Select the correct answer.
2 answers:
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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:
Examples of reciprocating motion in daily life are;
1) The needles of a sewing machine
2) Electric powered reciprocating saw blade
3) The motion of a manual tire pump
Explanation:
A reciprocating motion is a motion that consists of motion of a part in an upward and downwards or in a backward and forward (↔) direction repetitively
Examples of reciprocating motion in daily life includes the reciprocating motion of the needles of a sewing machine and the reciprocating motion of the reciprocating saw and the motion of a manual tire pump
In a sewing machine, a crank shaft in between a wheel and the needle transforms the rotary motion of the wheel into reciprocating motion of the needle.
Answer:
The heat is transferred is at the rate of 752.33 kW
Solution:
As per the question:
Temperature at inlet, = 273 + 20 = 293 K
Temperature at the outlet, = 273 + 200 = 473 K
Pressure at inlet,
Pressure at outlet,
Speed at the outlet,
Diameter of the tube,
Input power,
Now,
To calculate the heat transfer, , we make use of the steady flow eqn:
where
= specific enthalpy at inlet
= specific enthalpy at outlet
= air speed at inlet
= specific power input
H and H' = Elevation of inlet and outlet
Now, if
and H = H'
Then the above eqn reduces to:
(1)
Also,
Area of cross-section, A =
Specific Volume at outlet,
From the eqn:
Now,
Also,
Now, using these values in eqn (1):
Now, rate of heat transfer, q:
q = mQ =