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Isotropic Turbine Work

  1. Jun 26, 2016 #1
    I'm doing some back of the envelope calculations for the potential of a turbine thermal generator using ammonia as a working fluid. I've never done thermodynamics before so I'm looking for a reality check.

    Isotropic turbine work done by a unit mass is given as h2 - h1 or simply dh between the the state of the fluid before and after the turbine. The temperature of the gas at each state is the temperature at the vaporizer/condenser and the pressure is the vapor pressure of the gas at those temperatures.

    I'm assuming I can substitute h2 and h1 in my turbine work equation for the specific enthalpy of a saturated gas hg at each temperature from an engineering lookup table for ammonia, and that this will give me the ideal energy per gram for my turbine subsystem.
     
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  3. Jun 26, 2016 #2

    Twigg

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    One thing you might have overlooked: was the ammonia at the input of the turbine saturated or superheated? Was the ammonia at the turbine exit saturated gas or a saturated liquid-vapor mixture? You need to look at the temperatures and pressures and compare them to a phase diagram to be sure. It would be very unlikely for a real turbine to take in saturated gas and spit out more saturated gas somewhere else on the boundary of ammonia's vapor dome. In general, you could have a superheated or saturated vapor going in and possibly a liquid-vapor saturated mixture leaving the turbine. Then the temperature of the fluid at the outlet isn't enough information to fix that state. You would need to equate the entropy of the fluid before and after the turbine, to solve for the quality of the liquid-vapor mixture leaving the turbine and entering the condenser. (I'm assuming there's a condenser downstream of this turbine because what you're describing sounds a lot like an ideal Rankine cycle).
     
  4. Jun 26, 2016 #3
    Yes it is a Rankine cycle (sorry for not specifying)

    Like I said this is back of the envelope to see if the theory works. I don't have a prototype model to measure. What I'm interested in is the maximum theoretical output for an idealized system. An upper limit on reality basically.

    The temperature difference in the working fluid is only 10-15K so I hadn't considered super heating. If I did superheat the input gas, would it be a valid approximation to assume that enthalpy scales linearly between vaporization and 50K of superheating? My lookup table only gives superheating in increments of 50K but the relationship looks roughly linear.

    For a liquid-vapor mixture wouldn't the production of a liquid in the turbine do negative work? If not how would I find the ideal liquid-vapor ratio? As far as calculating the enthalpy for the resulting fluid I'd assume that the specific enthalpy would by hf*liquid mass fraction + hg*gas mass fraction.
     
  5. Jun 26, 2016 #4

    Twigg

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    As far as superheating goes, linear interpolation of those 50K increments in enthalpy values should work well.

    As far as the liquid-vapor mixture, I don't remember off the top of my head which is bigger, the enthalpy of saturated ammonia vapor or liquid. For water I think producing liquid might decrease your efficiency. That being said, I mentioned it because you need to consider condensation in the turbine to ensure you don't violate the second law. In setting (W/m) = dh, you've already assumed there's no heat exchange between your turbine and the surroundings. Since it's an entropic process, the entropy generated per unit time per unit mass by your turbine is equal to the entropy per unit mass of the turbine outlet minus the entropy per unit mass of the inlet. If the entropy of the pure saturated ammonia vapor at your outlet is less than the entropy of the superheated gas at the inlet, then you would be violating the second law, and you won't be calculating a theoretical maximum anymore. If the entropy of your saturated gas at the outlet is higher than the entropy at the input, then your turbine won't be ideal anymore and you'll lose power every cycle. You can fix the entropy algebraically using either of the mass fractions of liquid or vapor phase ammonia (they both add to 1, so they can be used interchangeably). Then you'll have the true theoretical max.

    Better explanation and math to follow.
     
  6. Jun 26, 2016 #5

    Twigg

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    Two general equations govern the turbine:

    1. Steady-State Energy Balance (ignoring kinetic and potential effects):

    ## \frac{\dot{E}}{\dot{m}} = \frac{\dot{Q_{t}}}{\dot{m}} - \frac{\dot{W_{t}}}{\dot{m}} + (h_{i} - h_{o}) = 0##

    2. Steady-State Entropy Balance:

    ## \frac{\dot{S}}{\dot{m}} = \frac{1}{\dot{m}} \int \frac{\dot{\delta Q}}{T} + \frac{\dot{\sigma}}{\dot{m}} + (s_{i} - s_{o}) = 0##

    Where ##\dot{\sigma}## is the rate at which entropy is generated by the turbine, and the integral is over the entire boundary of the turbine system. The lower case h and s variables represent the enthalpy and entropy of the fluid per unit mass, respectively. Let me know if you need clarification on any of the other symbols (stating them all is a pain).

    There is no heat being added from the surroundings to the turbine (you're not igniting fuel or injecting heat from a hot reservoir into the turbine volume), so ##\dot{Q}## is only going to be zero or negative. If it's negative, that means you're losing energy that you would otherwise be getting off as heat (some of the work that the turbine could have made is instead being siphoned off as waste heat given to the surroundings). So for an ideal, maximum efficiency turbine, ##\dot{Q} = 0##. This gives you ##\frac{\dot{W}}{\dot{m}} = h_{o} - h_{i}##. Also, it tells you, based on the entropy balance, that ##\frac{\dot{\sigma}}{\dot{m}} = s_{o} - s_{i}##.

    The second law tells you that entropy is never destroyed, so ##\dot{\sigma} \geq 0##. At this point, you can take it on faith that entropy generation kills efficiency. But if you want the mathematical proof, just consider the thermodynamic relation for the enthalpy per unit mass, ##h_{o} - h_{i} = \int_{p_{i}}^{p_{o}} v dp - \int_{s_{i}}^{s_{o}} T ds##. The enthalpy difference (which remember is what's driving your turbine output shaft power) is optimized when ##s_{i} = s_{o}##.

    From what I've gathered, the parameters you have fixed are the inlet temperature (##T_{i}##), the outlet temperature (##T_{o}##), and the inlet pressure {##p_{i}##). Just to be on the safe side, let's assume that your outlet fluid could be a saturated liquid-vapor mixture. So now, there is a new unknown: the mass fraction of the vapor phase, a.k.a. the quality factor of the saturated mixture: ##x = \frac{m_{g}} {m_{g} + m_{f}}##.

    Now you can write the outlet enthalpy and entropy in terms of the quality factor:

    ##h_{o} = xh_{g}(T_{o}) + (1-x)h_{f}(T_{o})##
    ##s_{o} = xs_{g}(T_{o}) + (1-x)s_{f}(T_{o})##

    The inlet state is fixed by T_{i} and p_{i}, so you can get h_{i} and s_{i} from superheated ammonia vapor tables. Now you can do some algebra to solve for the quality factor of the outlet mixture by using the constraint that ##s_{i} = s_{o}## for a maximally efficient turbine.

    ##x = \frac{s_{o} - s_{f}(T_{o})}{s_{g}(T_{o}) - s_{f}(T_{o})} ##

    With the quality factor, you can now find the outlet enthalpy from the 2-phase mixture equations above, and you're in business.

    P.S.: If when you do this calculation, by some funky planetary alignment you find a quality factor of 1, then you will have hit the one in a million odds of accidentally designing an ideal turbine system that spits out pure saturated ammonia vapor. Similarly, if the quality factor turns out to be greater than 1, then you know what's coming out of the turbine is actually superheated vapor, and you can solve the problem instead by using the outlet temperature and outlet entropy to fix that outlet state. You'd just return to the superheated ammonia tables and use more linear interpolation to find a value of enthalpy for superheated ammonia vapor that gives the correct entropy and temperature.
     
  7. Jun 26, 2016 #6
    So I gather that the p1 and T1 are fixed by the vaporizer. You're sure that the temperature at the condenser fixes T2 and not p2?
     
  8. Jun 26, 2016 #7

    Twigg

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    The temperature at the inlet of the condenser would (ideally, assuming no pipe heat losses) fix the outlet temperature of the turbine.

    If you'd like me or another forum-goer to verify that your state fixing is all in order, it'd be best if you sent a labeled diagram, so we know what your ##T_{1}##'s and ##T_{2}##'s mean. Any givens you have would be helpful too.
     
  9. Jun 26, 2016 #8
    Sorry about the confusion.

    We could use this as a diagram: https://upload.wikimedia.org/wikipedia/commons/0/00/Rankine_cycle_layout.png

    The temperature of the condenser itself should determine the temperature of the condensate T1 as well as making the pressure in the condenser subsystem equal to the vapor pressure at that temperature. I would assume that this would fix the pressure of the fluid entering the condenser p4 but that T4 the temperature of the gas entering the condenser would be dependent on the performance of the turbine instead of the temperature of the condenser.
     
  10. Jun 26, 2016 #9

    Twigg

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    I understand what you're saying now. I did not mean that T1 fixes T4. I meant that the temperature of the condenser inlet equals the temperature of the turbine outlet (T4 = T4). It's fair to assume that the pressure at the condenser outlet fixes the pressure at the turbine outlet, for an ideal system, and yes, it makes sense to use the vapor pressure at T1 to fix p1, provided you don't cool past the saturated liquid line on the other end of the vapor dome and enter the compressed liquid regime (unlikely).
     
  11. Jun 26, 2016 #10
    Ok, so I use the the isentropic relation T4/T3 = (p4/p3)^((k-1)/k) to solve for the new T4 and I should be able to use your formulas to find first the mixture quality factor x and then h4.
     
  12. Jun 26, 2016 #11

    Twigg

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    It's true that T4 depends on the performance of the turbine, but the isentropic relation you mentioned is intended for ideal gases, so I wouldn't recommend it for finding the pressure of the two phase mixture. As a matter of fact, we're both overthinking this. If you know T1, you know the p1 from a vapor pressure lookup table. Ideally, p4 = p1. Since at the turbine outlet saturated vapor and liquid phases of ammonia are coexisting in equilibrium, p4 fixes T4, since T4 is just the boiling point of ammonia at p4.
     
  13. Jun 26, 2016 #12
    So Starting with a know T3 and p3 I find the specific entropy going in. Using known T4 and p4 I find the mixture quality factor that corresponds to that ideally unchanged specific entropy. I find the specific enthalpy of the initial gas h3, then the specific enthalpy of the final liquid-gas mixture h4 using the formula h=xhg + (1-x)hf. Finally h3 - h4 gives me ideal specific work for the turbine.

    If I can use a linear model for entropy of a super heated gas between 0K and 50K of heating then I should be good.
     
    Last edited: Jun 26, 2016
  14. Jun 27, 2016 #13

    Twigg

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    Sounds right, but you don't need 4 givens (T3,p3,T4,p4). You only need p3, T3, and either T4 OR p4.
     
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