Answer:
That's a really nice question sadly I don't know the answer I'm replying to you cuz I'm tryna get points so... Sorry
Answer:
35.7 kg lid we put
Explanation:
given data
temperature = 105 celcius
diameter = 15 cm
Patm = 101 kPa
to find out
How heavy a lid should you put
solution
we know Psaturated from table for temperature is 105 celcius is
Psat = 120.8 kPa
so
area will be here
area = 
..................1
here d is diameter
put the value in equation 1
area = 
area = 0.01767 m²
so net force is
Fnet = ( Psat - Patm ) × area
Fnet = ( 120.8 - 101 ) × 0.01767
Fnet = 0.3498 KN = 350 N
we know
Fnet = mg
mass = 
mass = 
mass = 35.7 kg
so 35.7 kg lid we put
Answer: c. The Professional Engineers Act and Board Rules
Explanation:
The reference source may be consulted to answer questions regarding the Professional Engineers Act is the The Professional Engineers Act and Board Rules.
The Professional Engineers Act and Board Rules is an Act that was established in order to regulate the qualifications for professional engineered, register them and also make sure that their conducts and behavior are looked into.
Pretty sure it’s Cenozoic
Answer:
[a]. 0.49.
[2]. 0.536
[c]. 4.15 kmol; 5.84 kmol.
Explanation:
Without mincing words let's dive straight into the solution to the question above.
[a].
The initial external reflux ratio, LD that must be used = [(0.85 - 0.57)/ 0.85 - 0]/ [ 1 - (0.85 - 0.57)/ 0.85 - 0].
The initial external reflux ratio, LD that must be used = 0.329/ 1- 0.329 = 0.49.
[b].
The final external reflux ratio that must be used = [ 0.85 - 0.13/ 0.85 - 0]/ [ 1 - 0.85 - 0.13/ 0.85 - 0].
Hence, the final external reflux ratio that must be used =0.847/ 1 - 0.847 = 5.536.
[c].
The amount of distillate product that is withdrawn:
4 = 0.85 H(t) + 0.8 - 0.08.
H(t) = 4.15 kmol, and the value of Wfinal = 5.84 kmol.
