Answer:
Part 1: It would be a straight line, current will be directly proportional to the voltage.
Part 2: The current would taper off and will have negligible increase after the voltage reaches a certain value. Graph attached.
Explanation:
For the first part, voltage and current have a linear relationship as dictated by the Ohm's law.
V=I*R
where V is the voltage, I is the current, and R is the resistance. As the Voltage increase, current is bound to increase too, given that the resistance remains constant.
In the second part, resistance is not constant. As an element heats up, it consumes more current because the free sea of electrons inside are moving more rapidly, disrupting the flow of charge. So, as the voltage increase, the current does increase, but so does the resistance. Leaving less room for the current to increase. This rise in temperature is shown in the graph attached, as current tapers.
Answer:
The following statements are true:
A. For flows over a flat plate, in the laminar region, the heat transfer coefficient is decreasing in the flow direction
C. For flows over a flat plate, the transition from laminar to turbulence flow only happens for rough surface
E. In general, turbulence flows have a larger heat transfer coefficient compared to laminar flows 6.
Select ALL statements that are TRUE
B. In the hydrodynamic fully developed region, the mean velocity of the flow becomes constant
D. For internal flows, if Pr>1, the flows become hydrodynamically fully developed before becoming thermally fully developed
Explanation:
Answer:
B A and C
Explanation:
Given:
Specimen σ
σ
A +450 -150
B +300 -300
C +500 -200
Solution:
Compute the mean stress
σ
= (σ + σ)/2
σ
= (450 + (-150)) / 2
= (450 - 150) / 2
= 300/2
σ = 150 MPa
σ
= (300 + (-300))/2
= (300 - 300) / 2
= 0/2
σ = 0 MPa
σ
= (500 + (-200))/2
= (500 - 200) / 2
= 300/2
σ = 150 MPa
Compute stress amplitude:
σ
= (σ - σ)/2
σ
= (450 - (-150)) / 2
= (450 + 150) / 2
= 600/2
σ = 300 MPa
σ
= (300- (-300)) / 2
= (300 + 300) / 2
= 600/2
σ = 300 MPa
σ
= (500 - (-200))/2
= (500 + 200) / 2
= 700 / 2
σ = 350 MPa
From the above results it is concluded that the longest fatigue lifetime is of specimen B because it has the minimum mean stress.
Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.
In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.
