Find out the percent exergy loss

[yaro99]

Homework Statement

Reproduce the results in Figure 4.20 (shown in post). [Exergy loss in a liquefaction system].

Relevant Equations

Energy Balance: dE_cv/dt = Q_cv - W_cv + SIGMA [mdot_i (h_i + V_i^2+gzi)] - [mdot_e (h_e + V_e^2+gze)]

Exergy Balance: dE/dt = (1- T_0/T_j)*Q_j - W_cv +SIGMA [ (m_i*e_fi) - (m_e*e_fe) ] - Edot)

Entropy: (Q_dot,cv/m_dot)_int,rev = int|1-2(T ds)

Q_dot,cv/m_dot = eta * (Q_dot,cv/m_dot)_int,rev

Problem, with state values, and pie chart (Fig 4.20) showing answers:

^ This shows the system in question (Kapitza Liquefaction System). Methane gas enters into the compressor (c), then goes through the first heat exchanger (HX1). Some of it (z) gets routed to the expander (exp). Afterwards, the rest goes to the second heat exchanger (HX2), then to the Expansion Valve (JT). Then, the liquefied methane is collected (state 7), and the rest is routed back, first to HX2, then HX1, before going right back to the compressor, with a certain amount being supplied between state 13 and state 1.

Using Table 4-2, here are the results I obtained:

1. Compressor

2. Heat Exchanger 1

3. Heat Exchanger 2

4. Expansion Valve

5. Expander

Results)

My thermodynamics is a pretty rusty, so forgive me if I'm making a simple mistake, but I just haven't been able to find sources to help me out on this one.

I was wondering about the control volumes around each unit, namely the heat exchangers, the expander, and the expansion valve (is there work, heat transfer, both, or neither?). I'm sorry if this seems obvious but hours of research online and in my textbooks have not helped too much.

There are two main things I'm confused about: 1) My expansion valve analysis ends up with zero work (as the enthalpies are the same in both states), and there's no way this can be true, and 2) My exergy analysis on the expander results in negative exergy, which is not possible.

 

[BvU]

Hello yaro,

You'll forgive me if I don't drudge through all your material (if you wanted me to, you would have $\LaTeX$ed it, I suppose ?)

edit: quote is cut off somehow. So reproduce

Relevant Equations:: Energy Balance: $${dE_{cv}\over dt} = Q_{cv} - W_{cv} + \sum \left [\dot m_i (h_i + V_i^2+gz_i)\right ] - \left [\dot m_e (h_e + V_e^2+gz_e)\right ]$$

Exergy Balance: $${dE\over dt} = (1- T_0/T_j)*Q_j - W_{cv} +\sum\left[ (m_i*e_{fi}) - (m_e*e_{fe}) \right] - \dot E)$$

Entropy: $$\left({\dot Q_{cv}\over\dot m}\right)_{int,rev} = \int 1-2(T ds) \\

{\dot Q_{cv}\over\dot m} = \eta * \left({\dot Q_{cv}\over \dot m}\right)_{int,rev}$$

[edit] the example is worked out in the book, so I suggest you ask specifically which step you don't follow, in stead of dumping the whole lot.

(nit picking: with this notation, energy and exergy are bound to be mixed up erroneously ! But I see the book is guilty too.)and instead ask about your confusion

There are two main things I'm confused about: 1) My expansion valve analysis ends up with zero work (as the enthalpies are the same in both states), and there's no way this can be true, and 2) My exergy analysis on the expander results in negative exergy, which is not possible.

1) $h$ includes $pV$ work, so why this confusion ?
2) why not ? See (4-24) : $h$ doesn't change, but $s$ surely does !

[edit] sorry, mixed up expander and valve. in the former, $h$ changes too.

 

[yaro99]

Thank you for your response.

Sorry for not using LATEX, it's been years since I've used it. I will start using it now. Thanks for applying it to my Relevant Equations.

I actually didn't realize it was worked out in the book, it was a few pages before my particular problem.

After looking over the book, I still have a couple questions. First, regarding your answers:

1) $h$ includes $pV$ work, so why this confusion ?

Does this mean I need the pressure and volume to calculate the work? If so, I only have pressure and not volume, so I would not be able to solve for work.

1) why not ? See (4-24) : $h$ doesn't change, but $s$ surely does !

So, because the valve is isenthalpic, there is no work, but there is still exergy destruction due to entropy? If I understand correctly, one can solve for exergy destruction using either an exergy balance, or an entropy balance, and the value will be the same in both cases.
In the latter case, one would calculate the entropy and plug it into this equation:
$E_d=T_0\sigma$

Additional question:
1) Compressor analysis: The book has calculated the compressor work as follows:

$-\dot{W}_c = \frac{1}{\eta_c}*(h_2-h_1)*\dot{m}$

However, when I calculated the problem, I accounted for heat transfer, since there must be heat leaving the system when work is done. Using the second entropy equation given above, and pluging that into the energy balance (first equation above), I got:

$-\dot{W}_c = \frac{1}{\eta_c}*(h_2-h_1)*\dot{m}+\frac{1}{\eta_c}*(T_1)*(s_1-s_2)*\dot{m}$

Is there no heat transfer from the compressor? Why?

 

[BvU]

Pinging @Chestermiller

 

[Chestermiller]

Pinging @Chestermiller

Sorry. Don't know much about energy.

 

[BvU]

1) My expansion valve analysis ends up with zero work (as the enthalpies are the same in both states), and there's no way this can be true

Please explain why you think so.

Does this mean I need the pressure and volume to calculate the work? If so, I only have pressure and not volume, so I would not be able to solve for work.

This is about your 1), we are talking about the expansion valve (throttling valve) here, right ? There is no non-flow work in or out.

For an adiabatic valve there is no change in enthalpy (ideal gas, $\mu = 0$), $h_{in}\cong h_{out}$. As you see in table 2, $h_6 = h_5$, so the valve is adiabatic, but $T_6 < T_5$ which means ideal gas is not the case. So with $h = u+pV$ you could check the change in internal energy. However, $u$ is not given.

To recover $V$ you need more properties or an equation of state. Non ideal gas means $pV = Z \,RT$ as equation of state where $Z$ is the thermodynamic compressibility factor so as collateral you get $Z$.

All you can do is check that $\Delta e = -T_0 \Delta s$. And see that it's a relatively small loss.

So, because the valve is isenthalpic, there is no work, but there is still exergy destruction due to entropy? If I understand correctly, one can solve for exergy destruction using either an exergy balance, or an entropy balance, and the value will be the same in both cases.
In the latter case, one would calculate the entropy and plug it into this equation:
$E_d=T_0\sigma$

if $\sigma = \Delta s$: yes. Because $\Delta h = 0$.
Advice: keep the number of symbols to a minimum and don't memorize more than necessary. In my case, all I know is Exergy $\equiv h - h_0 - T_0(s-s_0)$.

2) My exergy analysis on the expander results in negative exergy, which is not possible

Please explain why you think so. Exergy is never created but always destroyed.

Now

Additional question:
1) Compressor analysis: The book has calculated the compressor work as follows:

Easier if we keep numbering and call this 3) to avoid confusion.

Compressor does work to

However, when I calculated the problem, I accounted for heat transfer, since there must be heat leaving the system when work is done.

If the compressor is reasonably insulated, not much heat loss occurs. Thermodynamically the heat you mention leaves with the outgoing fluid. This is accounted for in the enthalpy flow. An ideal (reversible) compression step is isentropic and yields a $\displaystyle {W_{isentropic} = \Delta h_{is} = h_{is}-h_{in} }$. A real one has $\displaystyle {\eta_c \equiv {h_{is} - h_{in} \over h_{out}- h_{in} } \ }$ with $\eta_c<1$, so $h_{out} > h_{is}$ . In other words: more work an a higher enthalpy at the same pressure, in still other words: more heat going out with the fluid.

Very nice concept, this enthalpy !

[edit] fixed the W_isentropic expression

 

[BvU]

Sorry. Don't know much about energy.

Your modesty does you credit, but I adorned the post with "skeptical". It's all thermodynamics and you are the expert on that.

 

[Chestermiller]

Your modesty does you credit, but I adorned the post with "skeptical". It's all thermodynamics and you are the expert on that.

I'll try again, neglecting the part about exergy.

 

[Chestermiller]

Ther enthalpy change in the expansion valve is zero, so the expansion valve is operating adiabatically. However, even though the pressure and temperature are constant, there is a phase change, resulting in the formation of liquid. The enthalpy per unit mass of stream 6, given in the table, is actually a weighted average of the enthalpy per unit mass of the vapor (-400.4) and the enthalpy per unit mass of the liquid (-911.5). From streams 7 and 8, the mass fraction liquid in stream 6 is 85.2%, and the mass fraction vapor is 14.8%.

Actually, from the pressure-enthalpy diagram for methane (below), it appears that stream 5 is a compressed liquid. When this liquid is flashed in the expansion valve, a small amount flashes to vapor (14.8%), and the remainder stays liquid. The result is also a lower temperature (-161.6 vs -141.6) as a result of the heat of vaporization.

 

[Chestermiller]

The heat balance on HX1 at steady state should read:
$$0=\dot{Q}+m_{2,3}(h_2-h_3)+m_{12,13}(h_{12}-h_{13})$$So, $$0=\dot{Q}+(1)(-40.01+190.8)+(0.9652)(-211.6+23.18)$$This yields $$\dot{Q}=31.07\ kJ/kg$$
This means that the heat exchange is not operating adiabatically, and it is gaining some heat through the shell of the exchanger.

Assuming that the heat transfer from the surroundings occurs at the surrounding air temperature (298 K), the entropy balance on the exchange should read $$0=\frac{\dot{Q}}{T_{surr}}+m_{2,3}(s_2-s_3)+m_{12,13}(s_{12}-s_{13})+\dot{\sigma}_{gen}$$Plugging in the numbers from the table gives a rate of entropy generation of $$\dot{\sigma}_{gen}=0.0833\ \frac{kJ}{kg.K}$$

 

[Chestermiller]

Analysis of expander:

Assuming that the expander is operating adiabatically, the energy balance on the expander reads:

$$0=-W_{exp}+m_{9,10}(h_9-h_{10})$$

And the entropy balance on the expander reads: $$0=m_{9,10}(s_9-s_{10})+\dot{\sigma}$$

 

[Chestermiller]

The previous post assumes that the expander is perfectly adiabatic and that all the entropy generation takes place within the process fluid. However, another reasonable assumption would be that the expander exchanges heat with the surroundings, but that the process fluid experiences no entropy generation. In this case, all the entropy generation would be confined to the shell of the expander where finite conductive heat transfer is occurring. In this case, the energy balance on the expander would be $$0=\dot{Q}-\dot{W}+m_{9,10}(h_9-h_{10})$$ where Q is the heat flow through through the shell. The overall entropy balance on the expander would be: $$0=\frac{\dot{Q}}{T_{surr}}+m_{9,10}(s_9-s_{10})+\dot{\sigma}$$and the entropy balance on the shell would be:
$$0=\frac{\dot{Q}}{T_{surr}}-\frac{\dot{Q}}{T_{eff}}+\dot{\sigma}$$where $T_{eff}$ is the effective average temperature of the process fluid at the interface with the shell, and would lie between -33.64 C (240 K) and -161.6 C (112 K). If we subtract the previous two equations, we obtain: $$\dot{Q}=-T_{eff}m_{9,10}(s_9-s_{10})$$In addition, we then also find that $$\dot{W}=m_{9,10}[(h_9-h_{10})-T_{eff}(s_9-s_{10})]$$and$$\dot{\sigma}=-m_{9,10}(s_9-s_{10})\left[1-\frac{T_{eff}}{T_{surr}}\right]$$

So, under this assumption, the work done by the expander on the surrounding would be greater and the entropy generation would be less.

 

[BvU]

Thanks @Chestermiller. It appears @yaro99 hasn't been seen since Wednesday 3:23 AM and therefore has not seen your #11 and #12.
I haven't had a reply on #6 (relatively much work) so I am slightly not amused. There better be a good reason for this playing stumm.

 

[BvU]

Further comments ( I seem to like this exercise)

• What's labeled 'Compressor' in Fig. 4-9 is a compressor with a cooler to bring the temperature back to 25 $^\circ$C. No matter, the exergy analysis is over a control volume that 'excludes the compressor' ? But Fig. 4-20 includes the compressor !
• Table 4-2 should show that stream 11 is the sum of 8 and 10, so 0.8004 kg/kg, not 0.9652. That influences the heat exchange ('cold recovery') considerably !
• The source is Handbook of Liquefied Natural Gas, Saeid Mokhatab et al. I can't see how they reference enthalpy. When I reconstruct the flowsheet (with the stream 11 error) in the freeware Coco flowsheeting program from Amsterchem, the exergies come out fairly OK.

 

[Chestermiller]

• Table 4-2 should show that stream 11 is the sum of 8 and 10, so 0.8004 kg/kg, not 0.9652. That influences the heat exchange ('cold recovery') considerably !

If this is an error in the mass balance, then there must be an error in a stream or streams somewhere else (because the overall mass entering is equal to the overall mass leaving). The question is where?

 

[Chestermiller]

I found the error in their flow sheet. The overall mass balance is not satisfied. In the table, the flow rates of streams 7 and 8 are switched. As the diagram now stands, the makeup feed is not equal to the product flow rate.

 

[BvU]

Except that (with the physical properties as found in Coco/TEA), the flash does come up with a vapour flow of 0.0361 and a liquid flow of 0.1983 (per kg gas in stream 2).
Table 4-2 doesn't have a feed flow rate, but I suppose you refer to the 1-0.9652 -- we agree.

 

[Chestermiller]

Except that (with the physical properties as found in Coco/TEA), the flash does come up with a vapour flow of 0.0361 and a liquid flow of 0.1983 (per kg gas in stream 2).
Table 4-2 doesn't have a feed flow rate, but I suppose you refer to the 1-0.9652 -- we agree.

That's very compelling. So the other alternative would be for streams 11, 12, & 13 to be 0.8004. With that change, I checked, and HX1 would now become adiabatic. I think that clinches it.