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The rectangular tank shown here, with its top at ground level, is used to catch runoff water. Assume that the water weighs 
62.5lb//ft^(3).
a. How much work does it take to empty the tank by pumping the water back to ground level once the tank is full?
b. If the water is pumped to ground level with a (10/11)-horsepower (hp) motor (work output 
500ft-lb//sec), how long will it take to empty the tank (to the nearest minute)?
c. Show that the pump in part (b) will lower the water level 
10ft (halfway) during the first 18 minutes of pumping.
d. What are the answers to parts (a) and (b) in a location where water weighs 
62.31lb//ft^(3)?62.61lb//ft^(3)?
a. Set up an integral to find the work done.

W=

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How much work does it take to empty the tank?

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b. How long will it take to empty the tank?

◻ minutes (Round to the nearest minute as needed.)
c. How much work does it take to lower the water level halfway?

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The rectangular tank shown here, with its top at ground level, is used to catch runoff water. Assume that the water weighs 62.5lb/ft3 62.5 \mathrm{lb} / \mathrm{ft}^{3} .\newlinea. How much work does it take to empty the tank by pumping the water back to ground level once the tank is full?\newlineb. If the water is pumped to ground level with a (1010/1111)-horsepower (hp) motor (work output 500ftlb/sec) 500 \mathrm{ft}-\mathrm{lb} / \mathrm{sec}) , how long will it take to empty the tank (to the nearest minute)?\newlinec. Show that the pump in part (b) will lower the water level 10ft 10 \mathrm{ft} (halfway) during the first 1818 minutes of pumping.\newlined. What are the answers to parts (a) and (b) in a location where water weighs 62.31lb/ft3?62.61lb/ft3? 62.31 \mathrm{lb} / \mathrm{ft}^{3} ? 62.61 \mathrm{lb} / \mathrm{ft}^{3} ? \newlinea. Set up an integral to find the work done.\newlineW= \mathrm{W}= \newline \square \newlineHow much work does it take to empty the tank?\newline \square \newline \square \newlineb. How long will it take to empty the tank?\newline \square minutes (Round to the nearest minute as needed.)\newlinec. How much work does it take to lower the water level halfway?\newline \square \newline \square

Full solution

Q. The rectangular tank shown here, with its top at ground level, is used to catch runoff water. Assume that the water weighs 62.5lb/ft3 62.5 \mathrm{lb} / \mathrm{ft}^{3} .\newlinea. How much work does it take to empty the tank by pumping the water back to ground level once the tank is full?\newlineb. If the water is pumped to ground level with a (1010/1111)-horsepower (hp) motor (work output 500ftlb/sec) 500 \mathrm{ft}-\mathrm{lb} / \mathrm{sec}) , how long will it take to empty the tank (to the nearest minute)?\newlinec. Show that the pump in part (b) will lower the water level 10ft 10 \mathrm{ft} (halfway) during the first 1818 minutes of pumping.\newlined. What are the answers to parts (a) and (b) in a location where water weighs 62.31lb/ft3?62.61lb/ft3? 62.31 \mathrm{lb} / \mathrm{ft}^{3} ? 62.61 \mathrm{lb} / \mathrm{ft}^{3} ? \newlinea. Set up an integral to find the work done.\newlineW= \mathrm{W}= \newline \square \newlineHow much work does it take to empty the tank?\newline \square \newline \square \newlineb. How long will it take to empty the tank?\newline \square minutes (Round to the nearest minute as needed.)\newlinec. How much work does it take to lower the water level halfway?\newline \square \newline \square
  1. Determine Tank Dimensions and Water Weight: Determine the dimensions of the tank and the weight of the water. Assume the tank is a rectangular prism with dimensions 20ft20 \, \text{ft} (length) ×10ft\times 10 \, \text{ft} (width) ×20ft\times 20 \, \text{ft} (depth). The weight of water is given as 62.5lb/ft362.5 \, \text{lb/ft}^3.
  2. Calculate Volume of Water: Calculate the volume of water in the tank when full. Volume = length×width×depth=20ft×10ft×20ft=4000ft3\text{length} \times \text{width} \times \text{depth} = 20 \, \text{ft} \times 10 \, \text{ft} \times 20 \, \text{ft} = 4000 \, \text{ft}^3.
  3. Calculate Total Weight of Water: Calculate the total weight of the water in the tank. Total weight =volume×density=4000ft3×62.5lb/ft3=250,000lb.= \text{volume} \times \text{density} = 4000 \, \text{ft}^3 \times 62.5 \, \text{lb/ft}^3 = 250,000 \, \text{lb}.
  4. Set up Integral for Work Done: Set up the integral for work done to pump the water to ground level. Work WW = 020(weight of water per layer×distance to pump)dx\int_{0}^{20} (\text{weight of water per layer} \times \text{distance to pump}) \, dx, where xx is the depth from the top.
  5. Simplify Integral: Simplify the integral. Assume each layer of water is 1ft1 \, \text{ft} thick, so the weight of each layer is 62.5lb/ft3×200ft2=12,500lb62.5 \, \text{lb/ft}^3 \times 200 \, \text{ft}^2 = 12,500 \, \text{lb}. The distance to pump each layer is xftx \, \text{ft}. W=020(12,500lb×xft)dxW = \int_{0}^{20} (12,500 \, \text{lb} \times x \, \text{ft}) \, dx.
  6. Calculate Integral: Calculate the integral. W=12,500lb020xdx=12,500lb×[x22]020=12,500lb×(200)=2,500,000ft-lb.W = 12,500 \, \text{lb} \int_{0}^{20} x \, dx = 12,500 \, \text{lb} \times \left[\frac{x^2}{2}\right]_{0}^{20} = 12,500 \, \text{lb} \times (200) = 2,500,000 \, \text{ft-lb}.

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