Answer:
The velocity at R/2 (midway between the wall surface and the centerline) is given by (3/4)(Vmax) provided that Vmax is the maximum velocity in the tube.
Explanation:
Starting from the shell momentum balance equation, it can be proved that the velocity profile for fully developedblaminar low in a circular pipe of internal radius R and a radial axis starting from the centre of the pipe at r=0 to r=R is given as
v = (ΔPR²/4μL) [1 - (r²/R²)]
where v = fluid velocity at any point in the radial direction
ΔP = Pressure drop across the pipe
μ = fluid viscosity
L = pipe length
But the maximum velocity of the fluid occurs at the middle of the pipe when r=0
Hence, maximum veloxity is
v(max) = (ΔPR²/4μL)
So, velocity at any point in the radial direction is
v = v(max) [1 - (r²/R²)]
At the point r = (R/2)
r² = (R²/4)
(r²/R²) = r² ÷ R² = (R²/4) ÷ (R²) = (1/4)
So,
1 - (r²/R²) = 1 - (1/4) = (3/4)
Hence, v at r = (R/2) is given as
v = v(max) × (3/4)
Hope this Helps!!!
Answer:
a. The final temperature is 69.8°C
b. The final pressure is 2.67bar
c. The amount of entropy produced is 0.38568kJ/K
Explanation:
Pls, check the attached files for detail step by step explanation as typing the solution here can not be explicit.
Answer:
0.05 mg / gallon
Explanation:
mass of chemecila coming in per minute = 50*10 = 500 mg/min
at a time t min , M = mass of chemical = 500*t mg
conecntartion of chemecal = 500t/10000 = 0.05 mg / gallon
Answer:D
Explanation:
Take longer time to retrieve than long term memory, involves transient modifications in the function of pre existing synapses, such as channel modifications.
Answer:
it is not possible to place the wires in the condui
Explanation:
given data
total area = 2.04 square inches
wires total area = 0.93 square inches
maximum fill conduit = 40%
to find out
Can it is possible place wire in conduit conduit
solution
we know maximum fill is 40%
so here first we get total area of conduit that will be
total area of conduit = 40% × 2.04
total area of conduit = 0.816 square inches
but this area is less than required area of wire that is 0.93 square inches
so we can say it is not possible to place the wires in the conduit
