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zalisa [80]
3 years ago
9

Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2/s is being discharged by an 8-mm-diameter, 42-m-long horiz

ontal pipe from a storage tank open to the atmosphere. The height of the liquid level above the center of the pipe is 4 m. Disregarding the minor losses, determine the flow rate of oil through the pipe. Solve this problem by making the assumption that since the velocity through the pipe is so small, the pressure at the pipe entrance is nearly the same as the hydrostatic pressure at that location.
Engineering
1 answer:
sladkih [1.3K]3 years ago
8 0

Answer:

The flow rate of oil through the pipe is 1.513E-7 m³/s.

Explanation:

Given

Density, ρ = 850 kg/m³

Kinematic viscosity, v = 0.00062 m²/s

Diameter, d = 8-mm = 0.008m

Length of horizontal pipe, L = 42-m

Height, h = 4-m.

We'll solve the flow rate of oil through the pipe by using Hagen-Poiseuille equation.

This is given as

∆P = (128μLQ)/πD⁴

Where ∆P = Rate of change of pressure

μ = Dynamic Viscosity

Q = Flow rate of oil through the pipe.

First, we need to determine the dynamic viscosity and the rate of change in pressure

Dynamic Viscosity, μ = Density (ρ) * Kinematic viscosity (v)

μ = 850 kg/m³ * 0.00062 m²/s

μ = 0.527kg/ms

Then, we calculate the rate of change of pressure.

Assuming that the velocity through the pipe is so small;

∆P = Pressure at the bottom of the tank

∆P = Density (ρ) * Acceleration of gravity (g) * Height (h)

Taking g = 9.8m/s²

∆P = 850kg/m³ x 9.8m/s² x 4m

∆P = 33320N/m²

Recall that Hagen-Poiseuille equation.

∆P = (128μLQ)/πD⁴ --- Make Q the subject of formula

Q = (πD⁴P)/(128μL)

By substituton;

Q = (π * 0.008⁴ * 33320)/(128 * 0.527 * 42)

Q = 0.00000015133693643099

Q = 1.513E-7 m³/s.

Hence, the flow rate of oil through the pipe is 1.513E-7 m³/s.

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Answer:

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Let r = Initial radius of second cylinder = 16mm.

Now, we can proceed....

For the two cylinders to have the same deformed hardness,

They must be deformed by the same percentage.

To get the deformed radius of the second cylindrical specimen, we'll need to calculate the percentage deformation of specimen 1.

This is given by

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Ad = 225π

So, %deformation = %d =(400π - 225π)/400π

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= 0.4375

%d = 43.75%

Now, we'll solve for the deformed radius of the second cylindrical specimen.

Note that, the two specimens have the same % deformation (to have the same hardness).

So, we have

rd - ro = 1 - %d

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ro = original radius = 16

rd/ro = 1 - %d becomes

rd / 16 = 1 - 43.75%

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Increasing the magnification and decreasing the field view

we are given:

a(t)=4*t^2-)

where t= time in seconds, and a(t) = acceleration as a function of time.

and

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x(2) = -20 )

where x(t) = distance travelled as a function of time.

need to find x(4).

from (1), we express x(t) by integrating, twice.

velocity = v(t) = integral of (1) with respect to t

v(t) = 4t^3/3 - 2t + k1     )

where k1 is a constant, to be determined.

integrate (4) to find the displacement x(t) = integral of (4).

x(t) = integral of v(t) with respect to t

= (t^4)/3 - t^2 + (k1)t + k2   )   where k2 is another constant to be determined.

from (2) and (3)

we set up a system of two equations, with k1 and k2 as unknowns.

x(0) = 0 - 0 + 0 + k2 = -2   => k2 = 2   )

substitute (6) in (3)

x(2) = (2^4)/3 - (2^2) + k1(2) -2   = -20

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Answer:

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The cost associated with the heat loss through the windows for an 8-hour flight is:

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For polycarbonate windows: $262.08

For soda-lime glass windows: $1,238.016 (least efficient)

Explanation:

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\frac{dQ}{dt}=\dot Q =-kS\nabla \vec{T}

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To obtain the thermal conductivity of the soda-lime glass we use the graphic attached to this answer (In this case for soda-lime glass k₃₀₀=0.992w/m·K).

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Cost=C_{hc}\cdot \dot Q \cdot t \cdot n=1\frac{\$}{Kwh} \cdot \dot Q \cdot 8h \cdot 130

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