Fluid fills the space between two parallel plates. The differential equation that describes instantaneous fluid velocity for uns
teady flow with the fluid moving parallel to the walls is
rho. ζu/ζx = μ. ζ^2 u/ζy^2
The lower plate is stationary and the upper plate oscillates in the x-direction with a frequency ω and an amplitude in the plate velocity of U. Use the characteristic dimensions to normalize the differential equation and obtain the dimensionless groups that characterize the flow.
rho. ζu/ζx = μ. ζ^2 u/ζy^2
The lower plate is stationary and the upper plate oscillates in the x-direction with a frequency ω and an amplitude in the plate velocity of U. Use the characteristic dimensions to normalize the differential equation and obtain the dimensionless groups that characterize the flow.
1 answer:
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Answer:
0.2 kcal/mol is the value of for this reaction.
Explanation:
The formula used for is:
where,
= Gibbs free energy for the reaction
= standard Gibbs free energy
R =Universal gas constant
T = temperature
Q = reaction quotient
k = Equilibrium constant
We have :
Reaction quotient of the reaction = Q = 46
Equilibrium constant of reaction = K = 35
Temperature of reaction = T = 25°C = 25 + 273 K = 298 K
R = 1.987 cal/K mol
1 cal = 0.001 kcal
0.2 kcal/mol is the value of for this reaction.
Answer:
drag coefficient for the parachute is Cd = 1.84177
Explanation:
given data
diameter = 8.0 m
average vertical speed = 3 m/s
total weight of load = 200 N
to find out
drag coefficient for the parachute
solution
we will apply here drag coefficient formula that is express as here
drag coefficient Cd = ......................1
here w is total load and A is area and v is speed
and here ρ = 1.22 kg/cu
put here value we get
Cd =
Cd = 1.84177
drag coefficient for the parachute is Cd = 1.84177