The first law of thermodynamics for open systems

In summary: In my judgment, that relationship is not correct. The correct relationship is $$\dot{m}C_p\Delta T=\dot{n}(-\Delta H)$$where ##\dot{m}## is the total mass flow rate of propane and air entering the system (or reaction products leaving the system), ##\dot{n}## is the molar flow rate of propane entering, and ##-\Delta H## is the molar heat of combustion of propane.
  • #1
Gaby
7
2
Homework Statement
A fire is created in a room. The height of the door is 1.83 m and its
width is 0.74 m. The neutral plane is established at the height
1.0 m. The fire is sustained by propane (C3H8) with a mass flow
rate of 1.25 X10-3 kg/s and the oxidiser is air (79% vol. nitrogen
N2, 21% vol. oxygen O2). The air flows through the door opening
with an average velocity of Vin=0.7 m/s and a temperature of
Tin=298K.

(iv) Determine the average temperature and velocity of
outflowing gases.

Apply the principle of energy conservation for a steady-state open
system. Neglect the difference in kinetic and potential energy of
inflowing and outflowing gases as compared to the rate of energy
transfer to the gases (i.e. due to the heat release rate by the fire).
Assume that
− the combustion is perfect,
− the entire heat of reaction is spent on heating of gases in the
room;
− the average specific heat of outflowing gases is
cp=1050 J/kg/K and the average molecular mass of the
outflowing gases is equal to that of the ambient air
Min=Mout=29 kg/kmol.
Relevant Equations
The first law of thermodynamics for open systems
(Please see enclosed pdf doc, sorry equation wont copy correctly into text box.)
Please see enclosed my calculations for parts 1-3.

I am stuck on the final part of of a 4 part question, any help or guidance would be very much appreciated, thanks.
 

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  • Heat Transfer & Thermofluids.pdf
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  • #2
You do the final part the same way you determined the adiabatic flame temperature, except that, for this system, the air flow is not stoichiometric: Q = 0 and ##\Delta h=0##
 
  • #3
Hi thanks for this. To find the average temperature do I need to find the heat release rate? ( hc x mass flow rate) & mass flow rate = volumetric flow rate x density; I ve been stumped on how to approach this question; I would be very grateful on any guidance on how to approach it, thanks.
 
  • #4
Gaby said:
Hi thanks for this. To find the average temperature do I need to find the heat release rate? ( hc x mass flow rate) & mass flow rate = volumetric flow rate x density; I ve been stumped on how to approach this question; I would be very grateful on any guidance on how to approach it, thanks.
In the mixture for this problem, there is probably excess air. The heat of reaction per unit mass of propane is known (what is it) and you know the mass flow rate and specific heat of the exit stream.
 
Last edited:
  • #5
Thanks!
 
  • #6
Chestermiller said:
In the mixture for this problem, there is probably excess air. The heat of reaction per unit mass of propane is known (what is it) and you know the mass flow rate and specific heat of the exit stream.
Hi Chestermiller, Are you able to expand this a little as I am still confused. Any help is much appreciated!
 
  • #7
BigRed42 said:
Hi Chestermiller, Are you able to expand this a little as I am still confused. Any help is much appreciated!
Sure. For this problem, what is the mass flow rate of the C3H8 and of the air?
 
  • #8
Chestermiller said:
Sure. For this problem, what is the mass flow rate of the C3H8 and of the air?

Thanks very much. Mdot for propane = 1.25 x 10 -3. I don't know the mass flow rate for the air. Sorry!
 
  • #9
BigRed42 said:
Thanks very much. Mdot for propane = 1.25 x 10 -3. I don't know the mass flow rate for the air. Sorry!
You can calculate the mass flow rate of the air from the information provided in the problem statement.
 
  • #10
Is it m=pVA
= 1.19 kg/s
 
  • #11
BigRed42 said:
Is it m=pVA
= 1.19 kg/s
What values did you use in this equation? I get a different result.
 
  • #12
Chestermiller said:
What values did you use in this equation? I get a different result.
A= 1.35m2
V=0.7m/s
p=1.184

Thanks!
 
  • #13
BigRed42 said:
A= 1.35m2
V=0.7m/s
p=1.184

Thanks!
Shouldn’t the area only be 0.74 because the air is only coming in in the bottom part of the window?
 
  • #14
Chestermiller said:
Shouldn’t the area only be 0.74 because the air is only coming in in the bottom part of the window?
Ahhh That would make sense. Thanks! So do I just use the mas flow rate and the other known values in the equation
T2= T1 + Qdot/ P.(Vdot).Cp?
 
  • #15
BigRed42 said:
Ahhh That would make sense. Thanks! So do I just use the mas flow rate and the other known values in the equation
T2= T1 + Qdot/ P.(Vdot).Cp?
You need to use the total mass flow rate of both propane and air. What do P and Vdot mean?
 
  • #16
Chestermiller said:
You need to use the total mass flow rate of both propane and air. What do P and Vdot mean?
Hi Chester, V(dot above the letter) Volume flow rate. Q(dot) Heat release rate. P is the pressure at 1atm

Thanks for your help.
 
  • #17
BigRed42 said:
Hi Chester, V(dot above the letter) Volume flow rate. Q(dot) Heat release rate. P is the pressure at 1atm

Thanks for your help.
In my judgment, that relationship is not correct. The correct relationship is
$$\dot{m}C_p\Delta T=\dot{n}(-\Delta H)$$where ##\dot{m}## is the total mass flow rate of propane and air entering the system (or reaction products leaving the system), ##\dot{n}## is the molar flow rate of propane entering, and ##-\Delta H## is the molar heat of combustion of propane.
 
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Related to The first law of thermodynamics for open systems

1. What is the first law of thermodynamics for open systems?

The first law of thermodynamics for open systems is a fundamental principle in physics that states that energy cannot be created or destroyed, only transferred or converted from one form to another. This law applies to systems that exchange both matter and energy with their surroundings.

2. How is the first law of thermodynamics applied to open systems?

In open systems, the first law of thermodynamics is applied by considering the energy and matter flows into and out of the system. The energy entering the system must be equal to the energy leaving the system, plus any changes in the internal energy of the system.

3. What is an example of an open system in everyday life?

An example of an open system in everyday life is a pot of boiling water on a stove. Energy (heat) and matter (steam) are entering and leaving the system, and the internal energy of the water is changing as it boils.

4. How does the first law of thermodynamics relate to the conservation of energy?

The first law of thermodynamics is a statement of the conservation of energy, which is a fundamental principle in physics. It states that energy can neither be created nor destroyed, only transformed from one form to another. This principle is reflected in the first law of thermodynamics for open systems.

5. Can the first law of thermodynamics be violated?

No, the first law of thermodynamics is a fundamental law of physics and has been experimentally verified countless times. It is a universal principle that applies to all physical systems, and there is no known way to violate it.

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