Answer:
i) SF = 0.83
ii) 10 psi
iii) 3.16 psi
iv) 0.0083
Explanation:
Constant load = 1000 Psi
mean resistance = 1200 psi
Probability of failure = 1.9 * 10^-3
<u>Determine </u>
<u>i) Safety factor </u>
S.F = 1 / (mean load / load design ) = 1 / ( 1200 / 1000 ) = 0.83
<u>ii) Standard deviation </u>
1.9 X 10^-3 = e ( -1/2 ( μ / б) ^3 / (2.5 * 6 )
0.004756 = e^- 20000 / б^3
hence std = 10 psi
<u>iii) Variance </u>
=
= 3.16 psi
<u>iv) coefficient of variation </u>
Cv = std / mean resistance
= 10 / 1200 = 0.0083
<u />
The heat transfer which is in steady state, the heat transfer rate to the wall is equal to the wall.
<u>Explanation:</u>
- The convection transfer of heat to the wall is

- Here,
is the temperature of solid surface,
is the temperature of moving fluid stream which is adjacent of solid surface, h is the heat transfer coefficient. - The coefficient of convection heat transfers outer surface contains 3 times to the inner surface which experience smaller drop of temperature for 3 times that compares to inner surface.
- Hence, the temperatures outer surface get close to the surroundings of air temperature.
Answer:
I believe it is a civil drafter
Explanation:
O*NET site
Answer:
2.83 kg
Explanation:
Given:
Volume, V = 0.8 m³
gage pressure, P = 200 kPa
Absolute pressure = gage pressure + Atmospheric pressure
= 200 + 101 = 301 kPa = 301 × 10³ N/m²
Temperature, T = 23° C = 23 + 273 = 296 K
Now,
From the ideal gas equation
PV = mRT
Where,
m is the mass
R is the ideal gas constant = 287 J/Kg K. (for air)
thus,
301 × 10³ × 0.8 = m × 287 × 296
or
m = 2.83 kg
Answer:
Given:
P₁ = 500 kPa
T₁ = 860 K
P₂= 100 kPa
T₂ = 460 K
Let's take entropy properties of T1 and T2 from ideal properties of air,
at T = 860K, s(T₁) = 2.79783 kJ/kg.K
at T = 460K, s(T₂) = 2.13407 kJ/kg.K
using entropy balance equation:
![\frac{\sigma _cv}{m} = s(T_2)- s(T_1) - R In [\frac{P_2}{P_1}]](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Csigma%20_cv%7D%7Bm%7D%20%3D%20s%28T_2%29-%20s%28T_1%29%20-%20R%20In%20%5B%5Cfrac%7BP_2%7D%7BP_1%7D%5D%20)
= - 0.2018 kJ/kg. K
In this case the entropy is negative, which means the value of exit temperature is not correct, beacause entropy should always be positive(>0).