Question

can you please help i cant seem to find the answer

Answer 1

ANSWERS

1. 13 m³

2. v2 = 86.25 m/s

3. -35.8 atm

EXPLANATION

1. The volume of water flowing in a 8min period is

$V=v_1\cdot t\cdot A_1$

Where v1 = 13.8m/s, t = 8min and A1 is the section of the pipe:

$A_1=\pi\cdot r^2=\frac{\pi}{4}d^2_1$

So the volume is (remember to transform the time from minutes to seconds and the diameter from centimeters to meters)

$V=13.8\frac{m}{s}\cdot8\min \cdot\frac{60s}{1\min}\cdot\frac{\pi}{4}\cdot0.05^2m^2$$V=13.8\frac{m}{s}\cdot480s\cdot0.00196m^2$$V\approx13m^3$

2. The speed in the left section of the pipe is

$v_2=v_1\cdot\frac{A_1}{A_2}$

This is because the flow rate is constant along the pipe (Q = v*A).

As shown in the previous item the area is

$A=\frac{\pi}{4}\cdot d^2$

So when dividing both areas the factor π/4 cancels out. Therefore the speed is

$v_2=v_1\cdot\frac{d^2_1}{d^2_2}$

In this case we can use centimeters for the diameters because in the division the units cancel out

$v_2=13.8m/s\cdot\frac{5^2\operatorname{cm}}{2^2\operatorname{cm}}=13.8m/s\cdot\frac{25}{4}=86.25m/s$

3. Using Bernoulli's equation:

$p_1+\frac{1}{2}\cdot\rho\cdot v^2_1+\rho\cdot g\cdot h_1=p_2+\frac{1}{2}\cdot\rho\cdot v^2_2+\rho\cdot g\cdot h_2_{}$

The pipe is horizontal so h1 = h2. Also, the pipe is open on the right side p1 = p0, where p0 is atmosferic pressure. The third term on each side of the equation cancels out because they are equal:

$p_0+\frac{1}{2}\cdot\rho\cdot v^2_1=p_2+\frac{1}{2}\cdot\rho\cdot v^2_2$

Solving for p2:

$p_2=p_0+\frac{1}{2}\cdot\rho\cdot v^2_1-\frac{1}{2}\cdot\rho\cdot v^2_2$

1/2 and ρ are common factors:

$p_2=p_0+\frac{1}{2}\cdot\rho(v^2_1-v^2_2)$

Atmosferic pressure is given 1.01x10⁵ Pa, ρ is also given 1000 kg/m³. The speeds are v1 = 13.8m/s and v2 = 86.25 m/s. The pressure in the left section of the pipe is

$p_2=1.01\times10^5Pa+\frac{1}{2}\cdot1000\frac{\operatorname{kg}}{m^3}\cdot(13.8^2-86.25^2)\frac{m^2}{s^2}$$p_2=-3.52\times10^6Pa$

Thus, the guage pressure in the left section of the pope is:

$(-35.2\times10^5Pa-1.01\times10^5Pa)=-3.62\times10^6Pa\approx-35.8\text{atm}$