Answer:
a) 
b) 
Explanation:
From the properties of Super-heated Refrigerant 134a Vapor at 
, 
; we obtain the following properties for specific enthalpy and specific entropy.
So; specific enthalpy 
specific entropy 
Also; from the properties of saturated Refrigerant 134 a vapor (liquid - vapor). pressure table at 
; we obtain the following properties:
Given that the power input to the compressor is 2 hp;
Then converting to Btu/hr ;we known that since 1 hp = 2544.4342 Btu/hr
2 hp = 2 × 2544.4342 Btu/hr
2 hp = 5088.8684 Btu/hr
The steady state energy for a compressor can be expressed by the formula:
By neglecting kinetic and potential energy effects; we have:
b) To determine the entropy generation; we employ the formula:
In a steady state condition 
Hence;
![\sigma _c = [200 \ lb/hr (0.2157 -0.2315) \ Btu/lb .^0R - \dfrac{(-3730.8684 \ Btu/hr)}{(40^0 + 459.67^0)^0R}]](https://tex.z-dn.net/?f=%5Csigma%20_c%20%3D%20%5B200%20%5C%20lb%2Fhr%20%280.2157%20-0.2315%29%20%5C%20Btu%2Flb%20.%5E0R%20%20-%20%5Cdfrac%7B%28-3730.8684%20%5C%20Btu%2Fhr%29%7D%7B%2840%5E0%20%2B%20459.67%5E0%29%5E0R%7D%5D)
![\sigma _c = [(-3.16 ) \ Btu/hr .^0R + (7.4667 ) Btu/hr ^0R}]](https://tex.z-dn.net/?f=%5Csigma%20_c%20%3D%20%5B%28-3.16%20%29%20%5C%20Btu%2Fhr%20.%5E0R%20%20%2B%20%287.4667%20%29%20Btu%2Fhr%20%5E0R%7D%5D)
Answer:
Both Brass and 1040 Steel maintain the required ductility of 20%EL.
Explanation:
Solution:-
- This questions implies the use of empirical results for each metal alloy plotted as function of CW% and Yield Strength.
- So for each metal alloy use the attached figures as reference and determine the amount of CW% required for a metal alloy to maintain a Yield Strength Y = 345 MPa.
- Left Figure (first) at Y = 345 MPa ( y -axis ) and read on (x-axis):
1040 Steel --------> 0% CW
Brass ---------------> 22% CW
Copper ------------> 66% CW
The corresponding ductility (%EL) for cold Worked metal alloys can be determined from the right figure. Using the %CW for each metal alloy determined in first step and right figure to determine the resulting ductility.
- Right Figure (second) at respective %CW (x-axis) read on (y-axis)
1040 Steel (0% CW) --------> 25% EL
Brass (22% CW) -------------> 21% EL
Copper (66% CW) ----------> 4% EL
We see that both 1040 Steel and Brass maintain ductilities greater than 20% EL at their required CW% for Yield Strength = 345 MPa.
Answer:
(a) Aluminum Indium Gallium Phosphide (AlInGaP). Band gap = 1.81eV ≈ 2eV
(b) Gallium Nitride (GaN). Band Gap = 3.4eV
(c) Aluminium Gallium Arsenide (AlGaA). Band Gap = 1.42eV ≈ 2.16eV
(d) Zinc Selenide (ZnSe). Band Gap = 2.82eV
(e) Gallium Phosphide (GaP). Band Gap = 2.24eV
Explanation:
LED's are semi-conducting materials that convert electrical energy to light energy. The light color emitted from the LED depends on the semi-conducting material and other compositions.
The band gap of the semi-conductor determines its wavelength. High band gap semi-conductors emit lower wavelengths which means greater power(UV semi-conducting macterials fall under this category).
Answer: 1766.667 Ω = 1.767kΩ
Explanation:
V=iR
where V is voltage in Volts (V), i is current in Amps (A), and R is resistance in Ohms(Ω).
3mA = 0.003 A
Rearranging the equation, we get
R=V/i
Now we are solving for resistance. Plug in 0.003 A and 5.3 V.
R = 5.3 / 0.003
= 1766.6667 Ω
= 1.7666667 kΩ
The 6s are repeating so round off to whichever value you need for exactness.
Answer:
Throughout the clarification section elsewhere here, the definition of the concern is mentioned.
Explanation:
- The stress-strain curve provides designers with a long list of critical parameters that are needed for utility development. Including capacity, longevity, elasticity, apparent viscosity, tension electricity, resilience, as well as flexural strength, a load-pressure assignment gave us several mechanical households at a certain point of operation. It also assists in manufacturing.
- During which the overarching force can inform us about the maximum energy either workload the substance will experience, which could also be inferred within the action of the stress-strain. The dynamic properties can be seen by pre maximum activity because it will be before even the maximum yield intensity as well as the posted maximum would display plastic behavior however after the peak becomes achieved, the natural frequencies continue to decline.
