Answer is an increase in pressure will cause an decrease in the pressure
Answer:
Those products are generally called Work in Process WIP
Explanation:
Work in process (WIP), or work in progress (WIP), goods in process, or in-process inventory in a manufacturing industry/company refer to the company's partially finished goods waiting for completion and eventual sale or the value of these items.
These items are either just being produced or require further processing (like purification, separation, packaging or handling) in a queue or a buffer storage.
Explanation:
Temperature range → 0 to 80'c
respective voltage output → 0.2v to 0.5v
required temperature range 20'c to 40'c
Where T = 20'c respective voltage
Therefore, Sensor output changes from 0.275v to 0.35volts for the ADC the required i/p should cover the dynamic range of ADC (ie - 0v to 3v)
so we have to design a circuit which transfers input voltage 0.275volts - 0.35v to 0 - 3v
Therefore, the formula for the circuit will be
The simplest circuit will be a op-amp
NOTE: Refer the figure attached
Vs is sensor output
Vr is the reference volt, Vr = 0.275v
choose R2, R1 such that it will maintain required ratio
The output Vo can be connected to voltage buffer if you required better isolation.
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!!!
