Answer:
C
Explanation:
the total resistance is equal to the total potential difference divided by the Current
Answer:
The required pumping head is 1344.55 m and the pumping power is 236.96 kW
Explanation:
The energy equation is equal to:

For the pipe 1, the flow velocity is:

Q = 18 L/s = 0.018 m³/s
D = 6 cm = 0.06 m

The Reynold´s number is:


Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941
The head of pipe 1 is:

For the pipe 2, the flow velocity is:

The Reynold´s number is:


The head of pipe 1 is:

The total head is:
hi = 1326.18 + 21.3 = 1347.48 m
The required pump head is:

The required pumping power is:

Answer:

Explanation:
First, we will find actual properties at given inlet and outlet states by the use of steam tables:
AT INLET:
At 4MPa and 350°C, from the superheated table:
h₁ = 3093.3 KJ/kg
s₁ = 6.5843 KJ/kg.K
AT OUTLET:
At P₂ = 125 KPa and steam is saturated in vapor state:
h₂ =
= 2684.9 KJ/kg
Now, for the isentropic enthalpy, we have:
P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K
Since s₂ is less than
and greater than
at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

Now, we will find
(enthalpy at the outlet for the isentropic process):

Now, the isentropic efficiency of the turbine can be given as follows:

Answer:
1. High friction
2. High extrusion temperature
Explanation:
Surface cracking on extruded products are defects or breakage on the surface of the extruded parts. Such cracks are inter granular.
Surface cracking defects arises from very high work piece temperature that develops cracks on the surface of the work piece. Surface cracking appears when the extrusion speed is very high, that results in high strain rates and generates heat.
Other factors include very high friction that contributes to surface cracking an d chilling of the surface of high temperature billets.
Answer:

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:

](https://tex.z-dn.net/?f=0%2B%5Csum%20%5Cint_%7B0%7D%5E%7B4%7D%2020t%20%280.15m%29%20dt%20%3D0.46875%20%5Comega%20%2B%2030kg%5B%5Comega%280.15m%29%5D%280.15m%29)
And if we integrate the left part and we simplify the right part we have

And if we solve for
we got:
