Answer:
<em>the % recovery of aluminum product is 80.5%</em>
<em>the % purity of the aluminum product is 54.7%</em>
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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% = <em>80.5%</em>
% 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% = <em>54.7%</em>
Answer:
a)5.28 Å , b)3.73 Å , c)3.048 Å
Explanation:
the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.
Therefore, a particular unit cell consist only 1/8th part of an atom.
The lattice constant of a simple cubic primitive cell is 5.28 Å
We know formula of distance,
d = 
a)(100)
a=5.28 Å
Distance = 
=5.28 Å
b)(110)
Distance = 
= 3.73 Å
c)(111)
Distance= 
= 3.048 Å
Answer:
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A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined
Assumptions :
1. Steady operating conditions exist.
2. Kinetic and potential energy changes are negligible.
Properties: The specific heat of geothermal water ( 
[) is taken to be 4.18 kJ/kg.ºC.
Analysis (a) We need properties of isobutane, we can obtain the properties from EES.
a. Turbine
P
=
b. Pump
c. 
Answer:
526.5 KN
Explanation:
The total head loss in a pipe is a sum of pressure head, kinetic energy head and potential energy head.
But the pipe is assumed to be horizontal and the velocity through the pipe is constant, Hence the head loss is just pressure head.
h = (P₁/ρg) - (P₂/ρg) = (P₁ - P₂)/ρg
where ρ = density of the fluid and g = acceleration due to gravity
h = ΔP/ρg
ΔP = ρgh = 1000 × 9.8 × 7.6 = 74480 Pa
Drag force over the length of the pipe = Dynamic pressure drop over the length of the pipe × Area of the pipe that the fluid is in contact with
Dynamic pressure drop over the length of the pipe = ΔP = 74480 Pa
Area of the pipe that the fluid is in contact with = 2πrL = 2π × (0.075/2) × 30 = 7.069 m²
Drag Force = 74480 × 7.069 = 526468.1 N = 526.5 KN
