Given:
• Height of ladder = 9.7 m
,
• Diameter of water hose = 2.7 inch
,
• Diameter of pump outlet = 3.47 inch
,
• Guage of water pump = 236.49 kPa = 236.49 x 10³ Pa
,
• Density of water = 1000 kg/m³
Let's find the speed of the water jet emerging from the nozzle.
To find the speed of the water jet, apply Bernoulli's equation.
We have:
[tex]\frac{1}{2}p(v^2_N-v^2_p)=p_{pumpguage}-pgh[/tex]
Rewrite the formula for VN:
[tex]\begin{gathered} v^2_N-v^2_p=(\frac{2}{p})p_{pumpguage}-2gh_{}_{} \\ \\ v^2_N-(\frac{A_N}{A_P})^2v^2_p=(\frac{2}{p})p_{pumpguage}-2gh \\ \\ v^2_N-(\frac{r^4_N}{r^4P_{}})^{}v^2_p=(\frac{2}{p})p_{pumpguage}-2gh \\ \\ \\ v_N=\sqrt[]{\frac{(\frac{2}{p})p_{pumpguage}-2gh}{1-(\frac{r^4_N}{r^4_p})}} \\ \end{gathered}[/tex]
• VN is the speed of the water jet from nozzle.
,
• g is acceleration due to gravity = 9.8 m/s²
,
• h is the height of ladder = 9.7 m
,
• rN is the radius of nozzle = diameter of noozle/2 = 2.7/2 = 1.35 inches
,
• rp is the radius of the pump outlet = diameter of pump outlet/2 = 3.47/2 = 1.735 inches
,
• p is density of water = 1000 kg/m³
Input values into the formula and solve for vN:
[tex]\begin{gathered} v_N=\sqrt[]{\frac{(\frac{2}{1000})(236.49\times10^3)-2(9.8)(9.7)}{1-(\frac{1.35}{1.735})^4}} \\ \\ v_N=21.1315\text{ m/s} \end{gathered}[/tex]
Therefore, the speed of the water jet from the noozle is 21.1315 m/s
ANSWER:
21.1315 m/s
