Answer:64.10 Btu/lbm
Explanation:
Work done in an isothermally compressed steady flow device is expressed as
Work done = P₁V₁ In { P₁/ P₂}
Work done=RT In { P₁/ P₂}
where P₁=13 psia
P₂= 80 psia
Temperature =°F Temperature is convert to °R
T(°R) = T(°F) + 459.67
T(°R) = 55°F+ 459.67
=514.67T(°R)
According to the properties of molar gas, gas constant and critical properties table, R which s the gas constant of air is given as 0.06855 Btu/lbm
Work = RT In { P₁/ P₂}
0.06855 x 514.67 In { 13/ 80}
=0.06855 x 514.67 In {0.1625}
= 0.06855 x 514.67 x -1.817
=- 64.10Btu/lbm
The required work therefore for this isothermal compression is 64.10 Btu/lbm
Both of the technicians are correct.
<u>Explanation:</u>
The thrust angle is defined as the angle formed by the imaginary line drawn perpendicular to the rear axis center line. It is used in alignment of the four wheels.
It is also used to determine the direction the front wheels are pointing. While the scrub radius is the intersecting point of the vertical center line of front tires with the imaginary line drawn from the steering knuckles.
Thus both the technicians are saying correct. The thrust angle and scrub radius are used to determine the alignment of wheels, if there is any misalignment in wheels then it needs to make it correct to prevent accidents.
Answer:
Explanation:
Given
Temperature of solid 
Einstein Temperature 
Heat Capacity in the Einstein model is given by
![C_v=3R\left [ \frac{T_E}{T}\right ]^2\frac{e^{\frac{T_E}{T}}}{\left ( e^{\frac{T_E}{T}}-1\right )^2}](https://tex.z-dn.net/?f=C_v%3D3R%5Cleft%20%5B%20%5Cfrac%7BT_E%7D%7BT%7D%5Cright%20%5D%5E2%5Cfrac%7Be%5E%7B%5Cfrac%7BT_E%7D%7BT%7D%7D%7D%7B%5Cleft%20%28%20e%5E%7B%5Cfrac%7BT_E%7D%7BT%7D%7D-1%5Cright%20%29%5E2%7D)
Substitute the values
Answer:
nothing much what class r u in
Answer:
it is not possible to place the wires in the condui
Explanation:
given data
total area = 2.04 square inches
wires total area = 0.93 square inches
maximum fill conduit = 40%
to find out
Can it is possible place wire in conduit conduit
solution
we know maximum fill is 40%
so here first we get total area of conduit that will be
total area of conduit = 40% × 2.04
total area of conduit = 0.816 square inches
but this area is less than required area of wire that is 0.93 square inches
so we can say it is not possible to place the wires in the conduit
