Answer:
a) 4.21 min
b) 21.9666 hrs
c) 1.3657 Pc/hr
Explanation:
Given that;
Batch quantity = 300 units
time setup = 55min
unload/loading time = 0.75
processing time = 3.46
a) Average cycle time;
Average Cycle Time TC = loading time + processing time
TC = 0.75 + 3.46
Tc = 4.21 min
b) Time to complete the batch
Time to complete the batch Tb = setup time + process time + non operation time
Tb = (55min * 1hr/60min) + (300 * 3.46 * 1hr/60min) + (300 * 0.75 * 1hr/60min)
Tb = 0.9166 + 17.3 + 3.75
Tb = 21.9666 hrs
c) Average production rate
Average production rate Rp = 1 / ( Tb / batch size)
we substitute
Rp = 1 / ( 21.9666 / 300 )
Rp = 1 / 0.7322
Rp = 1.3657 Pc/hr
<h3><u>
Answer:</u></h3>
When a welder must certify for their appropriate welder's certifications, all of the samples are basically flat work. Simple tack welds to deep fill welds are required.
<h3><u>
Explanation:</u></h3>
If you are welding two pieces of metal together, having the work as flat as possible allows for the best access for the weld to be proper. There are often more times then not that the work will not be in a flat position so if you are really just starting out, your practice welds should be made on flat work to get the skill necessary to weld well in other positions.
Problem Solvers
Explanation:
Engineers find problems in the world, and then they find solutions for them.
Answer:
The additive drag at flight condition will be found by the following equation
Area = A1 = 6m2
Da = Additive drag
Cda = Additive drag coefficient
P = prassure at altitude of 12Km
Po = Prassure at sea level
\gamma = Ratio of specific heat capacity
The formula of additive drag is given below
D = Cda q A
q = Dynamic prassure , A = cross sectional area
q =( \gamma/ 2) P0 M02
D = Cda ( \gamma/2) po Mo2 A
Cda=0.32
\gamma=1.4
M=0.8
p0= 101325pa
D = 0.32 (1.4/2)(101325pa)(0.6)2 6
D = 49025N
Explanation:
The additive drag at the flight conditions will be D= 49025N
Answer:
Concentration factor will be 1.2
So option (C) will be correct answer
Explanation:
We have given outer diameter D = 1.25 in
And inner diameter d = 1 in and fillet ratio r = 0.2 in
So 
ratio will be 
And 
ratio will be 
Now from the graph in shaft vs torsion the value of concentration factor will be 1.2
So concentration factor will be 1.2
So option (C) will be correct answer.
