Chapter 10, Problem 17P

(a)

To determine

The water gauge pressure.

Answer to Problem 17P

  $\text{9}{\text{.6×10}}^{\text{5}}{\text{N/m}}^{\text{2}}$

Explanation of Solution

Given:

Height of the tank, ${\mathrm{h}}_{\text{tank}}=5\text{m}$

Length of the pipe, ${\mathrm{h}}_{\text{pipe}}=110\text{m}$

Angle inclination, $\text{θ=58}°$

Formula used:

The gauge pressure is,

  ${\mathrm{P}}_{\text{G}}=\mathrm{\rho }\mathrm{g}\mathrm{h}$

Here, $\mathrm{\rho }$ is the density, g is the acceleration due to gravity and h is the height.

Calculation:

The height is the height from the bottom of the hill to the top of the tank

  $\begin{array}{c}{\mathrm{P}}_{\text{G}}=\mathrm{\rho }\mathrm{g}\mathrm{h}\\ \text{=(1}{\text{.00×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}\text{)(9}\text{.8}{\text{m/s}}^{\text{2}}\text{)}\left[\text{5}\text{m+(110}\text{m)sin}{\text{58}}^{\text{o}}\right]\\ \text{=9}{\text{.6×10}}^{\text{5}}{\text{N/m}}^{\text{2}}\end{array}$

Conclusion:

Therefore, gauge pressure of the water tank is $\text{9}{\text{.6×10}}^{\text{5}}{\text{N/m}}^{\text{2}}$ .

(b)

To determine

The height up to which the water shoots.

Answer to Problem 17P

  $\text{98m}$

Explanation of Solution

Given:

Height of the tank, ${\mathrm{h}}_{\text{tank}}=5\text{m}$

Length of the pipe, ${\mathrm{h}}_{\text{pipe}}=110\text{m}$

Angular inclination, $\mathrm{\theta }=58°$

Formula used:

Write the expression for the height of the water shoot.

  $\mathrm{h}={\mathrm{h}}_{\text{tank}}\text{+(}{\mathrm{h}}_{\text{pipe}}\text{)}\sin \mathrm{\theta }$

Here, ${\mathrm{h}}_{\text{tank}}$ is the height of the tank,

  ${\mathrm{h}}_{\text{pipe}}$ is the height of the pipe,

  $\mathrm{\theta }$ represents the angular inclination of pipe from the horizontal surface.

Calculation:

Substitute the values in the above expression.

  $\begin{array}{c}\mathrm{h}=\text{5}\text{m+(110}{\text{m)sin58}}^{\text{o}}\\ \text{=5}\text{m+110}\text{m×0}\text{.848}\\ \text{=5}\text{m+93}\text{.28}\text{m}\\ \text{=98}\text{.28}\text{m}\end{array}$

Conclusion:

The water at high to shoot up to the top of the tank is $\text{98}\text{.28}\text{m}$ .