Question

Be as thorough as possible, step-by-step.

(10 pts) Consider a tank as depicted below whose boundary curved is described x y2 by = 1 and, although not depicted in the picture, has a spout atop that 2 3 extends 1 meter above the top of the tank. Cross sectional slices of this tank, parallel to the ground, are circular. If this tank is full of oil whose density is 700 kg/m3, then how much work is done in pumping this fluid out of the tank through the spout atop? 10 m

Answer 1

lom dy 22 2 tank out the Gluen: The boundary of tank is described by - - USEL • 1 metes spout atope extend above the top of To find : Work dare in pumping the fluid of tank through spaut atop Here, wosk has to be done against gravity which tries to pull the liquid down. Corsides a small cyclinder of with dy (shown in figuse) Many such cylinders, must be pecsent in the entire tanke. So, we aiunagine the work required to lift small cylinders of liquid up and out of the tank. And then we integrate it ouer entire volume. Wosk done = I force, distance, the force = ng (m be the mass of liquid) -

We know density = mass Voleeme Marss - density X Volume Mass. 8X Volume We will calculate mass of the small disco and then integrate it over the entire task element of the tank. Volume this where - Width. Area of circle of to X dy 5m Hy n the tank We know, 8²= x² + y² - And boundary of is given as - x² - y 4 - 1 = 14 y² 2 This gines 22 2 3 x² = 2121 2 Put in 5 p² = xity 2+ 24 + y² = 2+ 5y?

Put in ①, we get. Volume of this small cylinder 4 (2+ 5y ²) dy but in ③ Mass = Volume force 3 8x fx/2 + 5y ² dy hut in ③ we get. mg fx/ 2+ 5y 2g dy but from . Wosk done = f force distance. Distance liquid has to trand is (tit 1)m. He is the height of the tank and I'm is the Distance to travel: (H-y+1) hata work done grgl/2+54 ) [+ y + 1) dy height of spaut atan. 0 Mars is present from yod y =H.

we page $ (2457") (t-.yt) dys To find t use eq" A. ² = 2+ 24² r² = x + y2 = 2+ sy It is given at uppermost past- r=5m ,2 so (25) = 2+54 23x3 = 4² 5 3.714 m 5 3.714 W = pga - 5 ut 12 4-3.714 put in al lequ ®) 2H - 2y + 2+ & Hy 2 - 5y3t 5 ydy W = ggr [ 2y H g ² + 2y + 5y²H + 5 6 7 W= 700X9.8 X 3.14 2H2-H2 + 2H+ 5 H4+5 #3 W = Too x 9.8 * 3.14 H+ 2H+ &#+5 | CH-374) W = 212 x 106 Joules. 9 1.6