*(7ln19 - y)

Force = area*pressure = 2x*dy*62.4*32.17*(7ln19 - y)

x = e^(y/7)

Total pressure = int{0 to 7ln19} 2*e^(y/7)*dy*62.4*32.17*(7ln19 - y)

= int{0 to 7ln19} 4014.81*e^(y/7)*(7ln19 - y)*dy

= int{0 to 7ln19} 82749.66*e^(y/7)dy - int{0 to 7ln19} 4014.81y*e^(y/7)*dy

= 2.96*10^6 lbft^2/s^2

3) Pressure on a strip at height y of width dy = 62.4*32.17*(21 - y)

Area of strip = 2x*dy = (y+7)/2 *dy

Net force = int{0 to 21} 62.4*32.17*(21 - y)*(y+7)/2 *dy

= 3.09*10^6 lbft^2/s^2

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