Answer:
<em>the % recovery of aluminum product is 80.5%</em>
<em>the % purity of the aluminum product is 54.7%</em>
<em></em>
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:
Not knowing the units the tolerance is 0.02. I would presume mm but hopefully your question has more detail.
Explanation:
The tolerance is the portion after the main dimension (+/- 0.02). In our case we have bilateral tolerance since there is tolerance in both directions (positive and negative). If you were building a part the acceptable range would be 2.98 to 3.02 based on the tolerance provided.
Answer:
88750 N
Explanation:
given data:
plastic deformation σy=266 MPa=266*10^6 N/m^2
cross-sectional area Ao=333 mm^2=333*10^-6 m^2
solution:
To determine the maximum load that can be applied without
plastic deformation (Fy).
Fy=σy*Ao
=88750 N
Question
Determine the average water exit velocity
Answer:
53.05 m/s
Explanation:
Given information
Volume flow rate, 
Diameter d= 8cm= 0.08 m
Assumptions
- The flow is jet flow hence momentum-flux correction factor is unity
- Gravitational force is not considered
- The flow is steady, frictionless and incompressible
- Water is discharged to the atmosphere hence pressure is ignored
We know that Q=AV and making v the subject then
where V is the exit velocity and A is area
Area, 
where d is the diameter
By substitution
To convert v to m/s from m/s, we simply divide it by 60 hence
Explanation:
There are 8.35 pounds in a gallon of water. Water weighs 1 gram per cubic centimeter or 1 000 kilogram per cubic meter, i.e. density of water is equal to 1 000 kg/m³; at 25°C (77°F or 298.15K) at standard atmospheric pressure.
