Homework hints

Again, careful with the units! You will most likely need to convert liters to m³.

Thermo books can get a little sloppy with their language sometimes. It would be better to say "find the power per unit mass flow rate" than "find the work per unit mass". That is, W_dot/m_dot instead of W/m.

You will need conservation of energy for a closed system over a finite time, the ideal gas equation, and relationships for Δu and Δs for an ideal gas with constant specific heats. But wait, there's more! pdV work is present too!

A useful relation to remember: R = c_p - c_v.

Does this look almost exactly the same as 2-1? Yes! Will the answers be about the same? Yes! But will the tools be different? Yes, yes, yes! You now must use the ideal gas tables for air to find u and s^o (not s–remember that s^o is only a piece of s).

A useful relation to remember: R = c_p - c_v.

You already have experience finding property values. Here is more, 'cuz we want you to be perfect. (Well, perfecter. ☺)

Equally as important here is learning to identify the states on P-v and T-v diagrams. We have seen in class that skething these before applying conservation equations can save you muchísimo tiempo.

What is the purpose of a compressor? Does power go in or out?

What are you going to do with "adiabatic" and "reversible" given to you in the problem statement? Is this enough to fix the state of the stuff leaving the compressor (meaning you have two independent, intentive properties)?

Can you use ∫PdV to find work? Well yes you can! But the result will be trivial, yes?

Note that the authors are asking you to find heat transfer in units of B/lbm, which is more appropriately called "heat transfer per unit mass." This is important because you do not know what the mass is, nor can you ever solve for it! An extra variable m will be floating around in your solutions, then. (One trick you can use is to set m=1 lbm, and then everything is by definition per unit mass.)

Can you use ∫PdV to find work? Theoretically yes, but you have know idea what the function P=P(V) is, and so you can't solve that integral. Remember, this is a reversible process. What is the area under a T-s curve, then? Do you know the function T=T(s) is? (Yep.)

Also see hints in 3.35 about the whole "per unit mass" deal.

Since you are given an isentropic efficiency, you will need to write conservation of energy for a pretend compressor for which s₂=s₁ to find the pretend work/mass (power/mass flow rate) first. How does the real work/mass (power/mass flow rate) compare to that? Now that you know the real work/mass, how will you find the real exit temperature?

This is an ideal gas with constant specific heats and so you don't need any property tables here! Woo-hoo! Use a value of c_p = 1.004 kJ/kg·K to get the answers here.

You will need the relationships for Δh and Δs that you learned in ConAps and that we reviewed early in the quarter. If you are using a bunch of plug and chug formulas that make use of k then you most likely don't understand the concepts fully. We'll make use of k relations later in the quarter...

You will need to assume a value of T₀. 20°C is a typical choice. (Can I leave it as 20°C though?)

Remember that you must evaluate the properties of the stuff at T₀ and P₀, not the properties of the environment itself when finding exergy.

A throttling process is designed to greatly reduce pressure over a small distance in the direction of flow. Throttling valves have no moving parts, you don't plug them in, and the surface area is too small for there to be any heat tranfer.

A throttling valve

The rate of exergy coming in/going out is due to mass flow. Which term(s) in the Accounting of Exergy are those?

As is often the case, you will have to assume a T₀ and/or P₀. Typical values are around STP. I assumed 20°C for the answer provided here.

Since you do not have a mass gflow rate, you can oly calculate a heat transfer per unit massf low rate, q=Q_dot/m_dot. If you prefer, you can just make m_dot=1kg/s, and by definition everything is now per unit mass flow rate.

In the case of the waster heat, the exergy destruction is due to heat transfer through a finite temperature difference. In the resistive heating case, it is due to the Ohm's Law (electrical resistance is basically electric friction). Though you will find that the exergetic efficiency for the waste heat system is higher, that could change drastically as T_b increases.

1. 1.11×10⁵ kW
2. 2.87×10⁵ kW
3. 38.8%
4. 2335 kg/s

In some sense this is straight forward—just go around the cycle and analyze one component at a time and apply property relations. In another sense it is hard, because sometimes you will have to go on to the next component before you have everything you need. Try not to let that bug you. Eventually you will have all you need.

Speaking of which, note that the mass flow rates going through different parts of the cycle are different. For example, the high pressure turbine and the low pressure turbine have different flow rates. How might you deal with that? I would suggest assuming that the massflow rate through the first stage (high pressure) turbine is 1.0 kg/s. The mass flow rate going through the second stage (low pressure) is an unknown. When you eventually get to analyzing the open feedwater heater you will be able to get that number.

Also keep in mind that when you are calculating the efficiency, these different mass flow rates will need to be accounted for.

1. 2611 kJ/kg (or kW/[kg/s])
2. 34.7 %

I highly recommend you work in symbols only, and then automate the number crunching using the ideal gas property calculator. Just plug in the new values of P₂ and run with it. Remember, however, that the electronic property calculator is not the same set of tables as in your text. Namely, there is no s⁰ function in the electronic version. However, you shouldn't need it, should you?

The back work ratio is nothing more than the ratio of pump power in to power out (not net power out). Note that these are larger for gas vapor cycles than for vapor power cycles. Why is that?

Remember, each step in the cycle is made up of a finite time, closed system process. Apply CoE and property relations to each step, and go on to the next one.

This is a cold air standard cycle! "But it just says 'air standard,' you say. Yes! But notice how you are also given a constant value of c_v? Hence, cold air standard...

As in 9.3, each step in the cycle is made up of a finite time, closed system process. Apply CoE and property relations to each step, and go on to the next one. It is also useful to remember that for an ideal gas, R = c_p - c_v.

You can find W_2→3 with CoE, or you can use W_2→3 = m∫₂³Pdv where pv^k = constant. You should be able to tell me why!

Lastly, note that from 3 to 1, there is both Q and W.

1.  
2. 16.2%

1. 0.144 kg/min
2. 55.4 kW
3. 2.71
4. 88.8%

1.  
2. 43.65 lbm/lbmol
3.  
4. 100.2 ft³

I highly recommend you pick a closed system consisting of both gases before and then afer the valve in opened. ("Before" and "after"? Hmmm... Wonder what form of CoE I should use?)

You could try pdV work, but you don't know how p=p(V). And so, why not read the paragraph above?

In reality, the only new part of this problem compared to others in the past is that you have more work to do in finding your properties, since it is a mixture of ideal gases. Treat the mixture as an ideal gas with variable specific heats. And yes! you can use your magic electronic property tables, but use the program for ideal gases, and use it on a molar basis.

As long as the composition isn't changing (and it's not) any specific property can be found with

What would this be on a molar basis?

You'll need incroporate the the mixture composition into conservation of energy as well. Try to massage your conservation of energy equation to the point that you have separate expressions for the specific enthalpy change for both Ar and CO₂. Also note that you will NOT use property tables here, but rather assume constant specific heats to find the changes in h.

In part (c) don't forget that Dalton's Model requires partial pressures for specific entropy changes!

Keep in mind that φ=P_v/P_g. What temperature will you use for P_v? For P_g? Why?

If no condensation occurs, then ω₂=ω₁. If condensation does occur, that what does ω₁ - ω₂ give you?

Also helpful here is ω = 0.622[P_v/(P-P_v)] = 0.622[P_v/(P-φP_g)], where P is the plain old pressure (total pressure, mixture pressure, barmoetric pressure, whatever you wanna' call it.) This works for ω₁ and ω₂.

Now there is a really hard way to do this, which involves using the equations to find your properties instead of the pysch chart. (I.e., h_mix=h_a(T)+ωh_g(T) and h_w=h_f and so on.) That will give you an analytic solution to solve for ω explicitly. (And if you cheat and use the solutions manual, that is what you will do and probably not understand any of it.) But there is another way that is much easier and that does let you use the pysch chart. What is it?

Alrighty, then. Try this. Do conservation of mass for the water only and then substitute that result into conservation of energy. You will get a term in CoE that looks like this: (ω₂-ω₁)h_f. What are typical values of ω? How big would (ω₂-ω₁) be, then? And finally, how big is (ω₂-ω₁)h_f compared to h_mix1 and h_mix,2? That's right! Virtually negligible! Give that a shot...

Almost exactly the same as in-class example, but with different numbers

If you did not use the pysch chart, you would have to iterate! Also note that this is a case where the conservation of mass on dry air only is imporant. The rates of dry air at (1), (2), and (3) are all different!

You absolutely must—no two ways about it, can't even for a second consider not—balance the theoretical reaction first. Percent excess, percent theoretical, etc. are referenced to the theoretical reaction, and cannot be taken into account in a balanced reaction final reaction in one step.

You will have to go back to psychrometric basics to do part (b)—you cannot use the pysch chart. (You should be able to tell me why.)