Answer:
a) 3.607 m
b) 1.5963 m
Explanation:
See that attached pictures for explanation.
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!!!
Answer:
The correct answer is
option C. current to pneumatic (V/P)
Explanation:
A current to pneumatic controller is basically used to receive an electronic signal from a controller and converts it further into a standard pneumatic output signal which is further used to operate a positioner or control valve. These devices are reliable, robust and accurate.
Though Voltage and current to pressure transducers are collectively called as electro pneumatic tranducers and the only electronic feature to control output pressure in them is the coil.
Answer:
The maximum length is 3.897×10^-5 mm
Explanation:
Extension = surface energy/elastic modulus
surface energy = 1.05 J/m^2
elastic modulus = 198 GPa = 198×10^9 Pa
Extension = 1.05/198×10^9 = 5.3×10^-12 m
Strain = stress/elastic modulus = 27×10^6/198×10^9 = 1.36×10^-4
Length = extension/strain = 5.3×10^-12/1.36×10^-4 = 3.897×10^-8 m = 3.897×10^-8 × 1000 = 3.897×10^-5 mm