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

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!!!

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation:

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