Answer: D) All of the above
Explanation:
Burn rate can be affected by all of the above reasons as, variation in chamber pressure because the pressure are dependence on the burn rate and temperature variation in initial gain can affect the rate of the chemical reactions and initial gain in the temperature increased the burning rate. As, gas flow velocity also influenced to increasing the burn rate as it flowing parallel to the surface burning. Burn rate is also known as erosive burning because of the variation in flow velocity and chamber pressure.
Answer:
pipefitters design systems whereas plumbers maintain systems
Answer:
11.541 mol/min
Explanation:
temperature = 35°C
Total pressure = 1.5 * 1.013 * 10^5 = 151.95 kPa
note : partial pressure of water in mixture = saturation pressure of water at T = 35°c )
from steam table it is = 5.6291 Kpa
calculate the mole fraction of H
( YH
)
= 5.6291 / 151.95
= 0.03704
calculate the mole fraction of air ( Yair )
= 1 - mole fraction of water
= 1 - 0.03704 = 0.9629
Now to determine the molar flow rate of water vapor in the stream
lets assume N = Total molar flow rate
NH
= molar flow rate of water
Nair = molar flow rate of air = 300 moles /min
note : Yair * n = Nair
therefore n = 300 / 0.9629 = 311.541 moles /min
Molar flowrate of water
= n - Nair
= 311.541 - 300 = 11.541 mol/min
Answer:
I believe this is your question.
See the hand worked solution attached.
Answer:
work=281.4KJ/kg
Power=4Kw
Explanation:
Hi!
To solve follow the steps below!
1. Find the density of the air at the entrance using the equation for ideal gases

where
P=pressure=120kPa
T=20C=293k
R= 0.287 kJ/(kg*K)=
gas constant ideal for air

2.find the mass flow by finding the product between the flow rate and the density
m=(density)(flow rate)
flow rate=10L/s=0.01m^3/s
m=(1.43kg/m^3)(0.01m^3/s)=0.0143kg/s
3. Please use the equation the first law of thermodynamics that states that the energy that enters is the same as the one that must come out, we infer the following equation, note = remember that power is the product of work and mass flow
Work
w=Cp(T1-T2)
Where
Cp= specific heat for air=1.005KJ/kgK
w=work
T1=inlet temperature=20C
T2=outlet temperature=300C
w=1.005(300-20)=281.4KJ/kg
Power
W=mw
W=(0.0143)(281.4KJ/kg)=4Kw