A rectangular tank that is 4 meters long, 3 meters wide and 6 meters deep is fil

Question

A rectangular tank that is 4 meters long, 3 meters wide and 6 meters deep is filled with a rubbing alcohol that has density 786 kilograms per cubic meter. In each part below, assume that the tank is initially full, and that gravity is 9.8 meters per second squared. Your answers must include the correct units.

(a) How much work is done pumping all of the liquid out over the top of the tank?

units

(b) How much work is done pumping all of the liquid out of a spout 2 meters above the top of the tank?

units

(c) How much work is done pumping two-thirds of the liquid out over the top of the tank?

units

(d) How much work is done pumping two-thirds of the liquid out of a spout 2 meters above the top of the tank?

units

Explanation / Answer

Consider a slab of liquid in the tank that is x units from the bottom of the tank and has a height of delta x.

The volume of this slab is delta V=lwh since we have given l=4m ,w=3m and deep=6m
delta V= (4)(3)x =12x cubm^3

Then, since mass=density * volume =786 (kg/cubm^3)*12x( cubm^3)=9432x kg

delta F=9432x kg

Then, the amount of work done, delta W, in moving this will depend on the distance the slab of liquid has traveled.

a). The liquid, in this case, will need to travel 6-x meter to reach the top of the tank.

delta W =delta F d

delta W=9432x *(6-x)   

integrating this from 0 to 6 we get

work is done pumping all of the liquid out over the top of the tank is delta W=9432*18=169,776 kg-meter.

b). But in this case the liquid need to travel additional 2 meters to reach the liquid out of a spout 2 meters above the top of the tank

delta W=9432x *(6-x+4) = 9432x *(10-x)

integrating delta W from 0 to 10 we get delta W=9432*50=471,600 kg-meter

c). we want to pump out 2 meter of liquid.integrating the expression in a from 0 to 2 we get the work

delta W=9432x *(6-x)   

integrating the delta W=9432x *(6-x) from 0 to 2

the work is done pumping two-thirds of the liquid out over the top of the tank is  

delta W=9432*10=94320 kg-meter

d). In the (b) we have delta W=9432x *(6-x+4) = 9432x *(10-x)

integrating the delta W=9432x *(10-x)  from 0 to 2

we get delta W=9432*18=169,776 kg-meter.

the work is done pumping two-thirds of the liquid out of a spout 2 meters above the top of the tank is

delta W=9432*18=169,776 kg-meter.

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