physics calculaton question
Discussion
Its a long time since I studied physics, can someone point me in the right direction to solve this calculation?
How long does it take to boil 1l of water from room temperature ( assume 20degC) using a natural gas burner with flow rate of 4L /min?
How long for all water to evaporate?
What is the efficiency?
I just need to know which equations to solve this.
Thanks!
How long does it take to boil 1l of water from room temperature ( assume 20degC) using a natural gas burner with flow rate of 4L /min?
How long for all water to evaporate?
What is the efficiency?
I just need to know which equations to solve this.
Thanks!
MasterBlaster said:
Its a long time since I studied physics, can someone point me in the right direction to solve this calculation?
How long does it take to boil 1l of water from room temperature ( assume 20degC) using a natural gas burner with flow rate of 4L /min?
How long for all water to evaporate?
What is the efficiency?
I just need to know which equations to solve this.
Thanks!
To get you started IIRCHow long does it take to boil 1l of water from room temperature ( assume 20degC) using a natural gas burner with flow rate of 4L /min?
How long for all water to evaporate?
What is the efficiency?
I just need to know which equations to solve this.
Thanks!
Q=mc(dT)
Q = energy in joules
m = mass in kg (1 in your case)
c = specific heat capacity - for water ~ 4.179 J/(gK) or 4719 J/(kgK)
dT = change in temp (celsius or Kelvin) 80 in your case.
(I have to nip out now and gas energy stuff I'd have to look up, likely someone else might know)
HTH
Watch your units throughout! Gas flow rate is volumetric; this would normally be converted to mass using the density of the gas at the flowing conditions (which are not stated!)
Time to boil:
Calorific value of gas * flow rate = heat input rate (kJ/kg * kg/s = kJ/s)
Heat required to boil liquid = mCpdT (as said by JustALooseScrew = 1*4.179*80 = 334kJ)
Time taken = Heat required to boil liquid / heat input rate (kJ / kJ/s = s)
Time to evaporate:
Latent heat of vaporisation of water (approx. 2,200 kJ/kg at atmospheric pressure)
Time to evaporate = mass of water * Heat of vaporisation / heat input rate (kg * kJ/kg / kJ/s = s)
Efficiency:
Difficult to calculate. It will depend on how much of the flame misses the container, how much of the energy in the gas is wasted heating the air, how much energy transmitted into the container is lost to atmosphere, etc., but generally, efficiency = energy output / energy input
This makes sense at 04.00. Whether it will tomorrow is another matter…
Mike…
Time to boil:
Calorific value of gas * flow rate = heat input rate (kJ/kg * kg/s = kJ/s)
Heat required to boil liquid = mCpdT (as said by JustALooseScrew = 1*4.179*80 = 334kJ)
Time taken = Heat required to boil liquid / heat input rate (kJ / kJ/s = s)
Time to evaporate:
Latent heat of vaporisation of water (approx. 2,200 kJ/kg at atmospheric pressure)
Time to evaporate = mass of water * Heat of vaporisation / heat input rate (kg * kJ/kg / kJ/s = s)
Efficiency:
Difficult to calculate. It will depend on how much of the flame misses the container, how much of the energy in the gas is wasted heating the air, how much energy transmitted into the container is lost to atmosphere, etc., but generally, efficiency = energy output / energy input
This makes sense at 04.00. Whether it will tomorrow is another matter…
Mike…
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