(7%) Problem 1: The pump in a water tower lifts water with a...

(7%) Problem 1: The pump in a water tower lifts water with a density of ??w = 1.00 kg/liter from a filtered pool at the bottom of the tower up hT = 29 m to the top of the tower. The water begins at rest in the pool and comes to rest at the top of the tower. The pump is 100% efficient in lifting the water, and it lifts a volume of Vw = 7.2 liters of water up the tower every second. The water tower sits on a hill that is Te = 35 m above sea level. Let gravitational potential energy be zero at ground level. 33% Part (a) Calculate the power, Pp in Watts, of the pump. 33% Part (b) If the pump runs for 1.6 hours, calculate the increase in the potential energy, ?Uw in J, of the water in the tower. 33% Part (c) The water tower, when full, holds 10^6 liters of water. Calculate the gravitational potential energy of the water, Uw in J, of a full water tower relative to sea level.

Transcript

00:01 Hi everyone, so this is the figure given in the question.

00:06 Let's start by solving the section a.

00:11 So in section a, the typical tab, the typical slab between the planes, this plane y and the y plus delta y, this plane.

00:27 Let's find the volume of that area first.

00:30 So between the planes.

00:46 At y and y plus delta y so the volume of that delta v will be equal to the 10 feet into 12 feet into the height let's say it's delta way so that is 120 d delta y feet cube.

01:36 Okay, yeah.

01:37 Now the force f required to lift the slab is equal to its weight.

01:44 So the force required to lift the slab is equal to weight means force f will be equal to 62 .4 delta.

02:06 And so we just found the value of delta v.

02:12 In terms of delta y.

02:14 Let's substitute this 120 delta y here instead of delta v.

02:20 That gives us 62 .4 into 120 delta y lb.

02:32 So it's force.

02:36 Now the distance through which the force f, this force f must add is about wide feet.

02:45 So the work done lifting the slab, is about the back then to for lifting the slab will be the force multiplied the distance that means this force 62 .4 multiplied 120 delta y that is the force into the distance is delta y the distance is y the distance is y the distance is y so multiplied by y.

03:38 Now the work it takes to lift all the water is approximately yeah let's denote it.

03:48 So it's introduce it will be now the work it takes to lift all the water is approximately yeah let's write in work takes to lift all the water.

04:35 To avoid the confusion between these works that's why i am writing these that will be approximately summation of 0 to 20 delta this delta w okay that means that will be summation of 0 to 20 let's substitute the while of delta y that is 62 .4 multiplied 120 y by delta y okay? the event will be filtered because that is force and distance.

05:32 Yeah, now this is the main sum for the functions.

05:38 4 into 120 over the interval 0 -220.

05:46 So the work of pumping the time empty is the limit of these sums.

05:56 So that means the work of pumping the tank empty.

06:21 Let's find that is the limit of the sum so that means w will be integral 0 to 2060 2 .4 into 120 y.

06:40 D .y...

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