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3 years ago

Answer the following questions briefly 1.Write the definitions of Comminution and Calcination

Engineering

1 answer:

VikaD [51]3 years ago

8 0

Answer:

Comminution is defined as the reduction of solid materials from one average particle size to a smaller average particle size through various methods which include crushing, grinding, cutting, vibrating etc,

Calcination is also defined as a thermal treatment process in the absence or limited supply of air or oxygen applied to ores and other solid materials to bring about a thermal(heat) decomposition.

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Wet steam at 15 bar is throttled adiabatically in a steady-flow process to 2 bar. The resulting stream has a temperature of 130°

cricket20 [7]

Answer:

$\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}$

Explanation:

The adiabatic throttling process is modelled after the First Law of Thermodynamics:

$m\cdot (h_{in} - h_{out}) = 0$

$h_{in} = h_{out}$

Properties of water at inlet and outlet are obtained from steam tables:

State 1 - Inlet (Liquid-Vapor Mixture)

$P = 1500\,kPa$

$T = 198.29\,^{\textdegree}C$

$h = 2726.9\,\frac{kJ}{kg}$

$s = 6.3068\,\frac{kJ}{kg\cdot K}$

$x = 0.967$

State 2 - Outlet (Superheated Vapor)

$P = 200\,kPa$

$T = 130\,^{\textdegree}C$

$h = 2726.9\,\frac{kJ}{kg}$

$s = 7.1776\,\frac{kJ}{kg\cdot K}$

The change of entropy of the steam is derived of the Second Law of Thermodynamics:

$\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}$

$\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}$

6 0

3 years ago

A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric

diamong [38]

Answer:

Explanation:

Given

$q_1=11.5\ nC$ charge is placed at $x=0\ cm$

another charge of $q_2=-1.2\ nC$ is at $x=3\ cm$

We know that Electric field due to positive charge is away from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to $q_2$ at some distance r from it

$E_2=\frac{kq_2}{r^2}$

Now Electric Field due to $q_1$ is

$E_1=\frac{kq_1}{(3+r)^2}$

Now $E_1+E_2=0$

$\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0$

$\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}$

$\frac{3+r}{r}=3.095$

thus $r=1.43\ cm$

Thus Electric field is zero at some distance r=1.43 cm right of $q_2$

3 0

3 years ago

Radioactive wastes are temporarily stored in a spherical container, the center of which is buried a distance of 10 m below the e

a_sh-v [17]

Answer:

Outside temperature =88.03°C

Explanation:

Conductivity of air-soil from standard table

K=0.60 W/m-k

To find temperature we need to balance energy

Heat generation=Heat dissipation

Now find the value

We know that for sphere

$q=\dfrac{2\pi DK}{1-\dfrac{D}{4H}}(T_1-T_2)$

Given that q=500 W

$500=\dfrac{2\pi 2\times .6}{1-\dfrac{2}{4\times 10}}(T_1-25)$

By solving that equation we get

$T_2$ =88.03°C

So outside temperature =88.03°C

6 0

3 years ago

The pressure of a gas in a rigid container is 125kpa at 300k, what we be the new pressure if the temperature increases to 900k

kipiarov [429]

Answer:

375 KPa

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 125 KPa

Initial temperature (T₁) = 300 K

Final temperature (T₂) = 900 K

Final pressure (P₂) =?

The new (i.e final) pressure of the gas can be obtained as follow:

P₁/T₁ = P₂/T₂

125 / 300 = P₂ / 900

Cross multiply

300 × P₂ = 125 × 900

300 × P₂ = 112500

Divide both side by 300

P₂ = 112500 / 300

P₂ = 375 KPa

Thus, the new pressure of the gas is 375 KPa

7 0

2 years ago

1.<br> se reception. It's<br> In remote areas, your GPS device<br> a good idea to have a _

marysya [2.9K]

GPS device details are given below.

Explanation:

Even a simple GPS unit has a wide range of settings and features. Because every unit’s operation varies, this article won’t provide step-by-step details. Read the owner's manual to familiarize yourself with it.. If you’d like additional help, you can also sign up for a GPS navigation class at an REI store.

Though steps vary, all GPS receivers do the following basic functions:

Display position: A GPS tells you where you are by displaying your coordinates; it also shows your position on its base map or topo map.

Record tracks: When tracking is turned on, a GPS automatically lays down digital bread crumbs, called “track points,” at regular intervals. You use those later to retrace your steps or to evaluate the path you traveled.

Navigate point-to-point: A GPS directs you by giving you the direction and distance to a location, or “waypoint.” You can pre-mark waypoints by entering their coordinates at home. In the field you can have the unit mark a waypoint at a place you'd like to return to, such as the trailhead or your campsite. A GPS unit provides the bearing and distance “as the crow flies” to a waypoint. Because trails don’t follow a straight line, the bearing changes as you hike. The distance to travel also changes (decreasing, unless you’re heading the wrong direction) as you approach your goal.

Display trip data: This odometer-like function tells you cumulative stats like how far you’ve come and how high you’ve climbed.

GPS and your computer: GPS units come with a powerful software program that lets you manage maps, plan routes, analyze trips and more. Invest the time to learn it and to practice using all of its capabilities.

5 0

3 years ago

Answer the following questions briefly 1.Write the definitions of *Comminution and *Calcination

Answer the following questions briefly 1.Write the definitions of Comminution and Calcination