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A blast furnace is a type of metallurgical furnace used for smelting to produce metals, typically iron and general steel, http://www.smallcapinvestor.com/smallcapinsights/china/2008-06-19-check_on_china_general_steel_holdings_inc. In a blast furnace, fuel and ore are continuously introduced and combined through the top of the furnace, while air (sometimes with oxygen enrichment) is forced into the bottom of the chamber, so that the chemical reactions occur throughout the furnace as the material flows downward. The end products are normally molten metal and slag phases tapped from the bottom, and flue gases leaving from the top of the furnace.

The Scotch tuyere (A tuyere is a tube, nozzle or pipe through which air is blown into a furnace or hearth.) is of cast iron, with a wrought-iron coil of pipe, for the cooling water, cast in the walls. The Lancashire tuyere is a hollow truncated cone with double walls, the cooling water flowing between them. An open spray tuyere is somewhat similar to the above, but the rear end is open (open tuyere), and the walls are kept cool by jets of water from perforated pipes inside. In the spreader tuyere, which is of a somewhat similar type, the water is distributed over a sheet to cool the upper part of the shell, a jet of water cooling the nose, and the water running back over the bottom to the outlet.

A vacuum or exhaust tuyere is one from which the water is sucked out, instead of being forced through, to lessen the leakage in the furnace if the tuyere breaks. The blast from the stoves is led into a large pipe (bustle pipe) surrounding the furnace and above the tuyeres with which it is connected. At a point near the top of the hearth is a hole for tap- ping the slag (slag notch, cinder notch, flushing hole, monkey), and lower down and to one side is the iron notch or tapping hole for tapping out the molten iron.

In modern furnaces the hearth is entirely within the furnace, and is called a closed hearth, the contents being tapped through a hole in the wall. This arrangement was first introduced by Lurmann, and was called a closed front (Lurmann front). "It was formerly the custom to have an open fore part (open front). In front of the furnace there was an arched-over opening extending from the furnace bottom to a little above the level of the tuyeres; the sides and roof of this opening formed a cavity known as the fore hearth.

A section of one of the fire-brick, called the dam, was Duquesne blast furnaces carried to the tuyere level; it formed the back of the fore hearth, and was supported by a stone (dam stone) or cast-iron dam plate; in the dam plate was a vertical slit which was stopped with loam, and which allowed of a tapping hole being made at any convenient level, while the excess of cinder ran off through a semicircular cinder notch. The arch above the dam was called the tymp; it was kept in position by a tymp plate (tymp stone) of cast iron, and generally cooled by running water.

In Lurmann's closed front arrangement the slag was tapped through a water-cooled tuyere (scoria block) situated in a water-cooled cast-iron plate (scoria plate). At the top of the furnace is the charging arrangement consisting usually of two bells, a large one below, and a smaller one above, each fitting in a hopper (bell and hopper, cup and cone), and so arranged that only one is lowered at a time, thus preventing the escape of gas. Various forms of ore distributing devices have been designed to insure the even disposition of the charge within the furnace.

Charging - The ore is kept in huge piles (stock pile) in a yard, and is handled by large traveling cranes (Ore Bridge, gantry crane). It is loaded into bins from which it is run into small hopper-bottom cars provided with scales for weighing, and from these is dumped into the buckets or boxes (skip) which are hoisted to the top of the furnace up an inclined track (skip bridge), and dumped automatically on the upper bell.

The limestone is similarly handled; the coke is measured by volume (occasionally weighed), the contents of the skip representing a definite weight. This method is known as skip charging. The older method, now seldom employed for new furnaces, is to weigh the materials in wheelbarrows which are raised to the top of the furnace by an elevator (hoist, lift), and dumped around the bell by hand (barrow charging). One complete unit or charge of ore, coke, and limestone is called a round.

Modern blast furnaces may have more than 30 tuyeres, through which the hot blast is forced into the furnace. They are usually constructed from copper and cooled with water to withstand the very high, extreme temperatures necessary to construct general steel and other products.

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Ore Car Load

PHYSICS: I need HELP with physics, someone please help a.s.a.p. Greatly appreciated thanks!?

A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected with a winch. The shaft is inclined at 30.0 degrees above the horizontal. The car accelerates uniformly to a speed of 2.20 m/s in 12.0 seconds and then continues at a constant speed. (a) What power must the winch motor provide when the car is moving at a constant speed? (b) What maximum power must the motor provide? (c) What total energy transfers out of the motor by work by the time the car moves off the end of the track, which is of lenght 1,250 m?

Power is the rate at which work is done. That is:

Power = work / time

Work is force times distance:

work = force × distance

Therefore:

Power = force × distance / time

Since distance/time = speed, you could also say:

Power = force × (distance/time)
Power = force × speed

> (a) What power must the winch motor provide when the car is moving at a constant speed?

Power = force × speed

They tell you that the speed is 2.20m/s, so you have that part.

The force equals how hard the winch is pulling. Since the car is no longer accelerating, the NET force on the car must be zero (zero acceleration ALWAYS means zero net force). That means the winch is pulling upslope exactly as hard as gravity is pulling downlsope; making the two opposing forces exactly cancel out.

Gravity always pulls downslope with a strength of (mg)sinθ (in this case, m=950kg and θ=30 degrees).

That means (since no acceleration), the winch's force exactly matches that in the opposite direction. Therefore:

Power = (force of winch) × speed
Power = (mg)sinθ × 2.20m/s
[During constant-speed phase]

> (b) What maximum power must the motor provide?

Power = force × speed

The maximum power happens when the force is maximum and the speed is maximum.

The maximum force (by the winch) happens while the car is _accelerating_. When the car is NOT accelerating, the winch's force is EQUAL to gravity; but in order to make the car _accelerate_, the winch must pull HARDER than gravity.

We can use the formula Fnet=ma, to determine how hard the winch is pulling during the acceleration.

The net force is:
Fnet = (F_winch − F_gravity)

Remember that F_gravity (downslope) is always: (mg)sinθ; so:

Fnet = (F_winch − (mg)sinθ)

Set this equal to "ma":

Fnet = ma
(F_winch − (mg)sinθ) = ma

Now figure out how much "a" is. Acceleration equals change in velocity divided by time. They tell you that the car took 12 seconds to go from zero to 2.20m/s; so:

a = Δv/t
a = (2.20m/s)/(12s)
a = 0.1833 m/s²

Combine with previous equation to get:

(F_winch − (mg)sinθ) = m(0.1833 m/s²)

Solve for F_winch:

F_winch = m(0.1833 m/s²) + (mg)sinθ

This is the maximum force provided by the winch.

Then:

(max power) = (max force) × (max speed)
(max power) = (m(0.1833 m/s²) + (mg)sinθ) × (max speed)
(max power) = (m(0.1833 m/s²) + (mg)sinθ) × (2.20m/s)

> (c) What total energy transfers out of the motor by work by the time the car moves off the end of the track, which is of lenght 1,250 m?

One way to calculate this is to figure out the work done during both phases (the acceleration phase and the constant-speed phase) and then add those two work amounts together. That would require you to figure out the distance traveled during each phase too.

But that's too hard. The easier way, is to notice how high the car got lifted against gravity. That height is:

h = (total length of slope) × sinθ

Whenever you lift something by a height of "h", and providing there's no friction, the amount of work done on it is:

total work = (weight) × (h) = mgh

So:

total work = mg × (total length of slope) × sinθ