Question

Water at a pressure of 3.6 atm at street level flows into an office building at a speed of 0.4 m/s through a pipe 6 cm in diameter.The pipes taper down to 2.8 cm in diameter by the top floor,28 m above.Calculate the water pressure in such a pipe on the top floor.(g=9.8 m/s square)

Answer 1

Water at a pressure of 3.6 atm at street level flows into an office building at a speed of $0.4\mathrm{m/s}$ through a pipe $6\mathrm{cm}$ in diameter. The pipes taper down to $2.8\mathrm{cm}$ in diameter by the top floor, 28 m above. Calculate the water pressure in such a pipe on the top floor. (g=9.8 m/s square)

Solution

According to Bernoulli's law applies:

$$P _ {1} + \frac {1}{2} * \rho * v _ {1} ^ {2} + \rho * g * y _ {1} = P _ {2} + \frac {1}{2} * \rho * v _ {2} ^ {2} + \rho * g * y _ {2}$$

$P_{1}$ with the initial pressure of 3,6 atm, the initial velocity $v_{1}$ of $0.4\frac{m}{s}$ and $y_{1} = 0$. $Y_{2}$ is

28 m and $P_{2}$ and $v_{2}$ are asked.

According to the law of continuity applies:

$$A _ {1} * v _ {1} = A _ {2} * v _ {2}$$

$A_{1}$ indicates the initial cross-section, $v_{1}$ is the initial velocity, $A_{2}$ is the cross-sectional end and $v_{2}$ is requested.

Solving this system gives the final speed

$$v _ {2} = \frac {A _ {1} * v _ {1}}{A _ {2}}$$ $$v _ {2} = \frac {0.03 ^ {2} * \pi * 0.4}{0.014 ^ {2} * \pi}$$ $$v _ {2} = 1,84 \frac {m}{s}$$

And for the final pressure:

$$P _ {2} = P _ {1} + \frac {1}{2} * \rho * v _ {1} ^ {2} - \left(\frac {1}{2} * \rho * v _ {2} ^ {2} + \rho * g * y _ {2}\right)$$ $$P _ {2} = 3.6 * 101325 + 0.5 * 1000 * 0.4 ^ {2} - 0.5 * 1000 * 1,84 ^ {2} - 1000 * 10 * 28 \mathrm{Pa}$$ $$P _ {2} = 83157,2 \mathrm{Pa}$$ $$P _ {2} = 0,82 \mathrm{atm}$$