A contractor is planning on including several skylights in each unit of a residential development. What type of worker would she
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Answer:
yield strength before cold work = 370 MPa
yield strength after cold work = 437.87 MPa
ultimate strength before cold work = 440 MPa
ultimate strength after cold work = 550 MPa
Explanation:
given data
AISI 1018 steel
cold work factor W = 20% = 0.20
to find out
yield strength and ultimate strength before and after the cold-work operation
solution
we know the properties of AISI 1018 steel is
yield strength σy = 370 MPa
ultimate tensile strength σu = 440 MPa
strength coefficient K = 600 MPa
strain hardness n = 0.21
so true strain is here ∈ = = 0.223
so
yield strength after cold is
yield strength =
yield strength =
yield strength after cold work = 437.87 MPa
and
ultimate strength after cold work is
ultimate strength =
ultimate strength =
ultimate strength after cold work = 550 MPa
Answer:
Nₐ₀-Nₐ = 1.33
Nₐ₀ = 2.5
Conversion X = 1.33/2.5 = <u>0.533</u>
Explanation:
A + 2B + 4C ⇒ 2X + 3Y
Given a stream containing 50% A, 25% B and 25% C, to get the limiting reactant, lets take a simple basis
Say stream is 10 moles, this give
A = 5moles
B = 2.5mole
C = 2.5moles
from the balanced equation above,
1mole of A ⇒ 4moles of C
∴ 5moles of A ⇒ (5x4)/1 ⇒ 20moles of C
also;
2mole of B ⇒ 4moles of C
∴ 2.5moles of B ⇒ (2.5x4)/2 ⇒ 5moles of C
so clearly from above reactant C is the limiting reactant.
<em>Note: To get conversion of a process, we must use the limiting reactant. this is because ones it is used up, the reaction comes to an end</em>
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Formula to obtain conversion is:
Conversion = (Amount of A used up)/(Amount of A fed into the system)
where, Nₐ₀-Nₐ = is the amount in moles of A used up
Nₐ₀ = amount in moles of A fed into the system
The next question is what mole of reactant C will give 0.1mole fraction of Y
Recall our basis = 10moles
<em>from conservation of mass law</em>, 10mole of product must come out which 0.1 moles fraction is Y
therefore amount Y in the product is = 0.1x10 = 1mole
if 3moles of Y ⇒ 4mole of C
∴ 1mole of Y ⇒ (1x4)/3 ⇒ 1.33moles of C
calculating the conversion of limiting reactant C that will give 0.1mole fraction of Y
Nₐ₀-Nₐ = 1.33
Nₐ₀ = 2.5
Conversion X = 1.33/2.5 = <u>0.533</u>
Answer:
The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²
Explanation:
Here we have the heat Q given as follows;
Q = 15 × 1075 × (1100 - ) = 10 × 1075 × (850 - 300) = 5912500 J
∴ 1100 - = 1100/3
= 733.33 K
Where
= Arithmetic mean temperature difference
= Inlet temperature of the gas = 1100 K
= Outlet temperature of the gas = 733.33 K
= Inlet temperature of the air = 300 K
= Outlet temperature of the air = 850 K
Hence, plugging in the values, we have;
Hence, from;
, we have
5912500 = 90 × A × 341.67
Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².