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:

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:

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

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

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

$P_{2} = 1.84 atm$

Answer 2

This question involves the concepts of Bernoulli's Theorem and continuity equation.  

a) Flow velocity at the top floor is "0.22 m/s".

b) Gauge pressure at the top floor is "1.86 atm".

a)

Applying the continuity equation to find the speed of the water at the top floor:

$A_1v_1=A_2v_2$

where,

A₁ = Area at bottom = $\frac{\pi d_1^2}{4}=\frac{\pi(0.05\ m)^2}{4} = 1.96\ x\ 10^{-3}\ m^2$

A₂ = Area at top floor = $\frac{\pi d_2^2}{4}=\frac{\pi(0.026\ m)^2}{4} = 0.53\ x\ 10^{-3}\ m^2$

v₁ = speed of flow at bottom = 0.06 m/s m/s

v₂ = speed of flow at top = ?

Therefore,

$v_2=\frac{(1.96\ x\ 10^{-3}\ m^2)(0.06\ m/s)}{0.53\ x\ 10^{-3}\ m^2}$

v₂ = 0.22 m/s

b)

Applying Bernoulli's Theorem to this situation:

$P_1+\rho gh_1+\frac{1}{2}\rho v_1^2=P_2+\rho gh_2+\frac{1}{2}\rho v_2^2$

where,

P₁ = Pressure at bottom = 3.8 atm = 385035 Pa

P₂ = Pressure at top floor = ?

$\rho$ = density of water = 1000 kg/m³

g = acceleration due to gravity = 9.81 m/s²

h₁ = height at bottom = 0 m

h₂ = height at top = 20 m

Therefore,

$385035\ Pa+(1000\ kg/m^3)(9.81\ m/s^2)(0\ m)+\frac{1}{2}(1000\ kg/m^3)(0.06\ m/s)^2=P_2+(1000\ kg/m^3)(9.81\ m/s^2)(20\ m)+\frac{1}{2}(1000\ kg/m^3)(0.22\ m/s)^2$

P₂ = 385035 Pa + 1.8 Pa - 196200 Pa - 24.2 Pa

P₂ = 188812.6 Pa = 188.81 KPa = 1.86 atm

Learn more about Bernoulli's Theorem here:

brainly.com/question/13098748?referrer=searchResults

The attached picture shows Bernoulli's Theorem.