Question

Water flows through a multisection pipe placed horizontally on the ground. The velocity is 3.0 m/s at the entrance and 2.1 m/s at the exit. What is the pressure difference between these two points?
a. 0.2 kPa
b. 2.3 kPa
c. 28 kPa
d. 110 kPa

Answer 1

Answer:

b. 2.3 kPa.

Explanation:

This situation can be modelled by Bernoulli's Principle, as there are no energy interaction throughout the multisection pipe and current lines exists between both ends. Likewise, this system have no significant change in gravitational potential energy since it is placed horizontally on the ground and is described by the following model:

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

Where:

$P_{1}$, $P_{2}$ - Pressures at the beginning and at the end of the current line, measured in kilopascals.

$\rho$ - Water density, measured in kilograms per cubic meter.

$v_{1}$, $v_{2}$ - Fluid velocity at the beginning and at the end of the current line, measured in meters per second.

Now, the pressure difference between these two points is:

$P_{1} - P_{2} = \rho \cdot \frac{v_{2}^{2}-v_{1}^{2}}{2}$

If $\rho = 1000\,\frac{kg}{m^{3}}$, $v_{1} = 3\,\frac{m}{s}$ and $v_{2} = 2.1\,\frac{m}{s}$, then:

$P_{1} - P_{2} = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \frac{\left(2.1\,\frac{m}{s} \right)^{2}-\left(3\,\frac{m}{s} \right)^{2}}{2}$

$P_{1} - P_{2} = -2295\,Pa$

$P_{1} - P_{2} = -2.295\,kPa$ (1 kPa is equivalent to 1000 Pa)

Hence, the right answer is B.