Question

Water at a gauge pressure of 3.8 atm at street level flows in to an office building at a speed of 0.06 m/s through a pipe 5.0 cm in diameter. The pipes taper down to 2.6cm in diameter by the top floor, 20 m above. Calculate the flow velocity and the gauge pressure in such a pipe on the top floor. Assume no branch pipe and ignore viscosity.

Answer 1

Answer:

The flow velocity in such a pipe at the top floor is v₂=  2.21m/s

The gauge pressure in such a pipe at the top floor is P₂= 1.8atm

..

Explanation:

The gauge pressure at the street:  P₁=3.8atm=3.8 *$10^{5}$ pa=

The velocity of water  v₁=0.60m/s

The diameter of pipe:  d₁=50mm=50 * $10^{-3}$ m

The diameter of pipe after taper:  d₂=2.6cm=2.6*$10^{-2}$m

By using the continuity equation,

A₁v₁=A₂v₂ - - - - - - - -(1)

Here,  A₁  is the cross sectional area of the pipe, and  

A₂  is the cross-sectional area of pipe by top floor.

Substitute the values in equation 1,

$\frac{\pi }{4}$d₁²v₁=$\frac{\pi }{4}$d₂²v₂

v₂=d₁²v₁ /d₂²

Substitute the values in above equation,

v₂=$\frac{0.6 * (50 *10^{-3} )^{2}}{(2.6 *10^{-2}) ^{2} }$

v₂=$\frac{1500*10^{-6} }{6.76 * 10 ^{-4} }$

v₂=2.21m/s

The flow meter is 2.21m/s

By using Bernoulli's principle,  

ρ(v  ₁²/2)+ρgh₁+P₁=ρ(v₂²/2)+ρgh₂+P₂

Here,  

ρ  is the density of the water,  

h₁ is the elevation of the pipe at the street,  

h₂ is the elevation of the pipe by the top floor.

Arrange the equation in terms of gauge pressure,

ρ(v  ₁²/2)  +  ρgh₁  +  P₁=  ρ(v₂²/2) +  ρgh₂  +  P₂

P₂=P₁+ (ρ/2) (v₁²-v₂²) - ρgh₂

Substitute the values in above equation,

P₂=$3.8*10^{5} -2262.05-196000$

P₂=181737.95Pa=1.8atm

The gauge pressure in a pipe at the top floor is  1.8atm

.

Answer 2

Answer:

a) Flow velocity = = 0.222 m/s

b) Gauge pressure = 1.84 atm

Explanation:

P₁ = 3.8 atm = 3.8 * 10⁵ Pa

v₁ = 0.06 m/s

d₁ = 5.0 cm

r₁ = 5/2 = 2.5 cm

At the gauge pressure, the water was at street level, h₁ = 0 m

d₂ = 2.6 cm

r₂ = 2.6/2 = 1.3 cm

h₂ = 20 m

Density of water, $\rho = 1000 kg/m^{3}$

Assumptions: Flow is steady and laminar, no viscosity, no branch pipe

a) calculate the flow velocity

To calculate the flow velocity, use the continuity equation

A₁v₁ = A₂v₂.............(1)

A₁ = πr²₁ = π(2.5)² = 6.25 π

A₂ = πr²₂ = π(1.3)² = 1.69π

Substituting the appropriate values into equation (1)

6.25 π * 0.06 = 1.69π * v₂

v₂ = 0.375/1.69

v₂ = 0.222 m/s

b) Calculate the gauge pressure

Using the Bernoulli equation:

$\frac{P_{1} }{\rho g} + \frac{v_{1} ^{2} }{2g} + h_{1} = \frac{P_{2} }{\rho g} + \frac{v_{2} ^{2} }{2g} + h_{2}$

$\frac{3.8 * 10^{5} }{1000 *9.8} + \frac{0.06 ^{2} }{2*9.8} + 0 = \frac{P_{2} }{1000*9.8} + \frac{0.22^{2} }{2*9.8} + 20$

$18.77322* 9800 = P_{2} \\P_{2} = 183977.6 Pa$

$P_{2} = 1.84 atm$