The answer is 2nd Step because the first step is to define the problem and third is to define your goals
Maximum shear stress in the pole is 0.
<u>Explanation:</u>
Given-
Outer diameter = 127 mm
Outer radius,
= 127/2 = 63.5 mm
Inner diameter = 115 mm
Inner radius,
= 115/2 = 57.5 mm
Force, q = 0
Maximum shear stress, τmax = ?
τmax ![= \frac{4q}{3\pi } (\frac{r2^2 + r2r1 + r1^2}{r2^4 - r1^4} )](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B4q%7D%7B3%5Cpi%20%7D%20%28%5Cfrac%7Br2%5E2%20%2B%20r2r1%20%2B%20r1%5E2%7D%7Br2%5E4%20-%20r1%5E4%7D%20%29)
If force, q is 0 then τmax is also equal to 0.
Therefore, maximum shear stress in the pole is 0.
To solve this problem it is necessary to apply the concepts related to density in relation to mass and volume for each of the states presented.
Density can be defined as
![\rho = \frac{m}{V}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7Bm%7D%7BV%7D)
Where
m = Mass
V = Volume
For state one we know that
![\rho_1 = \frac{m_1}{V}](https://tex.z-dn.net/?f=%5Crho_1%20%3D%20%5Cfrac%7Bm_1%7D%7BV%7D)
![m_1 = \rho_1 V](https://tex.z-dn.net/?f=m_1%20%3D%20%5Crho_1%20V)
![m_1 = 1.18*1](https://tex.z-dn.net/?f=m_1%20%3D%201.18%2A1)
![m_1 = 1.18Kg](https://tex.z-dn.net/?f=m_1%20%3D%201.18Kg)
For state two we have to
![\rho_2 = \frac{m_2}{V}](https://tex.z-dn.net/?f=%5Crho_2%20%3D%20%5Cfrac%7Bm_2%7D%7BV%7D)
![m_2 = \rho_2 V](https://tex.z-dn.net/?f=m_2%20%3D%20%5Crho_2%20V)
![m_1 = 7.2*1](https://tex.z-dn.net/?f=m_1%20%3D%207.2%2A1)
![m_1 = 7.2Kg](https://tex.z-dn.net/?f=m_1%20%3D%207.2Kg)
Therefore the total change of mass would be
![\Delta m = m_2-m_1](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%20m_2-m_1)
![\Delta m = 7.2-1.18](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%207.2-1.18)
![\Delta m = 6.02Kg](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%206.02Kg)
Therefore the mass of air that has entered to the tank is 6.02Kg
Answer:
The break force that must be applied to hold the plane stationary is 12597.4 N
Explanation:
p₁ = p₂, T₁ = T₂
![\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} }](https://tex.z-dn.net/?f=%5Cdfrac%7BT_%7B2%7D%7D%7BT_%7B1%7D%7D%20%3D%20%5Cleft%20%28%5Cdfrac%7BP_%7B2%7D%7D%7BP_%7B1%7D%7D%20%20%5Cright%20%29%5E%7B%5Cfrac%7BK-1%7D%7Bk%7D%20%7D)
![{T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} } = 280.15 \times \left (9 \right )^{\frac{1.333-1}{1.333} } = 485.03\ K](https://tex.z-dn.net/?f=%7BT_%7B2%7D%7D%7B%7D%20%3D%20T_%7B1%7D%20%5Ctimes%20%5Cleft%20%28%5Cdfrac%7BP_%7B2%7D%7D%7BP_%7B1%7D%7D%20%20%5Cright%20%29%5E%7B%5Cfrac%7BK-1%7D%7Bk%7D%20%7D%20%3D%20280.15%20%5Ctimes%20%5Cleft%20%289%20%20%5Cright%20%29%5E%7B%5Cfrac%7B1.333-1%7D%7B1.333%7D%20%7D%20%3D%20485.03%5C%20K)
The heat supplied =
× Heating value of jet fuel
The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s
The heat supplied =
·
= 20 kg/s
The heat supplied = 20*
= 21,350 kJ/s
= 1.15 kJ/kg
T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K
p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa
p₃ = p₂ = 855 kPa
T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K
T₄ = 1413.3 - 204.88 = 1208.42 K
![\dfrac{T_5}{T_4} = \dfrac{2}{1.333 + 1}](https://tex.z-dn.net/?f=%5Cdfrac%7BT_5%7D%7BT_4%7D%20%20%3D%20%5Cdfrac%7B2%7D%7B1.333%20%2B%201%7D)
T₅ = 1208.42*(2/2.333) = 1035.94 K
= √(1.333*287.3*1035.94) = 629.87 m/s
The total thrust =
×
= 20*629.87 = 12597.4 N
Therefore;
The break force that must be applied to hold the plane stationary = 12597.4 N.
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)