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 4.5 × 105 pa and a speed of 2.3 m/s. however, on the second floor, which is 4.0 m higher, the speed of the water is 4.4 m/s. the speeds are different because the pipe diameters are different. what is the gauge pressure of the water on the second floor?

the gauge pressure of the water on the second floor is Pg2 = 403765 Pa = 4.03 * 10⁵ Pa

Explanation:

According to Bernoulli's equation ( conservation of mechanical energy, therefore we neglect the friction inside the pipes)

P1 + ρ*g*h1 + 1/2*ρ*v1² = P2 + ρ*g*h2 + 1/2*ρ*v2²

where ρ= density of water , h=height , P= absolute pressure, g= gravity ( 9.8 m/s²)

since the gauge pressure Pg is

Pg = P - Pa (atmospheric pressure)

Pg1 +Pa + ρ*g*h1 + 1/2*ρ*v1² = Pg2 + Pa + ρ*g*h2 + 1/2*ρ*v2²

Pg1 + ρ*g*h1 + 1/2*ρ*v1² = Pg2 + ρ*g*h2 + 1/2*ρ*v2²

since our reference point to measure heights can be chosen, we choose h1=0 and h2= 4m. Thus

Pg1 + 1/2*ρ*v1² = Pg2 + ρ*g*h2 + 1/2*ρ*v2²

therefore

Pg2 = Pg1 + 1/2*ρ*v1² -ρ*g*h2 - 1/2*ρ*v2² =  Pg1 -ρ*g*h2 -1/2*ρ*(v2²-v1²)

replacing values and assuming ρ=1000 kg/m³

Pg2 = Pg1 -ρ*g*h2 -1/2*ρ*(v2²-v1²) = 4.5 * 10⁵ Pa - 1000kg/m³*9.8 m/s²*4m -1/2* 1000 kg/m³ * [(4.4 m/s)² - (2.3 m/s)² ] = 403765 Pa = 4.03 * 10⁵ Pa

Pg2 =4.03 * 10⁵ Pa

hypothesising or postulating