Question

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 3.30 105 Pa and a speed of 2.0 m/s. However, on the second floor, which is 4.4 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

Answer 1

Answer:

P₂ = 2.82755 * 10⁵ Pa

Explanation:

Parameters for the first floor:

Guage pressure, P₁ = 3.30 * 10⁵ Pa

Speed, v₁ = 2.0 m/s

Height of the first floor = h₁

Parameters of the second floor:

Guage Pressure, P₂ = ?

Speed = 3.5 m/s

height of the second floor = h₂

From the question, h₂ - h₁ = 4.4 m, i.e. h₁ - h₂ = - 4.4 m

Density of water, $\rho = 1000 kg/m^{3}$

According to the Bernoulli's equation

$P_{1} + 0.5 \rho (v_{1}) ^{2} + \rho g h_{1} = P_{2} + 0.5 \rho (v_{2}) ^{2} + \rho g h_{2}$

$P_{2} = P_{1} + 0.5 \rho((v_{1}) ^{2} - (v_{2} )^{2}) + \rho g(h_{1} - h_{2)$

$P_{2} = 3.30 * 10^{5} + 0.5*1000(2.0^{2} - 3.5^{2} ) + 1000*9.8(-4.4) \\P_{2} = 330000 - 4125 - 43120 \\P_{2} = 282755 Pa$

$P_{2} = 2.82755 * 10^{5} Pa$

Answer 2

The Gauge Pressure of water on the second floor is 2.287×$10^{5}$ Pa.

According to Bernoulli's principle

$P_{1}$ +  1/2ρ$v_{1} ^{2}$ + ρg$h_{1}$ = $P_{2}$ + 1/2ρ$v_{2} ^{2} }$ + ρg$h_{2}$

where the given data is

           $P_{1}$ = 3.30 ×$10^{5}$ Pa

           $v_{1}$ = 2.0 m/s

           $v_{2}$ = 3.5 m/s

           $h_{1}-h_{2}$ = 4.4 m

and density of water ρ = 1000kg/$m^{3}$

to find $P_{2}$ we put all the available data in the equation and calculate

⇒ 3.30 ×$10^{5}$ Pa + 0.5 × 1000×4 + 1000×9.8×(-4.4) - 05×1000×12.25 = $P_{2}$

        $P_{2}$ = 2.287×$10^{5}$ Pa. which is the gauge pressure on the second floor

Learn more about Bernoulli's Principle:

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