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:
7.15
Explanation:
Firstly, the COP of such heat pump must be measured that is,

Therefore, the temperature relationship, 
Then, we should apply the values in the COP.


The number of heat rejected by the heat pump must then be calculated.


We must then calculate the refrigerant mass flow rate.



The
value is 145.27 and therefore the hot reservoir temperature is 64° C.
The pressure at 64 ° C is thus 1849.36 kPa by interpolation.
And, the lowest reservoir temperature must be calculated.



the lowest reservoir temperature = 258.703 kpa
So, the pressure ratio should be = 7.15
Answer:
The pressure drop is 269.7N/m^2
Explanation:
∆P = ∆h × rho × g
∆h = 3.2cm = 3.2/100 = 0.032m, rho = 860kg/m^3, g = 9.8m/s^2
∆P = 0.032×860×9.8 = 269.7N/m^2
Answer:
The rate of entropy change of the air is -0.10067kW/K
Explanation:
We'll assume the following
1. It is a steady-flow process;
2. The changes in the kinetic energy and the potential energy are negligible;
3. Lastly, the air is an ideal gas
Energy balance will be required to calculate heat loss;
mh1 + W = mh2 + Q where W = Q.
Also note that the rate of entropy change of the air is calculated by calculating the rate of heat transfer and temperature of the air, as follows;
Rate of Entropy Change = -Q/T
Where Q = 30Kw
T = Temperature of air = 25°C = 298K
Rate = -30/298
Rate = -0.100671140939597 KW/K
Rate = -0.10067kW/K
Hence, the rate of entropy change of the air is -0.10067kW/K