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2 years ago

What are the two types of furnaces used in steel production?

Engineering

1 answer:

Kobotan [32]2 years ago

3 0

Explanation:

The two types of furnaces used in steel production are:

<u>Basic oxygen furnace </u>

In basic oxygen furnace, iron is combined with the varying amounts of the steel scrap and also small amounts of the flux in the Blast Furnace. Lance is introduced in vessel and blows about 99% of the pure oxygen causing rise in temperature to about 1700°C. This temperature melts scrap and the impurities are oxidized and results in the liquid steel.

<u>Electric arc furnace</u>

Electric arc furnace reuses existing steel. Furnace is charged with the steel scrap. It operates on basis of electrical charge between the two electrodes providing heat for process. Power is supplied through electrodes placed in furnace, which produce arc of the electricity through scrap steel which raises temperature to about 1600˚C. This temperature melts scrap and the impurities can be removed through use of the fluxes and results in the liquid steel.

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An amplifier with 40 dB of small-signal, open-circuit voltage gain, an input resistance of 1 MO, and an output resistance of 100

Vanyuwa [196]

convert 40db to standard gain

AL=10^40/20=100

calculate total voltage gain

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=83.33

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Pi=Vi^2/Ri Po=Vo^2/RL

power gain= Po/Pi

=13.90×10^6

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2 years ago

In order to protect yourself if you have a dispute with another drivers insurance company you should:

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Answer:

C. Get the names and addresses of witness to the crash

Explanation:

The best approach is to let your insurance company handle the dispute. Since that is not an option here, the best thing you can do is make sure you know who the witnesses are, so your insurance company can call upon them as needed.

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2 years ago

Write any four factors that may hamper the unity in diversity feature of Nepal and the Nepalese .

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Answer:

Explanation:

different religions

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2 years ago

Consider a 2D airfoil with a zero lift angle of attack of −4 degrees. The stall angle for this airfoil is 15 degrees, where the

zzz [600]

Answer:

Explanation:

Recquired Data

lift coefficient = 2.0

Distance = 20m

Area = 60m²

Force = 105 N

Density = 1kg/m³

we are recquired to find Maximum panding speed

The lift coefficient CL is defined by

CL≡L/qs=L/1/2рμ²S = 2L/рμ²S

where L, is the lift force, S, is the relevant surface area and q, is the fluid dynamic pressure, in turn linked to the fluid density rho ,, and to the flow speed u,

substituting the figure respectively

2.0 = 2 X 105 / 1 X μ² X 60

μ² = 2L /CL X P X S

μ² = 2 X 150 / 2.0 X 1 X 60

μ² = 300 / 120

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2 years ago

Under normal operating conditions, the electric motor exerts a torque of 2.8 kN-m.on shaft AB. Knowing that each shaft is solid,

77julia77 [94]

Answer:

Explanation:

The image attached to the question is shown in the first diagram below.

From the diagram given ; we can deduce a free body diagram which will aid us in solving the question.

IF we take a look at the second diagram attached below ; we will have a clear understanding of what the free body diagram of the system looks like :

From the diagram; we can determine the length of BC by using pyhtagoras theorem;

SO;

$L_{BC}^2 = L_{AB}^2 + L_{AC}^2$

$L_{BC}^2 = (3.5+2.5)^2+ 4^2$

$L_{BC}= \sqrt{(6)^2+ 4^2}$

$L_{BC}= \sqrt{36+ 16}$

$L_{BC}= \sqrt{52}$

$L_{BC}= 7.2111 \ m$

The cross -sectional of the cable is calculated by the formula :

$A = \dfrac{\pi}{4}d^2$

where d = 4mm

$A = \dfrac{\pi}{4}(4 \ mm * \dfrac{1 \ m}{1000 \ mm})^2$

A = 1.26 × 10⁻⁵ m²

However, looking at the maximum deflection in length $\delta$ ; we can calculate for the force $F_{BC$ by using the formula:

$\delta = \dfrac{F_{BC}L_{BC}}{AE}$

$F_{BC} = \dfrac{ AE \ \delta}{L_{BC}}$

where ;

E = modulus elasticity

$L_{BC}$ = length of the cable

Replacing 1.26 × 10⁻⁵ m² for A; 200 × 10⁹ Pa for E ; 7.2111 m for $L_{BC}$ and 0.006 m for $\delta$ ; we have:

$F_{BC} = \dfrac{1.26*10^{-5}*200*10^9*0.006}{7.2111}$

$F_{BC} = 2096.76 \ N \\ \\ F_{BC} = 2.09676 \ kN$ ---- (1)

Similarly; we can determine the force $F_{BC}$ using the allowable maximum stress; we have the following relation,

$\sigma = \dfrac{F_{BC}}{A}$

${F_{BC}}= {A}*\sigma$

where;

$\sigma =$ maximum allowable stress

Replacing 190 × 10⁶ Pa for $\sigma$ ; we have :

${F_{BC}}= 1.26*10^{-5} * 190*10^{6} \\ \\ {F_{BC}}=2394 \ N \\ \\ {F_{BC}}= 2.394 \ kN$ ------ (2)

Comparing (1) and (2)

The magnitude of the force $F_{BC} = 2.09676 \ kN$ since the elongation of the cable should not exceed 6mm

Finally applying the moment equilibrium condition about point A

$\sum M_A = 0$

$3.5 P - (6) ( \dfrac{4}{7.2111}F_{BC}) = 0$

$3.5 P - 3.328 F_{BC} = 0$

$3.5 P = 3.328 F_{BC}$

$3.5 P = 3.328 *2.09676 \ kN$

$P =\dfrac{ 3.328 *2.09676 \ kN}{3.5 }$

P = 1.9937 kN

Hence; the maximum load P that can be applied is 1.9937 kN

4 0

2 years ago