B1) 20 pts. The thickness of each of the two sheets to be resistance spot welded is 3.5 mm. It is desired to form a weld nugget
1 answer:
Answer:
I = 8975.95 A
Explanation:
Given :
Diameter of the nugget, D = 5.5 mm
Length of the nugget, l = 5.0 mm
Therefore, volume of the nugget,
=
Now, required melting energy, (M.E.) = 9.5 x V
= 9.5 x 118.73
= 1127.95 J
Given efficiency of the heat used from electric energy, η = 1/3
Then actual required energy = ME/η
= 1127.95 / (1/3) = 3383.84 J
We know that, electric energy (EE) =
where, I is current, R is resistance and t is time in seconds.
3383.75 =
I = 8975.95 A
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Answer:
5.6 mm
Explanation:
Given that:
A cylindrical tank is required to contain a:
Gage Pressure P = 560 kPa
Allowable normal stress = 150 MPa = 150000 Kpa.
The inner diameter of the tank = 3 m
In a closed cylinder there exist both the circumferential stress and the longitudinal stress.
Circumferential stress
Making thickness t the subject; we have
t = 0.0056 m
t = 5.6 mm
For longitudinal stress.
t = 0.0028 mm
t = 2.8 mm
From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value with the maximum thickness = 5.6 mm
Answer:
b) false
Explanation:
We know that Otto cycle is the ideal cycle for all petrol working engine.In Otto cycle all process are consider is ideal ,means there is no any ir-reversibility in the processes.
It consist four processes
1-2:Reversible adiabatic compression
2-3:Constant volume heat addition
3-4:Reversible adiabatic expansion
3-4:Constant volume heat rejection
Along with above 4 processes intake and exhaust processes are parallel to each other.From the P-v diagram we can see that all processes.
But actually in general we are not showing intake and exhaust line then it did not mean that in Otto cycle did not have intake and exhaust processes.
