Water circulates throughout a house in a hot-water heating system. Ifthe water is pumped at a speed of 0.50 m/s through a 4.0-cm-diameterpipe in the basement under a pressure of 3.0 atm, what will be thepressure in a 2.6-cm-diameter pipe on the second floor 5.0 m above ifthe flow speed is 1.2 m/s? Assume the pipes do not divide intobranches.

To solve this problem, we can use Bernoulli's equation, which states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume is constant for an incompressible, non-viscous fluid in steady flow.

The equation is: P1 + 1/2ρv1² + ρgh1 = P2 + 1/2ρv2² + ρgh2

Where: P1 and P2 are the pressures at the two points, v1 and v2 are the speeds of the fluid at the two points, h1 and h2 are the heights of the two points above some reference level, ρ is the density of the fluid, and g is the acceleration due to gravity.

Given: P1 = 3.0 atm = 3.0 * 1.013 * 10^5 Pa (converting atm to Pa), v1 = 0.50 m/s, h1 = 0, P2 = ?, v2 = 1.2 m/s, h2 = 5.0 m, ρ = 1000 kg/m³ (density of water), and g = 9.81 m/s².

Substituting these values into Bernoulli's equation, we get:

P1 + 1/2ρv1² + ρgh1 = P2 + 1/2ρv2² + ρgh2 3.0 * 1.013 * 10^5 Pa + 1/2 * 1000 kg/m³ * (0.50 m/s)² + 1000 kg/m³ * 9.81 m/s² * 0 = P2 + 1/2 * 1000 kg/m³ * (1.2 m/s)² + 1000 kg/m³ * 9.81 m/s² * 5.0 m

Solving for P2, we get:

P2 = P1 + 1/2ρv1² - 1/2ρv2² - ρgh2 P2 = 3.0 * 1.013 * 10^5 Pa + 1/2 * 1000 kg/m³ * (0.50 m/s)² - 1/2 * 1000 kg/m³ * (1.2 m/s)² - 1000 kg/m³ * 9.81 m/s² * 5.0 m

Calculating the above expression will give the pressure in the 2.6-cm-diameter pipe on the second floor.