Una tubería con secciones circulares se emplea para el traslado de agua entre el tanque central y un tanque auxiliar. El tanque auxiliar en el extremo de la tubería tiene 72.0 m3, se llena en 1.00 hora con un flujo constante de agua. En uno de los tramos de la tubería de sección A, tiene un radio de 0.025 m y presión manométrica de 2.0 atm; luego de recorrer 850.0 m y baja una altura de 15.0 m se llega a una sección B de radio 0.050 m. Encuentre:

El caudal del agua que se transporta en litros por segundo.

A) 20 L/s

B) 2.55 m/s

C) 10.20 m/s

D) 400.8 kPa

Explanation:

a)

In this problem, we know that the volume of the tank:

is filled in a time of

We also know that the volume flow rate of water in the pipe is constant in every point of the pipe, so it can be calculated directly from the two data above.

The volume of the tank, in liter, is

While the time, in seconds, is

Therefore, the volume flow rate in Liters per second is:

b)

The volume flow rate of water through the pipe can be also written as

where

A is the cross-sectional area of the pipe

v is the speed of the water

In section B, we have:

r = 0.050 m is the radius of section B

so, the cross-sectional area of section B is:

The volume flow rate in SI units is

Therefore, the speed of the water in section B is:

c)

As in part B), we know that the volume flow rate must remain constant through the entire pipe.

So, the volume flow rate in section A of the pipe is still

The radius of the pipe in section A is

Therefore, the cross-sectional area in section A of the pipe is

So, since we have

we can find the speed of water in section A:

d)

Here we want to find the gauge pressure in section B.

We know that:

is the pressure in section A

is the altitude of section A

is the speed of water in section A

is the speed of water in section B

We can write Bernoulli's equation:

where

is the water density

is the pressure in section B

And solving for pB, we find:

Which is

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg