Answer:
Explanation :
The given information to be listed can are Equipment Number, Equipment Type, Seat Capacity, Fuel Capacity, and Miles per Gallon.
Check the attached document for the solution.
Scientific notation is another way to write a number. In scientific notation, the letter E is used to mean "10 to the power of." For example, 1.314E+1 means 1.314 * 101 which is 13.14 . Scientific notation is merely a format used for input and output.
Explanation:
Note: Refer the diagram below
Obtaining data from property tables
State 1:
![\left.\begin{array}{l}P_{1}=1.25 \text { bar } \\\text { Sat - vapour }\end{array}\right\} \begin{array}{l}h_{1}=234.45 \mathrm{kJ} / \mathrm{kg} \\S_{1}=0.9346 \mathrm{kJ} / \mathrm{kgk}\end{array}](https://tex.z-dn.net/?f=%5Cleft.%5Cbegin%7Barray%7D%7Bl%7DP_%7B1%7D%3D1.25%20%5Ctext%20%7B%20bar%20%7D%20%5C%5C%5Ctext%20%7B%20Sat%20-%20vapour%20%7D%5Cend%7Barray%7D%5Cright%5C%7D%20%5Cbegin%7Barray%7D%7Bl%7Dh_%7B1%7D%3D234.45%20%5Cmathrm%7BkJ%7D%20%2F%20%5Cmathrm%7Bkg%7D%20%5C%5CS_%7B1%7D%3D0.9346%20%5Cmathrm%7BkJ%7D%20%2F%20%5Cmathrm%7Bkgk%7D%5Cend%7Barray%7D)
State 2:
![\left.\begin{array}{l}P_{2}=5 \text { bor } \\S_{2}=S_{1}\end{array}\right\} \quad h_{2}=262.78 \mathrm{kJ} / \mathrm{kg}](https://tex.z-dn.net/?f=%5Cleft.%5Cbegin%7Barray%7D%7Bl%7DP_%7B2%7D%3D5%20%5Ctext%20%7B%20bor%20%7D%20%5C%5CS_%7B2%7D%3DS_%7B1%7D%5Cend%7Barray%7D%5Cright%5C%7D%20%5Cquad%20h_%7B2%7D%3D262.78%20%5Cmathrm%7BkJ%7D%20%2F%20%5Cmathrm%7Bkg%7D)
State 3:
![\left.\begin{array}{l}P_{3}=5 \text { bar } \\\text { Sat }-4 q\end{array}\right\} h_{3}=71-33 \mathrm{kJ} / \mathrm{kg}](https://tex.z-dn.net/?f=%5Cleft.%5Cbegin%7Barray%7D%7Bl%7DP_%7B3%7D%3D5%20%5Ctext%20%7B%20bar%20%7D%20%5C%5C%5Ctext%20%7B%20Sat%20%7D-4%20q%5Cend%7Barray%7D%5Cright%5C%7D%20h_%7B3%7D%3D71-33%20%5Cmathrm%7BkJ%7D%20%2F%20%5Cmathrm%7Bkg%7D)
State 4:
Throttling process ![h_{4}=h_{3}=71.33 \mathrm{kJ} / \mathrm{kg}](https://tex.z-dn.net/?f=h_%7B4%7D%3Dh_%7B3%7D%3D71.33%20%5Cmathrm%7BkJ%7D%20%2F%20%5Cmathrm%7Bkg%7D)
(a)
Magnitude of compressor power input
![\dot{w}_{c}=\dot{m}\left(h_{2}-h_{1}\right)=\left(8 \cdot 5 \frac{\mathrm{kg}}{\min } \times \frac{1 \mathrm{min}}{\csc }\right)(262.78-234 \cdot 45)\frac{kj}{kg}](https://tex.z-dn.net/?f=%5Cdot%7Bw%7D_%7Bc%7D%3D%5Cdot%7Bm%7D%5Cleft%28h_%7B2%7D-h_%7B1%7D%5Cright%29%3D%5Cleft%288%20%5Ccdot%205%20%5Cfrac%7B%5Cmathrm%7Bkg%7D%7D%7B%5Cmin%20%7D%20%5Ctimes%20%5Cfrac%7B1%20%5Cmathrm%7Bmin%7D%7D%7B%5Ccsc%20%7D%5Cright%29%28262.78-234%20%5Ccdot%2045%29%5Cfrac%7Bkj%7D%7Bkg%7D)
![w_{c}=4 \cdot 013 \mathrm{kw}](https://tex.z-dn.net/?f=w_%7Bc%7D%3D4%20%5Ccdot%20013%20%5Cmathrm%7Bkw%7D)
(b)
Refrigerator capacity
![Q_{i n}=\dot{m}\left(h_{1}-h_{4}\right)=\left(\frac{g \cdot s}{60} k_{0} / s\right) \times(234 \cdot 45-71 \cdot 33) \frac{k J}{k_{8}}](https://tex.z-dn.net/?f=Q_%7Bi%20n%7D%3D%5Cdot%7Bm%7D%5Cleft%28h_%7B1%7D-h_%7B4%7D%5Cright%29%3D%5Cleft%28%5Cfrac%7Bg%20%5Ccdot%20s%7D%7B60%7D%20k_%7B0%7D%20%2F%20s%5Cright%29%20%5Ctimes%28234%20%5Ccdot%2045-71%20%5Ccdot%2033%29%20%5Cfrac%7Bk%20J%7D%7Bk_%7B8%7D%7D)
![Q_{i n}=23 \cdot 108 \mathrm{kW}\\1 ton of retregiration =3.51 k \omega](https://tex.z-dn.net/?f=Q_%7Bi%20n%7D%3D23%20%5Ccdot%20108%20%5Cmathrm%7BkW%7D%5C%5C1%20ton%20of%20retregiration%20%3D3.51%20k%20%5Comega)
![\ Q_{in} =6 \cdot 583 \text { tons }](https://tex.z-dn.net/?f=%5C%20Q_%7Bin%7D%20%3D6%20%5Ccdot%20583%20%5Ctext%20%7B%20tons%20%7D)
(c)
Cop:
![\beta=\frac{\left(h_{1}-h_{4}\right)}{\left(h_{2}-h_{1}\right)}=\frac{Q_{i n}}{\omega_{c}}=\frac{23 \cdot 108}{4 \cdot 013}](https://tex.z-dn.net/?f=%5Cbeta%3D%5Cfrac%7B%5Cleft%28h_%7B1%7D-h_%7B4%7D%5Cright%29%7D%7B%5Cleft%28h_%7B2%7D-h_%7B1%7D%5Cright%29%7D%3D%5Cfrac%7BQ_%7Bi%20n%7D%7D%7B%5Comega_%7Bc%7D%7D%3D%5Cfrac%7B23%20%5Ccdot%20108%7D%7B4%20%5Ccdot%20013%7D)
![\beta=5 \cdot 758](https://tex.z-dn.net/?f=%5Cbeta%3D5%20%5Ccdot%20758)
Answer:
a. Solid length Ls = 2.6 in
b. Force necessary for deflection Fs = 67.2Ibf
Factor of safety FOS = 2.04
Explanation:
Given details
Oil-tempered wire,
d = 0.2 in,
D = 2 in,
n = 12 coils,
Lo = 5 in
(a) Find the solid length
Ls = d (n + 1)
= 0.2(12 + 1) = 2.6 in Ans
(b) Find the force necessary to deflect the spring to its solid length.
N = n - 2 = 12 - 2 = 10 coils
Take G = 11.2 Mpsi
K = (d^4*G)/(8D^3N)
K = (0.2^4*11.2)/(8*2^3*10) = 28Ibf/in
Fs = k*Ys = k (Lo - Ls )
= 28(5 - 2.6) = 67.2 lbf Ans.
c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.
For C = D/d = 2/0.2 = 10
Kb = (4C + 2)/(4C - 3)
= (4*10 + 2)/(4*10 - 3) = 1.135
Tau ts = Kb {(8FD)/(Πd^3)}
= 1.135 {(8*67.2*2)/(Π*2^3)}
= 48.56 * 10^6 psi
Let m = 0.187,
A = 147 kpsi.inm^3
Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi
Ssy = 0.50 Sut
= 0.50(198.6) = 99.3 kpsi
FOS = Ssy/ts
= 99.3/48.56 = 2.04 Ans.