Question

Water (density p=1000 kg/m^3) is discharging from through a hole at the bottom of a graduated cylinder. The mass flow rate exiting the container may be approximated by: mass flow rate = C*p*A*(2*g*y)^(1/2) Where C is the dimensionless discharge coefficient , A is the area of the discharge hole, g is the acceleration due to gravity, and y is the water height in the container. The diameter of the graduated cylinder is 8 cm. For the following givens, estimate the time to empty half the water from the container

. C= .6 A=.04 cm^2 g= 9.8 m/s^2 y(initial)= 20cm

NEED PICTURE OF EXCEL OR SPREADSHEET OR MATLAB

I found normal calculations on other page of CHEGG.

Answer 1

mass flow rate, discharge through the hole

dm/dt = CpA ( 2gy)^1/2 --- (1)

radius of the cylinder R = 8cm

when the height of water in the cylinder changes by a dy

volume of the water flowing out dV= R² dy

mass of the water flowing out dm =p dV   ( p si the density of water)

= R² p -dy

replace dm , in mass flow eq. (1)

R² p dy/dt = CpA ( 2gy)^1/2

dy/y^1/2 = CA(2g)^1/2/R² dt

integrating on both sides , we get

2y^1/2 = k t ( k = CA(2g)^1/2/R² )

k = CA(2g)^1/2/R² = 0.6 *0.04 * (2*980)^1/2 /*8² = 0.005

y = k²/4 t² =

initially when t=0 y = 20 cm

y(t) =20 - 6.25e-6 t² cm

t= 1240 s for y to become half = 10 cm

we plot this in excel to see where y becomes half