Question

Fluid Mechanics: A water tank is a cylinder 4 m in height and 2 meters in diameter. The tank is full at time t_o. The tank sits on a platform 12 m tall. A water tap is located at the bottom center of the tank. The tap, when actuated, opens to a pipe 5 cm in radius.

a.       Write down Bernoulli’s equation. Identify the Pressure, Kinetic and Potential Energy terms. Show that each has the units of an energy density (Joules/Volume).

b.       Use Bernoulli’s equation to compare the difference in pressure between the surface of the water and the bottom of the tank near the edge? Assume the fluid is motionless near the edge.

c.       When a tap is opened water rushes in to a pipe 5 cm in radius. The pipe is straight and drops 12 m to ground level, but along the way slowly in radius to 2 cm. What is the speed of the water as it emerges from the pipe?

d.       Use the continuity equation to find the speed of the water as it enters the pipe 12 m above, i.e. bottom of the tank.

e.       What is the volume and mass flow rate of the stream of water?

f.        What is the pressure of the fluid as it enters the pipe, again at the bottom of the tank.

Answer 1

(a) Bernoulli's equation is given as following
P/pg + (V²/2g) + z = Constant
Where (P/pg) is pressure head
V²/2g is kinetic head
Z is potential head
(b) Now between the point A (surface of water) and point B (at thebottom edge)
at top the pressure will be atmospheric therfore the gauge pressure must be zero. Hence write the bernoulli equation between A and B
P_A/pg + V_A²/2g + Z_A = P_B/pg + V_B²/2g + Z_B
Now the
Velocity at A and B = 0 since the fluid is motionless
P_A/pg + 0 + 14 = P_B/pg + 0 + 12
P_B/pg - P_A/pg = 2
(P_B - P_A)/pg = 2
(P_B - P_A) = 2pg = 2*1000*9.81 = 19.62*10³ Pa
(c) Now apply the bernoullis between point A and point C
P_A/pg + V_A²/2g + Z_A = P_C/pg + V_C²/2g + Z_C
Pressure at point A and point C is atmospheric
Velocity at the point A = 0 m/s since fluid is motionless at point A
0 + 0 + 14 = 0 + (V_C²/2g) + 0
V_C²/2g = 14
V_C² = 14*(2g) = 14*2*9.81 = 274.68
V_C = 16.573 m/s
(d) According to continuity equation
Area*Velocity = Constant
A_BV_B = A_CV_C
(r_B²)V_B = (r_C²)V_C
(*5²)V_B = (*2²)*16.573
V_B = 2.652 m/s
(e) Volume flow rate = Area*Velocity = (A_BV_B) = ((r_B²))V_B = (*(0.05)²)2.652 = 0.0208 m³/s
Mass flow rate = p(volume flow rate) = 1000*0.0208 = 20.829 kg/s
(f) we know that the pressure difference is given in the part b
(P_B - P_A) = 19.62*10³ Pa
Since the pressure at the surface is atmospheric therefore pressure will be 0 .
P_B - 0 = 19.62 kPa
P_B = 19.62 kPa