Question

Water leaks out from an open tank through a hole of area 2 mm² in the bottom. Suppose water is filled up to a height of 80 cm and the area of cross-section of  the tank is 0.4 m². The pressure at the open surface and at the hole are equal to the atmospheric pressure. Neglect the small velocity of the water near the open surface in the tank. (a) Find the initial speed of water coming out of the hole. (b) Find the speed of water coming out when half of water has leaked out. (c) Find the volume of water leaked out during a time interval dt after the height remained is h. Thus find the decrease in height dh in terms of h and dt. (d) From the result of part (c) find the time required for half of the water to leak out.

Answer 1

(a) From the Bernoulli's theorem, 

P+ρgh+½ρv² = P'+ρgh'+½ρv'² 

Here P = P', h = 80 cm =0.80 m, h'=0, v =0, hence

 →ρgh =½ρv'² →v'² = 2gh 

→v' =√(2gh) =√(2*10*0.8) =√16 =4 m/s {Taking g =10 m/s²} 

 (b) When half the water has leaked out, h = 0.40 m 

Now v' = √(2gh) =√(2*10*0.40) =√8 m/s.  

(c) If at any instant t, the height remaining in the tank is h, then the volume of the water leaked in a small interval dt is dQ. 

dQ =Discharge rate*dt =Area*speed of water*dt 

dQ =A*v'*dt = (2 mm²)√(2gh)dt

And the decrease in height dh = The water leaked out in time dt divided by the open area of the tank 

→dh = dQ/(0.4 m²) 

→dh = (2 mm²)√(2gh)dt/(0.40 m²)

 →dh = (2 m² *10⁻⁶)√(2gh)dt/(0.40 m²) 

→dh = √(2gh)(2*10⁻⁶*10/4)dt

 →dh = √(2gh)*5*10⁻⁶dt 

(d) From above i.e. dh = √(2gh)*5*10⁻⁶dt 

→dt = 2*10⁵/√(2gh)dh 

Integrating between the limits for height from h to h/2 we get 

T = ∫dt =∫{2*10⁵/√(2gh)}dh 

={2*10⁵/√(2g)}∫h⁻1/2dh 

=[{2*10⁵/√(2g)}2√h] 

=[{2√2*10⁵/√g}√h], putting the limits H = h to H = h/2 we get 

T = [{2√2*10⁵/√g}{√h-√(h/2)] 

={(2√2*10⁵/√g)(√2-1)/√2}√h 

=2(√2-1)*10⁵*√h/√g 

Putting h = 0.80 m, g =10 m/s² T 

= 2*0.414*10⁵*√(0.8/10)/3600 hours 

=6.50 hours

Answer 2

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