A liquid of density 1237 kg/m² flows with
speed 1.84 m/s into a pipe of diameter 0.18 m.
The diameter of the pipe decreases to 0.05 m
at its exit end. The exit end of the pipe is
3.08 m lower than the entrance of the pipe,
and the pressure at the exit of the pipe 1.3 atm. Applying Bernoulli's principle what is the pressure P1 at the entrance end of the pipe?

The pressure P1 at the entrance end of the pipe = 444,001.4 Pa = 4.382 atm

Explanation:

The Bernoulli's principle is basically a law of conservation of energy, it theorizes that the sum of the energy due to pressure, kinetic energy and potential energy of a fluid is constant all through the points of the fluid motion.

Or that the losses due to pressure effects, kinetic effects and potential effects is equal to 0 all through the fluids motion.

It is given mathematically as

[(P₂ - P₁)/ρg] + [(v₂² - v₁²)/2g] + [z₂ - z₁] = 0

P₂ = Pressure at the exit = 1.3 atm = 1.3 × 101325 = 131,722.5 Pa

P₁ = Pressure at the entrance end = ?

ρ = density of the fluid = 1237 kg/m³

g = acceleration due to gravity = 9.8 m/s²

v₁ = velocity of the fluid at the entrance end = 1.84 m/s

v₂ = velocity of the fluid at the exit = ?

But we can obtain this using the continuity equation, the continuity equation stated mathematically is

ρ₁ A₁ v₁ = ρ₂A₂v₂

Since the density of the fluid is constant all through, the continuity equation becomes

A₁ v₁ = A₂ v₂

A₁ = cross sectional Area At the entrance end = (πd₁²/4)

A₂ = cross sectional Area At the exit = (πd₂²/4)

The continuity equation then further becomes

d₁² v₁ = d₂² v₂

0.18² × 1.84 = 0.05² × v₂

v₂ = 23.8464 m/s

z₂ = elevation of the exit end

z₁ = elevation of the entrance end

z₂ - z₁ = -3.08 m (minus sign because the exit end is lower than the entrance end)

[(P₂ - P₁)/ρg] + [(v₂² - v₁²)/2g] + [z₂ - z₁] = 0

Multiplying through by ρg, the Bernoulli equation becomes

[(P₂ - P₁)] + [ρ(v₂² - v₁²)/2] + ρg[z₂ - z₁] = 0

ΔP + [1237 (23.8464² - 1.84²)/2] + (1237×9.8×-3.08) = 0

ΔP + 349,616.522 - 37,337.608 = 0

ΔP + 312,278.914 = 0

ΔP = - 312,278.914 Pa

ΔP = (P₂ - P₁) = -312,278.914

P₂ = 131,722.5 Pa

P₁ = ?

131,722.5 - P₁ = -312,278.914

P₁ = 131,722.5 + 312,278.914 = 444,001.414 Pa

P₁ = (444,001.414) ÷ (101325) = 4.382 atm

Hope this Helps!!!

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

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