Aliquid of density 1370 kg/m 3 flows steadily through a pipe of varying diameter and height. at location 1 along the pipe, the flow speed is 9.67 m/s and the pipe diameter d 1 is 12.9 cm . at location 2, the pipe diameter d 2 is 16.7 cm . at location 1, the pipe is δ y = 9.85 m higher than it is at location 2. ignoring viscosity, calculate the difference δ p between the fluid pressure at location 2 and the fluid pressure at location 1.

ΔP = P₂ - P₁  = 132. 24 kPa

Explanation:

Given:

ρ = 1370 kg / m³ , v = 9.67 m / s  ,  d₁ = 12.9 cm , d₂ = 16.7 cm , Δy = 9.85 m

Using Bernoulli's equations to determine the difference ΔP  

P₂ + ρ * g * Z₂ + (ρ * V₂²) / 2 = P₁ + ρ * g * Z₁ + (ρ * V₁²) / 2

P₂ - P₁ = ρ * g * (Z₁ - Z₂) + [ ρ * (V₁² - V₂²) ] / 2

P₂ - P₁ = ρ * g * (Z₁ - Z₂) + ¹/₂ * ρ * V₁² * [ ( 1 - (d₁ / d₂) ⁴ ) ]

ΔP = 1370 kg / m³ * 9.8 m/s² * 9.85m  +  0.5 * 1370 kg / m³ * ( 1 - (12.9 cm / 16.7 cm )⁴ )

ΔP = 132247.9364 Pa

ΔP = 132. 24 kPa

P₂ - P₁=173.5kPa

Explanation:

The equation of continuity:  

A₁v₁=A₂v₂  

where A₁=πd₁²/4 and A₂=πd₂²/4

v₂=(A₁/A₂)v₁

v₂={(πd₁²/4)/(πd₂²/4)}v₁

v₂=(d₁²/d₂²)v₁

Use Bernoulli's equation

P₂+pgz₂+(pv₂²/2)=P₁+pgz₁+(pv₁²/2)

The difference between the fluid pressure at location 2 and the fluid pressure at location 1

P₂ - P₁=pg(z₁-z₂)+{p(v₁²-v₂²)}/2=pg(z₁-z₂)+1/2pv₁²(1-(d₁/d₂)⁴)

P₂ - P₁=(1.370×10³×9.8×9.85)+(1/2)(1.370×10³×(9.67)²){(1-(0.129m/0.167m)⁴}

P₂ - P₁=1.735×10⁵Pa

P₂ - P₁=173.5kPa

the results is a decrease in air temperature with an increase of altitude.

explanation:

probably not if net force is zero