What does WCS stand for? A. Western CAD System B. Worldwide Coordinate Sectors C. World Coordinate System D. Wrong CAD Settings
1 answer:
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Answer:
T(water)=50.32℃
T(air)=3052.6℃
Explanation:
Hello!
To solve this problem we must use the equation that defines the transfer of heat by convection, which consists of the transport of heat through fluids in this case water and air.
The equation is as follows!
Q = heat
h = heat transfer coefficient
Ts = surface temperature
T = fluid temperature
a = heat transfer area
The surface area of a cylinder is calculated as follows
Where
D=diameter=20mm=0.02m
L=leght=200mm)0.2m
solving
For water
Q=2Kw=2000W
h=5000W/m2K
a=0.01319m^2
Tα=20C
solving for ts
for air
Q=2Kw=2000W
h=50W/m2K
a=0.01319m^2
Tα=20C
Answer:
Explanation:
Given:-
- The diameter of the cylinder, d = 50 mm.
- The compressive load, F = 80 KN.
Solution:-
- We will form a 3-dimensional coordinate system. The z-direction is along the axial load, and x-y plane is categorized by lateral direction.
- Next we will write down principal strains ( εx, εy, εz ) in all three directions in terms of corresponding stresses ( σx, σy, σz ). The stress-strain relationships will be used for anisotropic material with poisson ratio ( ν ).
εx = - [ σx - ν( σy + σz ) ] / E
εy = - [ σy - ν( σx + σz ) ] / E
εz = - [ σz - ν( σy + σx ) ] / E
- First we will investigate the "no-restraint" case. That is cylinder to expand in lateral direction as usual and contract in compressive load direction. The stresses in the x-y plane are zero because there is " no-restraint" and the lateral expansion occurs only due to compressive load in axial direction. So σy= σx = 0, the 3-D stress - strain relationships can be simplified to:
εx = [ ν*σz ] / E
εy = [ ν*σz ] / E
εz = - [ σz ] / E .... Eq 1
- The "restraint" case is a bit tricky in the sense, that first: There is a restriction in the lateral expansion. Second: The restriction is partial in nature, such, that lateral expansion is not completely restrained but reduced to half.
- We will use the strains ( simplified expressions ) evaluated in " no-restraint case " and half them. So the new lateral strains ( εx', εy' ) would be:
εx' = - [ σx' - ν( σy' + σz ) ] / E = 0.5*εx
εx' = - [ σx' - ν( σy' + σz ) ] / E = [ ν*σz ] / 2E
εy' = - [ σy' - ν( σx' + σz ) ] / E = 0.5*εy
εx' = - [ σy' - ν( σx' + σz ) ] / E = [ ν*σz ] / 2E
- Now, we need to visualize the "enclosure". We see that the entire x-y plane and family of planes parallel to ( z = 0 - plane ) are enclosed by the well-fitted casing. However, the axial direction is free! So, in other words the reduction in lateral expansion has to be compensated by the axial direction. And that compensatory effect is governed by induced compressive stresses ( σx', σy' ) by the fitting on the cylinderical surface.
- We will use the relationhsips developed above and determine the induced compressive stresses ( σx', σy' ).
Note: σx' = σy', The cylinder is radially enclosed around the entire surface.
Therefore,
- [ σx' - ν( σx'+ σz ) ] = [ ν*σz ] / 2
σx' ( 1 - v ) = [ ν*σz ] / 2
σx' = σy' = [ ν*σz ] / [ 2*( 1 - v ) ]
- Now use the induced stresses in ( x-y ) plane and determine the new axial strain ( εz' ):
εz' = - [ σz - ν( σy' + σx' ) ] / E
εz' = - { σz - [ ν^2*σz ] / [ 1 - v ] } / E
εz' = - σz*{ 1 - [ ν^2 ] / [ 1 - v ] } / E ... Eq2
- Now take the ratio of the axial strains determined in the second case ( Eq2 ) to the first case ( Eq1 ) as follows:
... Answer
Answer:
a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s
Explanation:
Diffusion is governed by Arrhenius equation
I will be using R in the equation instead of k_b as the problem asks for molar activation energy
I will be using
and
°C + 273 = K
here, adjust your precision as neccessary
Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm
So:
and
You might notice that these equations have the form of
You can solve this equation system easily using calculator, and you will eventually get
After you got those 2 parameters, the rest is easy, you can just plug them all including the given temperature of 1180°C into the Arrhenius equation
And you should get D = 2.76*10^-16 m^/s as an answer for c)