Chapter 10, Problem 65GP

To determine

To Calculate: The range of the forces.

Answer to Problem 65GP

  $\text{150}\text{N}\le F\le \text{220}\text{N}$

Explanation of Solution

Given info:

Initial pressure

  $P_i=210\text{kPa}=210\text{kPa}\times \left(\frac{1000{\text{N/m}}^{\text{2}}}{1\text{kPa}}\right)=21\times {10}^4{\text{N/m}}^{\text{2}}$

Final pressure

  $P_f=310\text{kPa}=310\text{kPa}\times \left(\frac{{\text{1000N/m}}^{\text{2}}}{\text{1kPa}}\right)=31\times {10}^4{\text{N/m}}^{\text{2}}$

Diameter, $d=3\text{cm}=0.03\text{m}$

Formula used:

The continuity equation is

  $Av=\text{const}$

Where, A is the area and v is the velocity.

The force is calculated as

  $F=PA$

Here, P is pressure.

Calculation:

Firstly, calculate the area such that

  $A=\pi r^2=3.14\times {\left(1.5\times {10}^{-2}\text{m}\right)}^2$

And now use these values in the formula of force,and then the force is,

  $\begin{array}{c}{F}_i=P_i\times A\\ =3.14\times {\left(1.5\times {10}^{-2}\text{m}\right)}^2\times 21\times {10}^4{\text{N/m}}^{\text{2}}\\ =1.5\times {10}^2\text{N}\\ =150\text{N}\end{array}$

Where, $F_i$ is the initial force

Now, the final force is,

  $\begin{array}{c}{F}_f=P_f\times A\\ =3.14\times {\left(1.5\times {10}^{-2}\text{m}\right)}^2\times 31\times {10}^4{\text{N/m}}^{\text{2}}\\ =2.19\times {10}^2\text{N}\\ =220\text{N}\end{array}$

Conclusion:

The range of the force which needs to be applied to the pump handle from the beginning to end is $\text{150}\text{N}\le F\le \text{220}\text{N}$ .