Question

Water at 70 kPa flows through a horizontal 2 meter diameter pipe at a velocity of 71 m/s and an elevation of 54 meters. If the density of water is 1000 kg/m^3 and the specific weight is 9.8 kN/m^3, what is the total pressure? Round to the nearest kPa.

Answer 1

Answer:

The total pressure is 3120 kilopascals.

Explanation:

The total pressure of this water flow is determined by the Bernoulli's Principle, which is the sum of dynamic ($p_{d}$) and hydraulic pressure ($p_{h}$). That is:

$p_{T} = p_{d} + p_{h}$

$p_{T} = \frac{1}{2}\cdot \rho \cdot v^{2} + \gamma \cdot z + p$

Where:

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

$v$ - Flow velocity, measured in meters per second.

$\gamma$ - Specific weight, measured in newtons per cubic meter.

$z$ - Elevation, measured in meters.

$p$ - Static pressure, measured in pascals.

Given that $\rho = 1000\,\frac{kg}{m^{3}}$, $v = 71\,\frac{m}{s}$, $\gamma = 9800\,\frac{N}{m^{3}}$, $z = 54\,m$ and $p = 70000\,Pa$, the total pressure is:

$p_{T} = \frac{1}{2}\cdot \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(71\,\frac{m}{s} \right)^{2} + \left(9800\,\frac{N}{m^{3}} \right)\cdot (54\,m)+70000\,Pa$

$p_{T} = 3119700\,Pa$

$p_{T} = 3119,7\,kPa$ (1 kPa = 1000 Pa)

$p_{T} = 3120\,kPa$

The total pressure is 3120 kilopascals.