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liubo4ka [24]
3 years ago
5

Consider fully developed laminar flow in a circular pipe. If the viscosity of the fluid is reduced by half by heating while the

flow rate is held constant, how will the pressure drop change
Engineering
1 answer:
gladu [14]3 years ago
4 0

Answer:

The pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its original value.

Explanation:

For a fully developed laminar flow in a circular pipe, the flowrate (volumetric) is given by the Hagen-Poiseulle's equation.

Q = π(ΔPR⁴/8μL)

where Q = volumetric flowrate

ΔP = Pressure drop across the pipe

μ = fluid viscosity

L = pipe length

If all the other parameters are kept constant, the pressure drop across the circular pipe is directly proportional to the viscosity of the fluid flowing in the pipe

ΔP = μ(8QL/πR⁴)

ΔP = Kμ

K = (8QL/πR⁴) = constant (for this question)

ΔP = Kμ

K = (ΔP/μ)

So, if the viscosity is halved, the new viscosity (μ₁) will be half of the original viscosity (μ).

μ₁ = (μ/2)

The new pressure drop (ΔP₁) is then

ΔP₁ = Kμ₁ = K(μ/2)

Recall,

K = (ΔP/μ)

ΔP₁ = K(μ/2) = (ΔP/μ) × (μ/2) = (ΔP/2)

Hence, the pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its value.

Hope this Helps!!!

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Answer:

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A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric
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Answer:

Explanation:

Given

q_1=11.5\ nC charge is placed at x=0\ cm

another charge of q_2=-1.2\ nC is at x=3\ cm

We know that Electric field due to positive charge is away  from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to q_2 at some distance r from it

E_2=\frac{kq_2}{r^2}

Now Electric Field due to q_1 is

E_1=\frac{kq_1}{(3+r)^2}

Now E_1+E_2=0

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A manometer containing a fluid with a density of 60 lbm/ft3 is attached to a tank filled with air. If the gage pressure of the a
8090 [49]

Answer:

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And P(gage) = ρgh

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A diagrammatic representation of this setup is presented in the attached image.

Hope this helps!

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