Oil with a density of 890 kg/m3 moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is P1 = 2.00 ✕ 104 Pa,
and the pipe diameter is 7.00 cm. At another point
y = 0.30 m
higher, the pressure is
P2 = 1.25 ✕ 104 Pa
and the pipe diameter is 3.50 cm.

A tube is open at both its left and right ends. The tube starts at the left end, extends horizontally to the right, curves up and to the right, and extends horizontally to the right again. The right end is higher than its left end, and the change in height is labeled y. The pressure at the left end is labeled P1, and the pressure at the right end is labeled P2.

a- find the speed of the flow in the lower section (m/s)

b- find the speed of the flow in the higher section (m/s)

c- find the volume flow rate in the pipe (m^3/s)

(a) V₁ =  1.06 m/s

(b) V₂ = 4.24 m/s

(c) Q = 4.08 x 10⁻³ m³/s

Explanation:

(b)

The formula derived for Venturi tube can be used here:

P₁ - P₂ = (ρ/2)(V₂² - V₁²)

where,

P₁ - P₂ = Difference in Pressure = (2x 10⁴ Pa) - (1.25 x 10⁴ Pa) = 0.75 x 10⁴ Pa

ρ = Density of Oil = 890 kg/m³

V₂ = Velocity at Higher End = ?

V₁ = Velocity at Lower End = ?

Therefore,

0.75 x 10⁴ Pa = [(890kg/m³)/2](V₂² - V₁²)

V₂² - V₁² = (0.75 x 10⁴ Pa)/(445 kg/m³)

V₂² - V₁² = 16.85 m²/s²   equation (1)

Now, we will use continuity equation:

A₁V₁ = A₂V₂

where,

A₁ = Lower End Area = πd₁²/4 = π(0.07 m)²/4 = 3.848 x 10⁻³ m²

A₂ = Higher End Area = πd₂²/4 = π(0.035 m)²/4 = 9.621 x 10⁻⁴ m²

Therefore,

(3.848 x 10⁻³ m²)V₁ = (9.621 x 10⁻⁴ m²)V₂

V₁ = (9.621 x 10⁻⁴ m²)V₂/(3.848 x 10⁻³ m²)

V₁ = 0.25 V₂   equation (2)  

using this value in equation (1):

V₂² - (0.25 V₂)² = 16.85 m²/s²

0.9375 V₂² = 16.85 m²/s²

V₂² = (16.85 m²/s²)/0.9375

V₂ = √(17.97 m²/s²)

V₂ = 4.24 m/s

(a)

using the value of V₂ in equation (2):

V₁ = 0.25(4.24 m/s)

V₁ =  1.06 m/s

(c)

For fluid flow rate we use the following equation:

Flow Rate = Q = A₂V₂ = (9.621 x 10⁻⁴ m²)(4.24 m/s)

Q = 4.08 x 10⁻³ m³/s

v=ir = 3x15 =45 v

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:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .