Question

water at a gauge pressure of 3.8 atm at street flows into an office building at a speed of 0.60 m/s through a pipe 50 mm in diameter. The pipe tapers down to 2.6 cm in diameter by the top flooe, 18m above the street, where a faucet has been left open. calculate the flow velocity and the gauge pressure in such a pipe at the top floor.

Answer 1

A water at a gauge pressure of 3.8 atm at street flows into an office building at a speed of 0.60 m/s through a pipe 50 mm in diameter. The pipe tapers down to 2.6 cm in diameter by the top floor, 18m above the street, where a faucet has been left open. Calculate the flow velocity and the gauge pressure in such a pipe at the top floor.

Solution.

$$P _ {1} = 3. 8 \mathrm {a t m}, v _ {1} = 0. 6 0 \frac {\mathrm {m}}{\mathrm {s}}, d _ {1} = 5 0 \mathrm {m m} = 5 0 \cdot 1 0 ^ {- 3} \mathrm {m}, d _ {2} = 2. 6 \mathrm {c m} = 2. 6 \cdot 1 0 ^ {- 2} \mathrm {m},$$ $$h = 1 8 m, g = 9. 8 \frac {m}{s ^ {2}}, \rho = 1 0 0 0 \frac {k g}{m ^ {3}};$$ $$v _ {2} -? P _ {2} -?$$

For the water in the pipe:

$$v _ {1} S _ {1} = v _ {2} S _ {2};$$

$S_{1}$ - the sectional area of the pipe at the street;

$S_{2}$ - the sectional area of the pipe by the top floor;

$v_{1}$ - the flow velocity at the street;

$v_{2}$ - the flow velocity by the top floor.

$$S _ {1} = \pi \frac {d _ {1} ^ {2}}{4};$$ $$S _ {2} = \pi \frac {d _ {2} ^ {2}}{4}.$$ $$v _ {1} \pi \frac {d _ {1} ^ {2}}{4} = v _ {2} \pi \frac {d _ {2} ^ {2}}{4};$$ $$v _ {1} d _ {1} ^ {2} = v _ {2} d _ {2} ^ {2};$$

The flow velocity by the top floor:

$$v _ {2} = v _ {1} \frac {d _ {1} ^ {2}}{d _ {2} ^ {2}}.$$ $$v _ {2} = 0.60 \frac {m}{s} \cdot \frac {(50 \cdot 10^{-3} m)^2}{(2.6 \cdot 10^{-2} m)^2} = 2.22 \frac {m}{s}.$$

By Bernoulli's principle:

$$\frac {\rho v _ {1} ^ {2}}{2} + \rho g h _ {1} + P _ {1} = \frac {\rho v _ {2} ^ {2}}{2} + \rho g h _ {2} + P _ {2};$$

$\rho$ - the density of the water;

$g$ - the acceleration due to gravity;

$h_1$ - the elevation of the pipe at the street

$h_2$ - the elevation of the pipe by the top floor

$P_{1}$ - the gauge pressure at the street;

$P_{2}$ - the gauge pressure by the top floor.

$$h _ {1} = 0;$$ $$h _ {2} = h.$$ $$\frac {\rho v _ {1} ^ {2}}{2} + P _ {1} = \frac {\rho v _ {2} ^ {2}}{2} + \rho g h + P _ {2};$$ $$P _ {2} = P _ {1} + \frac {\rho}{2} (v _ {1} ^ {2} - v _ {2} ^ {2}) - \rho g h.$$

Converting the gauge pressure to $Pa$:

$$P _ {2} = 3.8 \ \mathrm{atm} \cdot \left(1.01 \cdot 10^{5} \frac {P a}{\mathrm{atm}}\right) = 3.838 \cdot 10^{5} P a.$$

The gauge pressure by the top floor:

$$\begin{array}{l} P _ {2} = 3.838 \cdot 10^{5} P a + \frac {1000 \frac {k g}{m^{3}}}{2} \left(\left(0.60 \frac {m}{s}\right)^{2} - \left(2.22 \frac {m}{s}\right)^{2}\right) - 1000 \frac {k g}{m^{3}} \cdot 9.8 \frac {m}{s^{2}} \cdot 18 m = \\ = 205115.8 P a. \\ \end{array}$$

Converting the gauge pressure to atm:

$$P _ {2} = \frac {205115.8 P a}{\left(1.01 \cdot 10^{5} \frac {P a}{\mathrm{atm}}\right)} = 2.03 \ \mathrm{atm}.$$

Answer:

The flow velocity in such a pipe at the top floor is $v_{2} = 2.22\frac{m}{s}$.

The gauge pressure in such a pipe at the top floor is $P_{2} = 2.03atm$.