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1 year ago

Advantage and disadvantages of Bessemer process of steel making

Engineering

1 answer:

Snowcat [4.5K]1 year ago

7 0

Advantages:

- allowed steel to be produced without fuel

- using the impurities of the iron to create the necessary heat

- reduced the costs of steel production

Disadvantages:

- could convert only a pig iron low in phosphorus and sulfur

- these elements could have been removed by adding a basic flux such as lime, but the basic sl*g produced would have degraded the acidic refractory lining of Bessemer's converter.

au revoir mon amour <3

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Determine the period of each of the following discrete-time signals (if a signal is not periodic, denote its period by infinity)

sergiy2304 [10]

Answer:

a) it is periodic

N = (20/3)k = 20 { for K =3}

b) it is Non-Periodic.

N = ∞

c) x(n) is periodic

N = LCM ( 5, 20 )

Explanation:

We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.

then the period of the signal is given as

N = ( 2π/w₀)K

k is least integer for which N is also integer

Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2

now

a) cos(2π(0.15)n)

w₀ = 2π(0.15)

Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3

so, it is periodic

N = (20/3)k = 20 { for K =3}

b) cos(2n);

w₀ = 2

Now, 2π/w₀ = 2π/2) = π

so, it is Non-Periodic.

N = ∞

c) cos(π0.3n) + cos(π0.4n)

x(n) = x1(n) + x2(n)

x1(n) = cos(π0.3n)

x2(n) = cos(π0.4n)

w₀ = π0.3

2π/w₀ = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3

∴ N1 = 20

AND

w₀ = π0.4

2π/w₀ = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5

∴ N² = 5

so, x(n) is periodic

N = LCM ( 5, 20 )

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2 years ago

In the flash distillation of salt water, the salt is totally nonvolatile (this is the equilibrium statement). Show a McCabe-Thie

Gala2k [10]

Answer:

attached below

Explanation:

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2 years ago

A spherical tank is being designed to hold 10 moles of carbon dioxide gas at an absolute pressure of 5 bar and a temperature of

lesya692 [45]

Answer:

r=0.228m

Explanation:

The equation that defines the states of a gas according to its thermodynamic properties is given by the general equation of ideal gases

PV=nRT

where

P=pressure =5bar=500.000Pa

V=volume

n=moles=10

R = universal constant for ideal gases = 8.31J / (K.mol)

T=temperature=80F=299.8K

solvig For V

V=(nRT)/P

$V=(\frac{(10)(8.31)(299.8)}{500000} )\\V=0.0498m^3$

we know that the volume of a sphere is

$V=\frac{4\pi r^3}{3} \\$

solving for r

$r=\sqrt[3]{ \frac{3 V}{4\pi } }$

solving

$r=\sqrt[3]{ \frac{3 (0.049)}{4\pi } }\\r=0.228m$

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3 years ago

Air at 25 C and 1 atm is flowing over a long flat plate with a velocity of 8 m/s. Determine the distance from the leading edge o

DIA [1.3K]

Answer:

Explanation:

Given data

Temprature of air= $25^{\circ}$

pressure =1 atm

velocity(V)=8m/s

From the table for air at $25^{\circ}$ and 1 atm

kinematic viscosity $\left ( \nu\right )=1.562\times 10^{-5} m^{2}/s$

Reynolds number for turbulent flow $=5\times 10^{5}$

and Re.no.= $\frac{V \times X}{\nu }$

therefore length where turbulent flows start is

X= $\frac{\left ( Re.no.\right )\times \nu }{V}$

X=0.976 m

Thickness of boundary layer is given by

$\delta _x=\frac{5X}{\sqrt{Re_x}}$

$\delta _x=\frac{5\times 0.976}{\sqrt{5\times 10^5}}$

$\delta _x=6.901mm$

for water

kinematic viscosity $\left ( \nu\right )=8.91\times 10^{-7} m^{2}/s$

Reynolds number for turbulent flow $=5\times 10^{5}$

and Re.no.= $\frac{V \times X}{\nu }$

therefore length where turbulent flows start is

X= $\frac{\left ( Re.no.\right )\times \nu }{V}$

X=0.055m

Thickness of boundary layer is given by

$\delta _x=\frac{5X}{\sqrt{Re_x}}$

$\delta _x=\frac{5\times 0.055}{\sqrt{5\times 10^5}}$

$\delta _x=0.3889mm$

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2 years ago

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A gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m^3, undergoes a constant-pressure expansion at

NNADVOKAT [17]

Answer:

Q = E + W = 0.25 + 0.4 = 0.29KJ

W = 0.4KJ

b) 2000KJ

Explanation:

The concept applied here is the first law of thermodynamics ,i.e the equation of the first law.

mathematically from first law,

dE = dQ - dW

where E is internal energy

Where energy may be transfereed as eitherf work (W) or heat (Q)

Work on the other hand can be at constant volume and pressure.

at constant pressure, W = Integral (pdV), with final volume(Vf) as the upper limit and initial volume(Vi) as the lower limit

Work = p(Vf -Vi)

attempting the first question, Vi = 0.1metre cube, Vf = 0.12metre cube, p = 2bar, p(atm) = 1bar, E = U = 0.25KJ

from W = p(Vf -Vi) = 2 x 100000 ( 0.12 - 0.1) = 400N/m = 0.4KJ

To get the heat transfer, from dE = dQ - dW

Q = E + W = 0.25 + 0.4 = 0.29KJ

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W = 100000 ( 0.12 - 0.1) = 2000KJ

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3 years ago

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