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Explanation:
Previous concepts
Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:
Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =
MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:
MO = H˙ O
Principle of Angular Impulse and Momentum
The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:
Solution to the problem
For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is ".
If we analyze the staritning point we see that the initial velocity can be founded like this:
And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:
And if we integrate the left part and we simplify the right part we have
And if we solve for we got:
Answer:
V = 0.30787 m³/s
m = 2.6963 kg/s
v2 = 0.3705 m³/s
v2 = 6.017 m/s
Explanation:
given data
diameter = 28 cm
steadily =200 kPa
temperature = 20°C
velocity = 5 m/s
solution
we know mass flow rate is
m = ρ A v
floe rate V = Av
m = ρ V
flow rate = V =
V = Av =
V =
V = 0.30787 m³/s
and
mass flow rate of the refrigerant is
m = ρ A v
m = ρ V
m = =
m = 2.6963 kg/s
and
velocity and volume flow rate at exit
velocity = mass × v
v2 = 2.6963 × 0.13741 = 0.3705 m³/s
and
v2 = A2×v2
v2 =
v2 =
v2 = 6.017 m/s
Answer:
Time of concentration, ⇒ 1280 min
Peak runoff rate, ⇒4.185 ff³/s
Explanation:
See detailed explanation
