Answer: The feed rate is
17,020kg/he and the rate is 13,520kg/h
Check the attachment for step by step explanation
Explanation:
You can use Hollomon's equation to estimate this.
Y_f is the flow stress.
K is the strenght coefficient or constant.
e is the strain
n is the strain hardening exponent.
You should be able to solve for e given the data in the problem statement.
Answer:
The smallest wire diameter that can be used is 1 cm
Explanation:
First, we find the smallest diameter using the criterion of maximum normal stress:
Max. Stress = 150 x 10^6 Pa = F/A
150 x 10^6 Pa = 12000 N/(πd²/4)
d² = (12000 N)(4)/(150 x 10^6 Pa)(π)
d = √1.0185 x 10^-4 m²
d = 0.010 m = 1 cm
Now, we find the smallest diameter using the criterion of maximum strain:
Max. Strain = Max. Change in Length/Original Length = 0.025 m/50 m
Max. Strain = 5 x 10^-4 mm/mm
Now,
Max. Strain = Stress/E = (F/A)/E = F/AE
using values:
5 x 10^-4 mm/mm = (12000 N)/(200 x 10^9 Pa)(πd²/4)
d =√(12000 N)(4)/(5 x 0^-4)(200 x 10^9 Pa)(π)
d = 0.012 m = 1.2 cm
Now, by comparison in both cases it can be noted that the smallest value of the diameter is <u>1 cm</u>, which is limited by maximum stress.
Answer:
Ts = 413.66 K
Explanation:
given data
temperature = 20°C
velocity = 10 m/s
diameter = 5 mm
surface emissivity = 0.95
surrounding temperature = 20°C
heat flux dissipated = 17000 W/m²
to find out
surface temperature
solution
we know that here properties of air at 70°C
k = 0.02881 W/m.K
v = 1.995 ×
m²/s
Pr = 0.7177
we find here reynolds no for air flow that is
Re = 
Re = 
Re = 2506
now we use churchill and bernstein relation for nusselt no
Nu = 
= 0.3 + ![\frac{0.62 Re6{0.5}Pr^{0.33}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\frac{2506}{282000})^{5/8}]^{4/5}](https://tex.z-dn.net/?f=%5Cfrac%7B0.62%20Re6%7B0.5%7DPr%5E%7B0.33%7D%7D%7B%5B1%2B%280.4%2FPr%29%5E%7B2%2F3%7D%5D%5E%7B1%2F4%7D%7D%20%5B1%2B%20%28%5Cfrac%7B2506%7D%7B282000%7D%29%5E%7B5%2F8%7D%5D%5E%7B4%2F5%7D)
h = 
0.3 + ![\frac{0.62*2506{0.5}0.7177^{0.33}}{[1+(0.4/0.7177)^{2/3}]^{1/4}} [1+ (\frac{2506}{282000})^{5/8}]^{4/5}](https://tex.z-dn.net/?f=%5Cfrac%7B0.62%2A2506%7B0.5%7D0.7177%5E%7B0.33%7D%7D%7B%5B1%2B%280.4%2F0.7177%29%5E%7B2%2F3%7D%5D%5E%7B1%2F4%7D%7D%20%5B1%2B%20%28%5Cfrac%7B2506%7D%7B282000%7D%29%5E%7B5%2F8%7D%5D%5E%7B4%2F5%7D)
h = 148.3 W/m².K
so
q conv = h∈(Ts- T∞ )
17000 = 148.3 ( 0.95) ( Ts - (20 + 273 ))
Ts = 413.66 K
C and B is the answer of the questions
