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An eddy current separator is to separate aluminum product from an input streamshredded MSW. The feed rate to the separator is 2,

blsea [12.9K]

Answer:

the % recovery of aluminum product is 80.5%

the % purity of the aluminum product is 54.7%

Explanation:

feed rate to separator = 2500 kg/hr

in one hour, there will be 2500 kg/hr x 1 hr = 2500 kg of material is fed into the machine

of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.

After the separation, 256 kg is collected in the product stream.

of this 256 kg, 140 kg is aluminium.

% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material

% recovery of aluminium = 140kg/174kg x 100% = 80.5%

% purity of the aluminium product = mass of aluminium in final product ÷ total mass of product collected in product stream

% purity of the aluminium product = 140kg/256kg

x 100% = 54.7%

8 0

3 years ago

A two-bus power system is interconnected by one transmission line. Bus 1 is a generator bus with specified terminal voltage magn

mixas84 [53]

Answer:

a) | v2 | , β2 ( load, bus voltage at bus 2 )

p1 , q1 ( slack, bus power at bus 1 )

b) q2 , β2

p1 and q1 ( slack, bus power at bus 1 )

Explanation:

Attached below is a schematic representation of the solution

a) Identify the variables in the solution vector assume Bus 2 is a load bus

The specified parameters are ; P2 and q2

while | v2 | and β2 are not specified

given that bus 2 is a load bus, bus 1 is a slack bus with ; | v1 | and β1 been specified while p1 and q1 are not specified

Hence the variables in the solution

= | v2 | , β2 ( load, bus voltage at bus 2 )

p1 , q1 ( slack, bus power at bus 1 )

b) Identify the variables in the solution vector ( assume Bus 2 is a PV bus )

specified at Bus 2 are ; | p2 | , | v2 |

unspecified : q2 , β2

Bus 1 ( still a slack bus )

specified parameter : | v1 | and β1

unspecified : p1 and q1

Hence the variables in the solution

= q2 , β2

p1 and q1 ( slack, bus power at bus 1 )

7 0

3 years ago

A satellite would have a mass of 270 kg on the surface of Mars. Determine the weight of the satellite in pounds if it is in orbi

koban [17]

Answer:

26 lbf

Explanation:

The mass of the satellite is the same regardless of where it is.

The weight however, depends on the acceleration of gravity.

The universal gravitation equation:

g = G * M / d^2

Where

G: universal gravitation constant (6.67*10^-11 m^3/(kg*s))

M: mass of the body causing the gravitational field (mass of Earth = 6*10^24 kg)

d: distance to that body

15000 miles = 24140 km

The distance is to the center of Earth.

Earth radius = 6371 km

Then:

d = 24140 + 6371 = 30511 km

g = 6.67*10^-11 * 6*10^24 / 30511000^2 = 0.43 m/s^2

Then we calculate the weight:

w = m * a

w = 270 * 0.43 = 116 N

116 N is 26 lbf

8 0

3 years ago

Free random points because im bored

tamaranim1 [39]

Answer:

thx for points B)

Explanation:

8 0

3 years ago

Read 2 more answers

A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas

ch4aika [34]

Answer:

gauge pressure is 133 kPa

Explanation:

given data

initial temperature T1 = 27°C = 300 K

gauge pressure = 300 kPa = 300 × 10³ Pa

atmospheric pressure = 1 atm

final temperature T2 = 77°C = 350 K

to find out

final pressure

solution

we know that gauge pressure is = absolute pressure - atmospheric pressure so

P (gauge ) = 300 × 10³ Pa - 1 × $10^{5}$ Pa

P (gauge ) = 2 × $10^{5}$ Pa

so from idea gas equation

$\frac{P1*V1}{T1} = \frac{P2*V2}{T2}$ ................1

so ${P2} = \frac{P1*T2}{T1}$

${P2} = \frac{2*10^5*350}{300}$

P2 = 2.33 × $10^{5}$ Pa

so gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 2.33 × $10^{5}$ - 1.0 × $10^{5}$

gauge pressure = 1.33 × $10^{5}$ Pa

so gauge pressure is 133 kPa

4 0

3 years ago