Answer:
The fluid level difference in the manometer arm = 22.56 ft.
Explanation:
Assumption: The fluid in the manometer is incompressible, that is, its density is constant.
The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.
And P(gage) = ρgh
ρ = density of the manometer fluid = 60 lbm/ft³
g = acceleration due to gravity = 32.2 ft/s²
ρg = 60 × 32.2 = 1932 lbm/ft²s²
ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³
h = fluid level difference between the two arms of the manometer = ?
P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²
1353.6 = ρg × h = 60 lbf/ft³ × h
h = 1353.6/60 = 22.56 ft
A diagrammatic representation of this setup is presented in the attached image.
Hope this helps!
Answer:
b. equal to the specific entropy of the gas at the inlet.
Explanation:
Isentropic process is the process in which the entropy of the system remains unchanged. The word isentropic is formed from the combination of the prefix "iso" which means "equal" and the word entropy.
If a process is completely reversible, without the need to provide energy in the form of heat, then the process is isentropic.
Answer:
Explanation:
The answer to the given problem is been solved in the fine attached below.
Answer:
Check the explanation
Explanation:
1. When a branch is taken with 5% frequency, the branch prediction scheme is "predict taken" because the process has to begin fetching and then execute at the target address.
2. When a branch is taken with 95% frequency, the branch prediction scheme is "predict not taken" because the branch outcome is definitely known.
3. When a branch is taken with 70% frequency, the prediction scheme is "dynamic prediction" the branch prediction can dynamically change during the program execution.
Answer:
A)
- Q ( kw ) for vapor = -1258.05 kw
- Q ( kw ) for liquid = -1146.3 kw
B )
- Q ( kj ) for vapor = -1258.05 kJ
- Q ( KJ ) for liquid = - 1146.3 KJ
Explanation:
Given data :
45.00 % mole of methane
55.00 % of ethane
attached below is a detailed solution
A) calculate - Q(kw)
- Q ( kw ) for vapor = -1258.05 kw
- Q ( kw ) for liquid = -1146.3 kw
B ) calculate - Q ( KJ )
- Q ( kj ) for vapor = -1258.05 kJ
- Q ( KJ ) for liquid = - 1146.3 KJ
since combustion takes place in a constant-volume batch reactor