To solve this problem we will use the Froude number that relates the Forces of Inertia with the Forces of Gravity. There will be jump in the downstream only if Froude Number (Fr) is greater than 1 at upstream. Our values are given as,
Then the velocity would be:
The number of Froude is given as,
Where,
V = Velocity
g = Gravity
D = Diameter
Replacing we have that
There will be no Jump, correct answer is B.
Answer:
Assumption:
1. The kinetic and potential energy changes are negligible
2. The cylinder is well insulated and thus heat transfer is negligible.
3. The thermal energy stored in the cylinder itself is negligible.
4. The process is stated to be reversible
Analysis:
a. This is reversible adiabatic(i.e isentropic) process and thus
From the refrigerant table A11-A13
sat vapor
m=
b.) We take the content of the cylinder as the sysytem.
This is a closed system since no mass leaves or enters.
Hence, the energy balance for adiabatic closed system can be expressed as:
ΔE
ΔU
)
workdone during the isentropic process
=5.8491(246.82-219.9)
=5.8491(26.91)
=157.3993
=157.4kJ
The classical motion for an oscillator that starts from rest at location x₀ is
x(t) = x₀ cos(ωt)
The probability that the particle is at a particular x at a particular time t
is given by ρ(x, t) = δ(x − x(t)), and we can perform the temporal average
to get the spatial density. Our natural time scale for the averaging is a half
cycle, take t = 0 → π/ ω
Thus,
ρ =
Limit is 0 to π/ω
We perform the change of variables to allow access to the δ, let y = x₀ cos(ωt) so that
ρ(x) =
Limit is x₀ to -x₀
Limit is -x₀ to x₀
This has as expected. Here the limit is -x₀ to x₀
The expectation value is 0 when the ρ(x) is symmetric, x ρ(x) is asymmetric and the limits of integration are asymmetric.
