Answer: The feed rate is
17,020kg/he and the rate is 13,520kg/h
Check the attachment for step by step explanation
The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb
<h3>The static equilibrium is given as:</h3>
F = P (Normal force)
Formula for moment at section
M = P(4 + 1.5/2)
= 4.75p
Solve for the cross sectional area
Area = 
d = 1.5

= 1.767 inches²
<h3>Solve for inertia</h3>

= 0.2485inches⁴
Solve for the tensile force from here

30x10³ = 
30000 = 14.902 p
divide through by 14.902
2013.15 = P
The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb
Read more on tensile force here: brainly.com/question/25748369
Answer:
power produced = 3098.52 kW
Explanation:
given data
insulated turbine = 100 bar
temperature = 400°C
pressure = 200 kPa
mass flow rate = 1.99 kg/s
solution
we use here steam table for At 100 bar and 400°C
h1 = 3096.5 KJ/Kg
and
at P2 = 200 Kpa
h2 = hf + 0.47 hg
h2 = 504.7 + 0.47 × 2201.6
h2 = 1539.452 KJ/Kg
so here
power produced is express as
power produced = m × (h1 - h2) .................1
power produced = 1.99 × ( 3096.5 - 1539.452 )
power produced = 3098.52 kW
Answer:
784.090909mph
Explanation:
1ft/s=0.681818 mph
1150ft/s=0.681818 x 1150=784.090909
Answer:
total time = 304.21 s
Explanation:
given data
y = 50% = 0.5
n = 1.1
t = 114 s
y = 1 - exp(-kt^n)
solution
first we get here k value by given equation
y = 1 -
...........1
put here value and we get
0.5 = 1 - e^{(-k(114)^{1.1})}
solve it we get
k = 0.003786 = 37.86 ×
so here
y = 1 -
1 - y =
take ln both side
ln(1-y) = -k ×
so
t =
.............2
now we will put the value of y = 87% in equation with k and find out t
t = ![\sqrt[1.1]{-\frac{ln(1-0.87)}{37.86*10^{-4}}}](https://tex.z-dn.net/?f=%5Csqrt%5B1.1%5D%7B-%5Cfrac%7Bln%281-0.87%29%7D%7B37.86%2A10%5E%7B-4%7D%7D%7D)
total time = 304.21 s