Points for free

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

Rsidential Solar Solution:a. A type of photovoltaic solar panel manufactured in China receives a rating of 250W. The rating proc

aksik [14]

Answer:

a) $\eta = 13.455\%$ , b) $E_{day} = 812.716\,kJ$ , c) $C_{month. total} = 19.505\, USD$ , d) $t = 40.588\,years$

Explanation:

a) The area of the solar panel is:

$A = (20\,ft^{2})\cdot (\frac{0.3048\,m}{1\,ft} )^{2}$

$A = 1.858\,m^{2}$

The energy potential is determined herein:

$\dot E_{o} = (1000\,\frac{W}{m^{2}} )\cdot (1.858\,m^{2})$

$\dot E_{o} = 1858\,W$

The efficiency of the solar panel is:

$\eta = \frac{\dot E}{\dot E_{o}}\times 100\%$

$\eta = \frac{250\,W}{1858\,W}\times 100\%$

$\eta = 13.455\%$

b) The energy generated by the solar panel is presented below:

$E_{day} = (0.135)\cdot (150\,\frac{W}{m^{2}} )\cdot (20\,ft^{2})\cdot \left(\frac{0.3048\,m}{1\,ft} \right)^{2}\cdot (6\,h)\cdot (\frac{3600\,s}{1\,h} )\cdot (\frac{1\,kJ}{1000\,J} )$

$E_{day} = 812.716\,kJ$

c) The energy generated per month and per panel is:

$E_{month} = 30\cdot E_{day}$

$E_{month} = 30 \cdot (812.716\,kJ)\cdot \left(\frac{1\,kWh}{3600\,kJ} \right)$

$E_{month} = 6.773\,kWh$

Monthly energy savings due to the use of 18 panels are:

$C_{month, total} = 18\cdot E_{month}\cdot c$

$C_{month, total} = 18\cdot (6.773\,kWh)\cdot (\frac{0.16\,USD}{1\,kWh} )$

$C_{month. total} = 19.505\, USD$

d) The payback of the solar energy system is:

$t = \frac{9500\,USD}{12\cdot (19.505\,USD)}$

$t = 40.588\,years$

6 0

3 years ago

A(n) 78-hp compressor in a facility that operates at full load for 2500 h a year is powered by an electric motor that has an eff

Jet001 [13]

Answer: $17,206.13

Explanation:

Hi, to answer this question we have to apply the next formula:

Annual electricity cost = (P x 0.746 x Ckwh x h) /η

P = compressor power = 78 hp

0.746 kw/hp= constant (conversion to kw)

Ckwh = Cost per kilowatt hour = $0.11/kWh

h = operating hours per year = 2500 h

η = efficiency = 93% = 0.93 (decimal form)

Replacing with the values given :

C = ( 78 hp x 0.746 kw/hp x 0.11 $/kwh x 2500 h ) / 0.93 = $17,206.13

5 0

3 years ago

Can you determine the critical distance along a flat surface?

Keith_Richards [23]

Explanation:

Consider a fluid of density, ρ moving with a velocity, U over a flat plate of length, L.

Let the Kinematic viscosity of the fluid be ν.

Let the flow over the fluid be laminar for a distance x from the leading edge.

Now this distance is called the critical distance.

Therefore, for a laminar flow, the critical distance can be defined as the distance from the leading edge of the plate where the Reynolds number is equal to 5 x $10^{5}$