Answer:
%Reduction in area = 73.41%
%Reduction in elongation = 42.20%
Explanation:
Given
Original diameter = 12.8 mm
Gauge length = 50.80mm
Diameter at the point of fracture = 6.60 mm (0.260 in.)
Fractured gauge length = 72.14 mm.
%Reduction in Area is given as:
((do/2)² - (d1/2)²)/(do/2)²
Calculating percent reduction in area
do = 12.8mm, d1 = 6.6mm
So,
%RA = ((12.8/2)² - 6.6/2)²)/(12.8/2)²
%RA = 0.734130859375
%RA = 73.41%
Calculating percent reduction in elongation
%Reduction in elongation is given as:
((do) - (d1))/(d1)
do = 72.14mm, d1 = 50.80mm
So,
%RA = ((72.24) - (50.80))/(50.80)
%RA = 0.422047244094488
%RA = 42.20%
90% of traffic crashes are due to driver error.
True
Answer: This is done by heating a large volume of quartz sand to temperatures as high as 1800˚C. The result is pure, isolated silicon, which is allowed to cool and then ground into a fine powder. To make silicone, this fine silicon powder is combined with methyl chloride and heated once again.
Explanation:
Answer:
Hello your question has some missing information below are the missing information
The refrigerant enters the compressor as saturated vapor at 140kPa Determine The coefficient of performance of this heat pump
answer : 2.49
Explanation:
For vapor-compression refrigeration cycle
P1 = P4 ; P1 = 140 kPa
P2( pressure at inlet ) = P3 ( pressure at outlet ) ; P2 = 800 kPa
<u>From pressure table of R 134a refrigerant</u>
h1 ( enthalpy of saturated vapor at 140kPa ) = 239.16 kJ/kg
h2 ( enthalpy of saturated liquid at P2 = 800 kPa and t = 60°C )
= 296.8kJ/kg
h3 ( enthalpy of saturated liquid at P3 = 800 kPa ) = 95.47 kJ/kg
also h4 = 95.47 kJ/kg
To determine the coefficient of performance
Cop = ( h1 - h4 ) / ( h2 - h1 )
∴ Cop = 2.49
Answer:
t = 2244.3 sec
Explanation:
calculate the thermal diffusivity
Temperature at 28 mm distance after t time = = 50 degree C
we know that
![\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]](https://tex.z-dn.net/?f=%5Cfrac%7B%2050%20-25%7D%7B300-25%7D%20%3D%20erf%20%5B%5Cfrac%7B28%5Ctimes%2010%5E%7B-3%7D%7D%7B2%5Csqrt%7B1.34%5Ctimes%2010%5E%7B-5%7D%5Ctimes%20t%7D%7D%5D)
from gaussian error function table , similarity variable w calculated as
erf w = 0.909
it is lie between erf w = 0.9008 and erf w = 0.11246 so by interpolation we have
w = 0.08073
![erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]](https://tex.z-dn.net/?f=erf%200.08073%20%3D%20erf%5B%5Cfrac%7B3.8245%7D%7B%5Csqrt%7Bt%7D%7D%5D)
solving fot t we get
t = 2244.3 sec
