tp review of basics
TRANSCRIPT
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Review Of Basics
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Stress and its types ?
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The force which tends to make one surface slide parallel to an adjacent
surface is called a shear force.
The force divided by the area (of the glue joint), the stress in the glue, is
called shear stress.
What is Shear Force & Shear Stress ?
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Difference between Pressure and Stress ?
-Pressure is the external force acting over a unit surface area of a material.-Stress is an internal force acting over a cross sectional area within the material.
-It may also be thought of as the internal resistive response of a
material to the applied external pressure.
-They both have the units of force / area.
Frictional Force : The Opposing ForceThe frictional force refer to the force that resists the relative motion of the fluid layers
and the solid surfaces. It usually resists the given material from sliding past each other.
Flux ?In transport phenomena , flux is defined as the rate of flow of a property per unit area, which
has the dimensions [quantity]·[time]−1·[area]−1.
For example, the magnitude of a river's current, i.e. the amount of water that flows through a
cross-section of the river each second, or the amount of sunlight that lands on a patch of ground
each second is also a kind of flux.
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Inter-relationship of Pressure, Stress, Momentum-Rate/Area,
Momentum-Flux
Stress =
Momentum= mass . velocity =
Momentum Rate/Area =
= Momentum-flux
So,
Pressure= Stress = = Momentum-flux
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Calculus Review :
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- What are functions ?
- One of the most important themes in engineering and other faculties as well
is the analysis of relationships between physical or mathematical quantities.
-Functions are used to describe the relationship between the variable quantities.-Examples, lets start with a simpler one :
S = vt
…means distance travelled depends upon the velocity and the travelling time
ie: S is function of v and t
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Ways to describe functions ?
-There are four ways to describe the functions:
-Numerically by tables
-Geometrically by graphs
-Algebraically by formulas
-Verbally
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For example, consider a case of heat transfer
through a solid metallic wall. T1 is lower temp.
at left end and T2 is higher temp. at right end.
Consider a plane wall 100 mm thick and of thermal
conductivity 100 W/m K. Steady-state conditions are
known to exist with T1 400 K and T2 600 K. Determine
the heat flux qx and the temperature gradient dT/dx
for the coordinate systems shown.
∆x = 100 mm= 0.1 m, k = 100 W/m.K, T1= 400K, T2= 600K
qx = ? & dT/dx = ?
qx = 100 ( 200/0.1) = 2 x 105 W/m2
dT/dx = 200/0.1 = 200
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Significance of Slopes in Transport Processes :
-For example, consider a case of heat transfer
through a solid metallic wall. T1 is lower temp. atleft end and T2 is higher temp. at right end.
-Slope of line T1T2, ie: ∆T/ ∆x
taken at any two points between T1 and T2 gives
us a change in temp. for a corresponding change
in distance i.e: rate of change of temp.
-Analogously for cases of momentum transfer and mass transfer the respective
slopes ie : ∆V / ∆x and ∆C / ∆x give us change in velocities and concentrations for
given distances.
- So once we know the rates of change (slopes) of temp., velocity or
concentration we could figure out heat transfer rate, momentum transfer rate
and mass transfer rates respectively.
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SLOPE:
-A line with positive slope rises uphill to the right
-And the one with negative slope falls downhill to the right
-The greater the absolute value of the slope, the more rapid the rise or fall.
-The slope of a vertical line is undefined. Since the run is zero for a vertical line,
we cannot evaluate the slope ratio m.
-Measure of Steepness Or Ratio of ‘Rise’ to its ‘Run’
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Average Rates of Change and Secant Lines :
-For straight lines, the rate of change (slope) is constant (always the same).
But unfortunately in practical engineering processes (transport processes)
this is not true all the time.
Rates of change may not necessarily be constant. Rather they vary interestingly).
These varying rates of change are reflected through curves instead of straight lines
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-So how to determine the varying rates of Change ?
-Consider a graph plotted for a chemical reaction. It gives the varying
concentration of the product being formed with respect to time.
-So by drawing secant on two desired points we get the average value of concentration-rate
-Concentration-rate between points 40 to 45 seconds & 20 to 25 seconds
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-How to determine the rate of change at a single point ie:
instantaneous rate of change ?
- A line that is tangent to the curve at point P gives us the instantaneous
rate of change at 23rd second.
-So at 23rd second the instantaneous rate of change of concentration is
16.7 gm-moles/litre/sec
-So we find instantaneous rates as limiting values of average rates ie:
by drawing tangent lines as a limiting position of secant lines.
-So these tangent lines lead to very important concept of ‘Limits’
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