Underground water is to be pumped by a 78 percent efficient 5-kw submerged pump to a pool whose free surface is 30 m above the underground water level. the diameter of the pipe is 7 cm on the intake side and 5 cm on the discharge side. take the density of water to be 1 kg/l = 1000 kg/m3. assume the elevation difference between the pump inlet and the outlet and the effect of the kinetic energy correction factors to be negligible

a. Q = 10.15 x 10⁻³ m³/s

b. Δp = 202.76 kPa

Explanation:

For solve the pipe system can use Bernoulli equations:

Patm / ρ + V₁² / 2 + g Z₁ + e pump / m = P atm / ρ + V₂² / 2 + g Z₂

e pump / ρ * Q = V₂² / 2  + g Z₂

a.

Q = V₂ * A₂

e pump / ρ * V₂ * A₂ = V₂² / 2  + g Z₂

Given: e pump = 0.78 * 5 kW = 3.9 kW, ρ = 1000 kg / m³ , A₂ = π * 5 ²cm /4 , Z₂ = 30 m

A₂ = 1.96 x 10 ⁻³ m²

V₂ = 5.18 m/s

Q = 10.15 x 10⁻³ m³/s

b.

A₁ = π * 7² cm / 4 = 3.85 x 10⁻³ m²

Δp = ρ * [ ( Q / A₁ )² / 2  -  (Q / A₂)² / 2 + e pump / ρ *Q]

Δp = 202.76 kPa

i don't understand is there any other way you can rephrase it plz let me know in the comments and i will do it.

explanation:

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.