Apump is used to send water through a hose, the diameter of which is 10 times that of the nozzle through which the water exits. if the nozzle is 1 m higher than the pump, and the water flows through the hose at 0.4 m/s, what is the gauge pressure of the water at the pump?

Use VFR1 = VFR2 to discover the velocity at in the hose VFR = A * V

D hose =10 * D nozzle, R hose = 5 * D nozzle

Area of a circle = πR^2

Area h=3.14*25*D^2 = 75.5D^2

(Radius=Diameter/2) area n = 3.14*(D^2/4) = .785D^2

Use VFR = VFR v2 = 0.4m/s

0.4*.785D^2 = 75.5*D^2* v1 D^2

= .314 =75.5*V1

v1 = 0.004m/s

Now we have the velocity, we can use Bernoulli's equation.

P1+ρgh1+ρV1^2 /2 = constant

There is no atmospheric pressure before so the P1= the gauge pressure at the pump, let’s call the height of the hose 0m and the height of the nozzle 1m so the is no ρgh1 Likewise, there is only atmospheric pressure at the nozzle which is 100000 PA, and lastly the density ρ of water is 1000 KG/M^3

Pg + 1000*.004^2/2 = 100000+1000*9.8*1+ 1000*0.4^2/2

Pg + .008= 100000+9800+80

Pg+.008= 109880

Pg=109880.008 PA

hello, i am papaguy.

i have found an answer to your question but use it as information.

information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

best of luck!

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