Answer:
The answer 30.07J/sec
Explanation:
GIVEN DATA:
temperature of steam = 40 degree celcius
mass flow rate = 63 kg/h
from saturated water tables,
from temperature 40 °, enthalpy of evaporation value is 2406 kj/kg
rate of heat transfer (Q) can be determine by using following relation
putting all value to get Q value
Q = 45 *2406
Q = 108270 kJ/h
Q = 30.07 kJ/sec
Answer:
moment of inertia = 4.662 * 10^6 
Explanation:
Given data :
Mass of machine = 400 kg = 400 * 9.81 = 3924 N
length of span = 3.2 m
E = 200 * 10^9 N/m^2
frequency = 9.3 Hz
Wm ( angular frequency ) = 2
= 58.434 rad/secs
also Wm =
------- EQUATION 1
g = 9.81
deflection of simply supported beam
t = 
insert the value of t into equation 1
W
=
make I the subject of the equation
I ( Moment of inertia about the neutral axis ) = 
I =
= 4.662 * 10^6 
Answer:
<u>note:</u>
<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>
Answer:
a. Use datum on shaft
b. Use datum on hex flat
c. Use datum on face below the head
d. Use datum on shaft
When these datum are used, they will prevent translation and rotation along axis which they act.