Answer:
sorry if wrong
Explanation:
One sheave means that you are using a single drum winder. They are the worst! Double drum winders control easier, brake better and are much more efficient. They save time ( two skips or cages) and can be clutched to perform faster shift transport. A single drum is slow, unbalanced and can be a nightmare if it trips out during hoisting. If the brake system is not perfect it can be a real hairy experience. For a runaway single drum, there is no counterbalance effect. It always runs to destruction. With a double drum, the driver still has a chance to control the winder to a certain extent and he has two sets of brakes to rely on. A single sheave could also mean a shaft with a single compartment. No second means of escape unless there are ladders or stairways. Not a very healthy situation.
Those are just a few points. I am sure much more can be said in favor of a double drum winder and two or more sheaves in the headgear. Most of the shafts I have worked at have multiple winders and up to ten compartments. They all have a small single drum service winder for emergencies and moves of personnel during shift times. They are referred to as the Mary - Annes. Apparently, the name originated in the U.K. where an aristocratic mine owner named the first such winder after his mistress.
57.5 m/s
I did 2.3/0.04
I’m not sure if it’s correct though
Answer:
COP = 3.828
W' = 39.18 Kw
Explanation:
From the table A-11 i attached, we can find the entropy for the state 1 at -20°C.
h1 = 238.43 KJ/Kg
s1 = 0.94575 KJ/Kg.K
From table A-12 attached we can do the same for states 3 and 4 but just enthalpy at 800 KPa.
h3 = h4 = hf = 95.47 KJ/Kg
For state 2, we can calculate the enthalpy from table A-13 attached using interpolation at 800 KPa and the condition s2 = s1. We have;
h2 = 275.75 KJ/Kg
The power would be determined from the energy balance in state 1-2 where the mass flow rate will be expressed through the energy balance in state 4-1.
W' = m'(h2 - h1)
W' = Q'_L((h2 - h1)/(h1 - h4))
Where Q'_L = 150 kW
Plugging in the relevant values, we have;
W' = 150((275.75 - 238.43)/(238.43 - 95.47))
W' = 39.18 Kw
Formula foe COP is;
COP = Q'_L/W'
COP = 150/39.18
COP = 3.828
Answer:
The original length of the specimen is found to be 76.093 mm.
Explanation:
From the conservation of mass principal, we know that the volume of the specimen must remain constant. Therefore, comparing the volumes of both initial and final state as state 1 and state 2:
Initial Volume = Final Volume
πd1²L1/4 = πd2²L2/4
d1²L1 = d2²L2
L1 = d2²L2/d1²
where,
d1 = initial diameter = 19.636 mm
d2 = final diameter = 19.661 mm
L1 = Initial Length = Original Length = ?
L2 = Final Length = 75.9 mm
Therefore, using values:
L1 = (19.661 mm)²(75.9 mm)/(19.636 mm)²
<u>L1 = 76.093 mm</u>
Answer:
I think that plane's name is
KLM ....Because you can see the mysterious Leter in the plane's
Forward and the back of the plane
