Answer:
For now the answer to this question is only for partial fraction. Find attached.
Answer:
1561.84 MPa
Explanation:
L=20 cm
d1=0.21 cm
d2=0.25 cm
F=5500 N
a) σ= F/A1= 5000/(π/4×(0.0025)^2)= 1018.5916 MPa
lateral strain= Δd/d1= (0.0021-0.0025)/0.0025= -0.16
longitudinal strain (ε_l)= -lateral strain/ν = -(-0.16)/0.3
(assuming a poisson's ration of 0.3)
ε_l =0.16/0.3 = 0.5333
b) σ_true= σ(1+ ε_l)= 1018.5916( 1+0.5333)
σ_true = 1561.84 MPa
ε_true = ln( 1+ε_l)= ln(1+0.5333)
ε_true= 0.4274222
The engineering stress on the rod when it is loaded with a 5500 N weight is 1561.84 MPa.
The air flow necessary to remain at the lower explosive level is 4515. 04cfm
<h3>How to solve for the rate of air flow</h3>
First we have to find the rate of emission. This is solved as
2pints/1.5 x 1min
= 2/1.5x60
We have the following details
SG = 0.71
LEL = 1.9%
B = 10% = 0.1 a constant
The molecular weight is given as 74.12
Then we would have Q as
403*100*0.2222 / 74.12 * 0.71 * 0.1
= Q = 4515. 04
Hence we can conclude that the air flow necessary to remain at the lower explosive level is 4515. 04cfm
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