Question

Tutorial lesson A cylindrical tank is being filled with water. The tank is initially empty but then water begins to flow into it at a rate of 74.00 kg/min. There is a small hole of radius r = 0.6000 cm at the bottom of the tank where water can escape. Because the flow rate of water leaving the hole is initially at a slower rate than water entering the tank, the water level rises. The average velocity of water leaving through the hole at the bottom of the tank is not constant and is a function of the height of the water level in the tank: v= 2gh where g is the gravitational constant of Earth, 9.81 m/s2 Additionally, the density of water at this temperature is 1.000 * 103 kg/m² What is the maximum height of the water level in the tank, if it is initially empty? Number 6.03 h If the radius of the tank is R = 0.3500 m, how long will it take the tank to reach 90.0% of this water level? Number 1.694 Incorrect

i got the hight correctly but not the time.

could you help me finding the time please

Answer 1

As requested in Question For Part 2 only.

Given:

Small hole Radius = 0.6 cm = 0.006 m ....................................................................( Given)

Tank Radius = 0.35 m

Density of Water = 1000 kg/m³

Earth Gravitational Constant (g) = 9.81 m/s²

Flow of water= 74 kg/min

As Velocity at hole is function of height in the tank v=$\sqrt{2gh}$

Solution:

Area of small hole = $\pi$ r²

Therefore Area of Small hole =  $\pi$*0.006*0.006 = 0.000113 m²

Rate of Flow Through Small hole =Area of Hole * Velocity through hole

Rate of Flow Through Small hole=0.000113*$\sqrt{2gh}$

Rate of Flow Through Small hole=0.0005*$\sqrt{h}$

Tank Radius = 0.35 m

Cross Sectional Area of Tank = $\pi$ R²

Cross Sectional Area of Tank = $\pi$ *0.35*0.35 =0.384 m²

Volume of Tank = Cross Sectional Area of Tank* height

Let the Height of tank be h

Volume of Tank = 0.384*h

As Flow of Water = 74 kg/min

1 hr = 60 min

Flow of Water = (74 kg/min)*(60 min/hr)*(1/1000 m³/kg)

Flow of Water= 5.55 m³/hr

Doing Volumetric Flow Balance for tank we have,

Rate of Flow in - Rate of Flow Out = Rate of Accumulation in Tank.

Rate of Flow in = 5.55 m³/hr

Rate of Flow Out = 0.0005*$\sqrt{h}$

Rate of Accumulation in Tank = Cross sectional area of Tank* rate of change in height of tank

Rate of Accumulation in Tank= 0.384 *dh/dt

Hence

5.55 - 0.0005*$\sqrt{h}$= 0.384 *dh/dt

Dividing by 0.384 on both sides we get

dh/dt=14.435-0.001302 $\sqrt{h}$

Let h=x²

i.e $\sqrt{h}$ =x

Differentiating on both sides we get

dh/dt =2*x *dx/dt

Substituting we get

2*x*dx/dt=14.435-0.001302*x

Dividing by 2 we get

x*dx/dt=7.2175-0.000651*x

Let 7.2175 be A and 0.000651 be B for better understanding

x*dx/dt=A-B*x

xdx/(A-B*x)=dt

Multiplying by B on both sides

B*x*dx/(A-B*x)=B*dt

Adding and subtracting A in numerator of LHS

(A+B*x-A)*dx/(A-B*x)=B*dt

Seperating into 2 parts we get

(A*dx/A-B*x) - (A-B*x)*dx/(A-B*x) =B*dt

(A*dx/A-B*x) -dx =B*dt

as x(0)=0......................( As tank is empty initially)

Integrating on both sides we get

A $\int_{0}^{x}$ dx/(A-B*x) -  $\int_{0}^{x}$dx = B$\int_{0}^{t}$dt

(-A/B) *ln(A-B*x/A) -x =B*t

(-A/B) *(ln(A-B*x)-ln(A)) -x =B*t

(-A/B) *(ln(A-B*x) + A*ln(A)/B - x =B*t

A/B= 7.2175/0.000651=11086.79

(-11086.79)*ln(7.2175-0.000651*x) +21913.13 -x =0.000651*t

Now 90% of 6.03 is 5.427

hence h=5.427

x=$\sqrt{h}$ =2.33

Now Substituting x as 2.323

we get

(-11086.79)*ln(7.2175-0.000651*2.33) +21913.13 -2.33 =0.000651*t

-21910.67+21913.13-2.33=0.000651*t

t=187.23 hr

Hence its going to take 187.23 hr to fill till 90% of level of steady state level.