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Homework Help: Water jacketed reactor flow rate

  1. Dec 12, 2013 #1
    1. The problem statement, all variables and given/known data[/b]
    (135) 1. Liquid hydrazine (N2H4) is injected into a jet combustion chamber at 400 K and burned with 100 % excess air that enters at 700 K. The combustion chamber is jacketed with water. The combustion products leave the jet exhaust at 900 K. In the test, 50 kmol of hydrazine are burned per hour. Water at 25 °C enters the water jacket for cooling at a mass flow rate of 40 kg/min. The reaction is

    N2H4(l) + O2(g) ---> N2(g) + H2O(g)

    Data: ΔH_f of N2H4(l) = 44.77 kJ/gmol; of Cp N2H4(l) = 139 J/gmol°C

    (95) a. What is the volumetric flow rate of water out of the cooling jacket (at 1 atm)?

    (20) b. The diameter for the water jacket inlet is 0.1 m, and the diameter for the jacket outlet is 0.03 m. The water jacket outlet is 2 m below the jacket inlet. Are the kinetic and potential energy changes for the water significant?

    (20) c. If the water flow were halted, what would be the temperature of the outlet gas from the reactor? The reactor is made of nickel, which melts at 1453 °C. Is this a problem? What design features should the reactor have to mitigate the effects of this possible accident?

    2. Relevant equations

    3. The attempt at a solution
    For this problem, it seems like not enough information is given. I should be given the molar heat capacity of water, nitrogen, and oxygen.

    What I want to do for this is to basically bring the reactants down to 298 K using heat capacity, then react it at 298 with the standard heat of reaction, then bring it back up to 900 K with heat capacity to get the enthalpy change of the reaction.

    Once I know that, I might have a chance at finding the temperature of the water leaving the jacket, but from there I don't know how I will find the density as a function of temperature to get the volumetric flow rate.

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    Last edited: Dec 12, 2013
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  3. Dec 12, 2013 #2


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    Water is a pretty common substance. I hear that heat capacity information has even been compiled into large books. In a pinch, though, the heat capacity of liquid water is a standard in the metric system.

    The information on oxygen and nitrogen can be found.

    I don't think you can bring the temp of the fuel and oxidizers down to 298 K. The fuel is injected at 400 K according to the problem statement.
  4. Dec 12, 2013 #3
    Hi SteamKing. What he's talking about here is using Hess's law to determine the overall enthalpy change. The 44.77 kJ/gmole must be the standard heat of formation of liquid hydrazine at 298 K, and the standard heats of formation of the other species in the reaction at 298 are all zero. So the heat of reaction at 298 is -44.77 kJ/gmole. So he has to take the reactants down to 25 C (mathematically) and then reheat the products up to the final temperature (mathematically) because the heat of reaction is only known at 25 C.

    Woopydalan, your first step is to find the heat load that must be transferred from the gas to the water in order for the products and excess reactants to exit the reactor at 900 K. In short, you have to find the ΔH for the reactor. Once you know the heat load, and you know that 40 kg/min are being used, you need to find the exit state of the water at 1 atm. It might be liquid, a combination of liquid and gas, or, most likely gas. Once you know the change in enthalpy of the water (which is the same as the change in enthalpy of the reactor gases), you should easily be able to find its final state.
  5. Dec 12, 2013 #4
    Is the heat of formation of water 0?? Also, this was given on an exam so all the data should be within the problem statement, which is why I wondered if my method was okay, since the heat capacity of water, oxygen, and nitrogen are not given. Also, why is the enthalpy change in the water the same as the other gases? The oxygen and part of the nitrogen should have a different enthalpy change because they start at 700 K then go to 900 K, whereas the hydrazine is fed at 400 K.
    Last edited: Dec 12, 2013
  6. Dec 12, 2013 #5
    Oops!!! No!!! I should have looked more carefully. Sorry. It is definitely not zero.

    Maybe the data were given in another problem statement, or in a separate data sheet for the exam?
    I was referring to the water in the water jacket, not the water in the product gas. Sorry for the confusion.

  7. Dec 12, 2013 #6
    No I know that, I should have clarified. I don't understand why the enthalpy change in the water in the water jacket is the same as the products from the reactor relative to what entered the reactor.
  8. Dec 12, 2013 #7
    You are familiar with the version of the first law for an open flow system operating at steady state, correct? In a flow system operating at steady state, the change in enthalpy between entrance and exit is equal to the heat flow from the surroundings. The surroundings in the case of the reactor flow is the jacket. The heat flow from the reactor gases is equal to the heat flow to jacket water. The change in enthalpy of jacket water between its entrance and exit is equal to the heat flow from its surroundings, which, in this case is the reactor. The two heat flow are equal in magnitude and opposite in sign.
  9. Dec 14, 2013 #8
    I have calculated the enthalpy change in the water, and now I am stuck as to how I can use the enthalpy to find the quality of the water. I would like to use a steam table, and I know that its at 25 C and 1 atm at the inlet, but not sure what the temperature at the outlet is. The steam table specifies a pressure and temperature for saturated water, so how can I use the steam table at 1 atm if I don't know the temperature, and it's probably not the temperature at 1 atm for saturated water on the steam table
    Last edited: Dec 15, 2013
  10. Dec 15, 2013 #9
    It might be. Your steam table must have the enthalpy per kg of water at 100 C, both for liquid water and for water vapor (probably relative to a reference state of liquid water at 25 C). Does the enthalpy per kg of your exit water stream fall between these two values? If so, you can calculate the fraction liquid and the fraction vapor in the exit water stream.

  11. Dec 15, 2013 #10
    Yes but how can I be certain that the temperature of the exit stream is the temperature at which my steam table has values calculated for saturated water at 1 atm? The problem doesn't give the inlet pressure, but gives the inlet temperature. Is it safe to assume the inlet pressure is 1 atm as well?
    Last edited: Dec 15, 2013
  12. Dec 15, 2013 #11
    If your calculated enthalpy lies between that of the saturated liquid and the saturated vapor at 100 C and 1 atm, then you have it. The problem statement says that the outlet pressure from the jacket is 1 atm. The enthalpy of liquid water (at the inlet) is pretty much independent of pressure, but not likely to differ much from the outlet pressure of the jacket. So, what is the final enthalpy per kg of the water exiting the jacket?

  13. Dec 15, 2013 #12
    I calculated the change in enthalpy of the water as 915 kJ/kg. I know the inlet temperature is 25 C, and the outlet pressure is 1 atm. I am still unclear as to how to proceed to find the quality of the water at the outlet stream.

    The steam table specifies the Temperature of saturated water at 1 atm as 373.14 K. IT gives the liquid enthalpy as 419.5 kJ/kg, and the vapor enthalpty as 2675.6 kJ/kg.

    The problem for me is that (1) how do I know my exit stream is saturated? (2) If I don't know the exit stream temperature, how can I be certain that I am using the correct part of the steam table (properties of saturated water), as opposed to maybe properties of gaseous water, or properties of liquid water. (3) How is all of this going to lead me to the quality of the exit stream?
    Last edited: Dec 15, 2013
  14. Dec 16, 2013 #13
    Those enthalpy values are relative to a liquid state at 0 C. In your table you will find that, for liquid water at 25 C, the enthalpy is 104.8 kJ/kg, so the enthalpy of the exit stream appropriate to your steam tables is 915 + 105 = 1020 kJ/kg.

    The enthalpy is a unique function of state. At 1 atm., there is only one set of conditions that corresponds to an enthalpy of 1020. It is either a superheated vapor (in which case the temperature is higher than 100 C, and unique), a subcooled liquid (in which case the temperature is below 100 C and unique), or it is a combination of saturated liquid and saturated vapor at 100 C (in which case the proportions of liquid and vapor are unique). Since in your case, the enthalpy is higher than that of the saturated liquid and lower than that of the saturated vapor, you should immediately be able to reach a conclusion regarding which of the three possibilities applies to your situation. So, what is your conclusion? We'll talk about the answer to question (3) after you give your answer to my question.
  15. Dec 16, 2013 #14
    Why would this be added, as opposed to subtracted?
  16. Dec 16, 2013 #15
    Because the 915 is the change relative to the starting point of 105.
  17. Dec 16, 2013 #16
    I think there is a misunderstanding on my part of how steam tables work as far as how their reference state works, and what the terms mean. On the SI units side of the table, it has tables listed for Properties of: Liquid Water, Gaseous Water, and Saturated water.

    On the AE units side of the table, it has tables listed as Properties of: Superheated Steam and Saturated Water.

    I think these terms are being used in a very technical sense and I don't know what they mean, reminiscent of how the word heat in science is used in a very specific manner.

    My impression is that liquid water means that the water is a liquid, and has no vapor coming off. I think if vapor is coming off, then it is by default saturated water (still not exactly clear what saturated water is). Superheated steam is just steam at 1 atm that is hotter than 100 C, and subcooled liquid is liquid water at temperatures less than 0 C. I am probably grossly misinformed here though.

    Not sure why the tables dont match up either, meaning why the AE side doesn't have 3 tables like the SI side, and have a new category called superheated steam
    Last edited: Dec 16, 2013
  18. Dec 16, 2013 #17
    For the most part you have it right. Except that subcooled water means liquid water at a pressure greater than its equilibrium vapor pressure (so no vapor is present). This is what you called liquid water. Superheated steam is just water vapor at a pressure lower than its equilibrium vapor pressure at the given temperature (so no liquid is present).

    There is nothing wrong with your tables. In your problem, the enthalpy per kg of the exit stream from the reactor jacket is 1020, and it is a combination of saturated liquid and saturated vapor at 100 C. The enthalpy per kg of saturated liquid at 100C is 419.5. The enthalpy per kg of saturated vapor at 100 C is 2675.6. Now all you need to do is to use these three enthalpy numbers to figure out what mass fraction of the effluent is saturated liquid and what mass fraction of the effluent is saturated vapor.
  19. Dec 16, 2013 #18
    So in other words, subcooled liquid means the quality is zero, and for superheated steam the quality is unity? Also on the steam tables, is the ''pressure'' supposed to be the vapor pressure of the water at a temperature, or is it saying if an external pressure specified is being applied to the water at a specific temperature, the enthalpy, specific volume, etc. will be the result.

    Also, how do you know the exit temperature is 100 C?
    Last edited: Dec 16, 2013
  20. Dec 16, 2013 #19
    I solved for the quality of the steam and found .734 mass fraction liquid, .266 mass fraction vapor.

    I think I've solved it, what's irking me though is that I am saying that the exit stream temperature has to be 100 C since I'm using saturated water at 1 atm with those enthalpy for steam and liquid. Is this to mean that saturated water cannot exceed 100 C at 1 atm?? I have no intuition for this stuff, it is very weird thinking about what happens to water at temperatures and pressures that I do not experience in everyday life

    Also with regards to the water at 25 C enthalpy, that is also saying that the pressure of the water is somewhere around 3 kPa with that calculation for enthalpy, who's to say the entering stream has a pressure at 3 kPa?? All thats given is its temperature.
    Last edited: Dec 16, 2013
  21. Dec 16, 2013 #20
    Almost, but not quite. It is possible to have saturated liquid with zero quality it there is no present vapor, and it is possible to have saturated vapor with quality of unity, if there is no saturated liquid present.
    Neither really. It is just the external pressure applied to the liquid, the vapor, or the combination of liquid and vapor. But, if both saturated liquid and saturated vapor are present, it does not uniquely determine the enthalpy, specific volume, etc. of the combination of liquid and vapor. To do that, the quality also needs to be known.
    Because the pressure is 1 atm, and the enthalpy of your combination of liquid and vapor lies between that of saturated liquid and saturated vapor at 1 atm. and 100 C. 100 C is the only temperature where this is possible. If the temperature were above 100 C at 1 atm. pressure, the enthalpy would be above that of the saturated vapor, and you would have superheated vapor. If the temperature were below 100C at 1 atm. pressure, the enthalpy would be below that of the saturated liquid, and you would have supercooled liquid.
  22. Dec 16, 2013 #21
    So saturated water with a quality of 0 or unity is basically at the bubble point or dew point, respectively? Also, while the definition of subcooled liquid and superheated steam is not that the quality is zero and unity, respectively, is it true that those are the respective qualities?
  23. Dec 16, 2013 #22


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  24. Dec 16, 2013 #23
    Yes. I have some well-intentioned advice. When you are learning a new area, it is very important to learn the terminology. Otherwise, it's very hard to discuss things with someone else and also to read the literature. Please to back to your text and solidify your ability to apply the terminology.

  25. Dec 16, 2013 #24
    okay I think my understandings are much better now than the beginning of this thread, thanks for the clarification. The only thing that is bothering me in this problem right now is from my steam table, the stream flowing in at 25 C has to be at 3 kPa, and this is a consequence of water being saturated.

    From my daily experience, I know that water exists at 25 C at 1 atm. This is incompatible with the steam table showing water at 25 C having to be at 3 kPa.

    I guess this means that water exists at 3 kPa and 25 C at the bubble point???? Are steam tables for saturated water only showing the temperature and pressure at which the liquid is at its bubble point?
    Last edited: Dec 16, 2013
  26. Dec 16, 2013 #25
    The water coming in is a supercooled liquid at 1 atm. It is not saturated. There is no vapor phase. It is not at the bubble point. Steam tables are not for saturated water only. They also incluce supercooled liquid and superheated vapor.
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