Answer:
Explanation:
a) for shifting reactions,
Kps = ph2 pco2/pcoph20
=[h2] [co2]/[co] [h2o]
h2 + co2 + h2O + co + c3H8 = 1
it implies that
H2 + 0.09 + H2O + 0.08 + 0.05 = 1
solving the system of equation yields
H2 = 0.5308,
H2O = 0.2942
B) according to Le chatelain's principle for a slightly exothermic reaction, an increase in temperature favors the reverse reaction producing less hydrogen. As a result, concentration of hydrogen in the reformation decreases with an increasing temperature.
c) to calculate the maximum hydrogen yield , both reaction must be complete
C3H8 + 3H2O ⇒ 3CO + 7H2( REFORMING)
CO + H2O ⇒ CO2 + H2 ( SHIFTING)
C3H8 + 6H2O ⇒ 3CO2 + 10 H2 ( OVER ALL)
SO,
Maximum hydrogen yield
= 10mol h2/3 molco2 + 10molh2
= 0.77
⇒ 77%
Answer:
Automatic transmissions should be in Drive and Manual transmissions should be in first gear.
Answer:
Problem 1 (10 points) In the first homework you were instructed to design the mechanical components of an oscillating compact disc reader. Since you did such a good job in your design, the company decided to work with you in their latest Blue-ray readers, as well. However, this time the task is that once the user hits eject button, the motor that spins the disc slows down from 2000 rpm to 300 rpm and at 300 rpm a passive torsional spring-damper mechanism engages to decelerate and stop the disc. Here, your task is to design this spring-damper system such that the disc comes to rest without any oscillations. The rotational inertia of the disc (J) is 2.5 x 10-5kg m² and the torsional spring constant (k) is 5 × 10¬³NM. Calculate the critical damping coefficient cc for the system. choice of the damper, bear in mind that a good engineer stays at least a factor of In your 2 away from the danger zone (i.e., oscillations in this case). Use the Runge Kutta method to simulate the time dependent angular position of the disc, using the value of damping coefficient (c) that calculated. you Figure 1: Blue-ray disc and torsional spring-damper system.
Answer:
17.658 kPa
Explanation:
The hydrostatic pressure of a fluid is the weight of a column of that fluid divided by the base of that column.
Also, the weight of a column is its volume multiplied by it's density and the acceleration of gravity:
Meanwhile, the volume of a column is the area of the base multiplied by the height:
Replacing:
The base cancels out, so:
The pressure depends only on the height of the fluid column, the density of the fluid and the gravity.
If you have two point at different heights (or depths in the case of objects submerged in water) each point will have its own column of fluid exerting pressure on it. Since the density of the fluid and the acceleration of gravity are the same for both points (in the case of hydrostatics density is about constant for all points, it is not the case in the atmosphere), we can write:
We do not know at what depth the man of this problem is, but it doesn't matter, because we know the difference in height of the two points of interes (h1 - h2) = 1.8 m. So:
