Water is circulating through a closed system of pipes in a two floor apartment. On the first floor, the water has a gauge pressure of 3.70 105 Pa and a speed of 2.4 m/s. However, on the second floor, which is 3.6 m higher, the speed of the water is 3.5 m/s. The speeds are different because the pipe diameters are different. What is the gauge pressure of the water on the second floor

[tex]\frac{P_{1}}{9810}[/tex] + [tex]\frac{V_{1} ^{2}}{19.62}[/tex] + [tex]h_{1}[/tex] = [tex]\frac{P_{2}}{9810}[/tex] + [tex]\frac{V_{2} ^{2}}{19.62}[/tex] + [tex]h_{2}[/tex] --------------(1)

Where

[tex]P_{1}[/tex] = Gauge pressure at inlet = 3.70105 pascal

[tex]V_{1}[/tex] = velocity at inlet = 2.4 [tex]\frac{m}{sec}[/tex]

[tex]P_{2}[/tex] = Gauge pressure at outlet = we have to calculate

[tex]V_{2}[/tex] = velocity at outlet = 3.5 [tex]\frac{m}{sec}[/tex]

[tex]h_{2} - h_{1}[/tex] = 3.6 m

Put all the values in equation (1) we get,

⇒ [tex]\frac{3.70105}{9810}[/tex] + [tex]\frac{2.4 ^{2}}{19.62}[/tex] = [tex]\frac{P_{2}}{9810}[/tex] + [tex]\frac{3.5 ^{2}}{19.62}[/tex] + 3.6

⇒ 0.294 = [tex]\frac{P_{2}}{9810}[/tex] + 0.6244 + 3.6

⇒ [tex]\frac{P_{2}}{9810}[/tex] = 0.294 - 0.6244 - 3.6

⇒ [tex]\frac{P_{2}}{9810}[/tex] = - 3.9304

⇒ [tex]P_{2}[/tex] = - 38557.224 pascal

This is the value of gauge pressure at outlet.