Answer:
The Process of printing
Explanation:
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A journeyman electrician with 16 years experience on-the-job was removing metal fish
tape from a hole at the base of a metal light pole. It was raining during the course of the
work. The fish tape became energized, electrocuting him. What could have been done to
prevent this accident? Select all that apply.
Explanation:
1. Ensure all circuits are de-energized before beginning work (29 CFR 1926.416(a)(3)).
2. Controls to be deactivated during the course of work on energized or de-energized
equipment or circuits must be tagged (29 CFR 1926.417(a)).
3. Employees must be instructed to recognize and avoid unsafe conditions associated with
their work (29 CFR 1926.21(b)(2)).
What is a shop air compressor
Answer:
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A gas stream (A) of 15.0 mol% H2, and the balance N2, is to be mixed with another gas stream
(B) containing 40.0 mol% H2, and the balance N2, to produce 100 kg/h of a 25 mol% H2, and the
balance N2 gas stream (C).
(a) Draw and fully label a flowchart of the mixing process.
(b) Calculate the average molecular weight of the product stream (C).
(c) Calculate the molar flow rates of the product stream (C) in kmol/h.
(d) Calculate the required flow rates of the feed mixtures A and B in kmol/h.
Answer:
A gas stream and other points are discussed below in details.
Explanation:
The method flow chart is displayed above.
B)Mole fraction of H2 in stream C=25%=0.25
Mole fraction of N2 in steams C=1-Mole fraction of H2=1-0.25=0.75
Average molecular weight of stream C=Mole fraction of H2*Molecular weight of H2+ Mole fraction of N2*Molecular weight of N2=0.25*2 kg/kmol+0.75*28 kg/kmol=21.5 kg/kmol
C) Mass flow rate of product C=100 kg/h
Molar flow rate of product C=Mass flow rate /Molecular weight=100 kg/h/21.5 kg/kmol=4.6512 kmol/h
D) Apply overall mole balance
Mole in=Mole out
Mole in=Mole of A+Mole of B=A+B
Let mole of A=A kmol/h
Mole of B=B kmol/h
Mole out=Mole of C stream=4.6512 kmol/h
A+B=4.6512 eq.1
Apply hydrogen mole balance
Mole of H2 in=Mole of H2 in A stream+ Mole of H2 in B stream=0.15 A+0.4 B
Mole of H2 out=Mole of H2 in C stream=0.25*4.6512=1.1628 kmol/h
1.1628 =0.15 A+0.4 B eq.2
Multiply eq.1 by 0.15
0.69768=0.15 A+0.15 B eq.3
Subtract eq.3 from eq.2
0.25 B=0.46512
B=1.86048 kmol/h
Substitute in eq1.
A=4.6512-1.86048=2.79042 kmol/h
Molar flow rate of A=2.79042 kmol/h
A molar flow rate of B=1.86048 kmol/h