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3 years ago

The smallest crystal lattice defects is a) cracks b) point defects c) planar defects d) dislocations.

Engineering

1 answer:

ss7ja [257]3 years ago

8 0

Answer:b) Point defects

Explanation: The point defect is the tiny defect that occurs in the lattice. It usually occurs when there is the atoms or ions missing in the lattice structure that creates a irregularity in the structure.The name point defect itself describes that the occurring defect is having a size of point thus is the smallest defect. Therefore option(b) is the correct option.

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What are the units or dimensions of the shear rate dv/dy (English units)? Then, what are the dimensions of the shear stress τ= μ

swat32

Answer:

1) Dimensions of shear rate is $[T^{-1}]$ .

2)Dimensions of shear stress are $[ML^{-1}T^{-2}]$

Explanation:

Since the dimensions of velocity 'v' are $[LT^{-1}]$ and the dimensions of distance 'y' are $[L]$ , thus the dimensions of $\frac{dv}{dy}$ become

$\frac{[LT^{-1}]}{[L]}=[T^{-1}]$ and hence the units become $s^{-1}$ .

Now we know that the dimensions of coefficient of dynamic viscosity $\mu$ are $[ML^{-1}T^{-1}]$ thus the dimensions of shear stress can be obtained from the given formula as

$[\tau ]=[ML^{-1}T^{-1}]\times [T^{-1}]\\\\[\tau ]=[ML^{-1}T^{-2}]$

Now we know that dimensions of momentum are $[MLT^{-1}]$

The dimensions of $Area\times time$ are $[L^{2}T]$

Thus the dimensions of $\frac{Moumentum}{Area\times time}=\frac{MLT^{-1}}{L^{2}T}=[MLT^{-2}]$

Which is same as that of shear stress. Hence proved.

4 0

3 years ago

Refrigerant-134a enters the expansion valve of a refrigeration system at 120 psia as a saturated liquid and leaves at 20 psia. D

Shkiper50 [21]

Solution :

$P_1 = 120 \ psia$

$P_2 = 20 \ psia$

Using the data table for refrigerant-134a at P = 120 psia

$h_1=h_f=40.8365 \ Btu/lbm$

$u_1=u_f=40.5485 \ Btu/lbm$

$T_{sat}=87.745^\circ F$

∴ $h_2=h_1=40.8365 \ Btu/lbm$

For pressure, P = 20 psia

$h_{2f} = 11.445 \ Btu/lbm$

$h_{2g} = 102.73 \ Btu/lbm$

$u_{2f} = 11.401 \ Btu/lbm$

$u_{2g} = 94.3 \ Btu/lbm$

$T_2=T_{sat}=-2.43^\circ F$

Change in temperature, $\Delta T = T_2-T_1$

$\Delta T = -2.43-87.745$

$\Delta T=-90.175^\circ F$

Now we find the quality,

$h_2=h_f+x_2(h_g-h_f)$

$40.8365=11.445+x_2(91.282)$

$x_2=0.32198$

The final energy,

$u_2=u_f+x_2.u_{fg}$

$=11.401+0.32198(82.898)$

$=38.09297 \ Btu/lbm$

Change in internal energy

$\Delta u= u_2-u_1$

= 38.09297-40.5485

= -2.4556

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