I need the answer pls

Advantage of a sheave wheel in a shaft headgear<br>

NikAS [45]

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.

5 0

3 years ago

The coefficient of performance of a reversible refrigeration cycle is always (a) greater than, (b) less than, (c) equal to the c

bezimeni [28]

Answer:

Option B is correct

Explanation:

Let

Higher temperature = $T_H$

Lower temperature = $T_L$

We know that COP is given by

$COP = \frac{T_L}{T_H-T_L}$

We see that COP is depends only on the temperature difference & Temperature difference is maximum for the Carnot cycle.

Therefore the COP of reversible refrigeration cycle is always less then the COP of an irreversible refrigeration cycle when each operates between the same two thermal reservoirs.

Therefore option B is correct

5 0

3 years ago

A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the

Olegator [25]

Answer:

119.35 mm

Explanation:

Given:

Inside diameter, d = 100 mm

Tensile load, P = 400 kN

Stress = 120 MPa

let the outside diameter be 'D'

Now,

Stress is given as:

stress = Load × Area

also,

Area of hollow pipe = $\frac{\pi}{4}(D^2-d^2)$

or

Area of hollow pipe = $\frac{\pi}{4}(D^2-100^2)$

thus,

400 × 10³ N = 120 × $\frac{\pi}{4}(D^2-100^2)$

or

D² = tex]\frac{400\times10^3+30\pi\times10^4}{30\pi}[/tex]

or

D = 119.35 mm

7 0

3 years ago

Read 2 more answers

Wind blows on the side of a fully enclosed hospital located on open flat terrain where V= 120 mi/h. Determine the external press

ss7ja [257]

Answer:

The external pressure is p = -21.9 psf or p = -8.85 psf

Explanation:

Given :

Velocity of wind, v = 120 mi / hr

$k_d = k_c =1$ (wind direction factor)

$k _{zt} = 1$ = topographical factor (for flat terrain)

$q_n$ = velocity pressure at height h

$\therefore q_n = 0.00256 k_z k_{zt} k_d v^2$

$q_n = 0.00256 \times k_z (1)(1)(120)^2$

$= 36.86 k_z$

But for height h = 30 ft, $k_z$ = 0.98 (from table)

$\therefore q_n = 36.86 \times 0.98$

= 36.16

Now, $\frac{L}{B}= \frac{200}{200} =1$ , so $C_p=-0.5$ (from table)

$p = q(G)(C_p)-q_n(GC_{pi})$

where, p = external pressure

G = 0.85 = gust factor (for typical rigid building)

$GC_{pi} = \pm 0.18$ (internal pressure co efficient)

Therefore putting the values,

$p = (36.13)(0.85)(-0.5)-(36.13)(\pm 0.18)$

p = -21.9 psf or p = -8.85 psf

4 0

4 years ago

List three main commodity areas that drive agriculture

Alborosie

Common agricultural commodities include dairy products, wheat, and coffee. You can invest in and “trade” these products virtually, without running a farm and purchasing and storing the items yourself.

3 0

1 year ago

I need the answer pls​

I need the answer pls