0
$\begingroup$

We all know the basic understanding of the adiabatic phenomenon: ideal gas expands and cools, compresses and heats up. Now by expand, let take a quick look at a scenario: let assume there is a tank filled with nitrogen gas at 300°C pressurized to 4500 psi, 468 g of the gas is vented through a tube into a large balloon and the tank pressure drops to 120 psi in 10 seconds. The balloon reaches a pressure of 20 psi through the process. Based on adiabatics, to what temperature does the gas cools by expanding through the tube and into the balloon or does it even cools at all?

$\endgroup$
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Feb 23 '19 at 15:16
-1
$\begingroup$

$$T=(300+273)\left(\frac{134.7}{4514.7}\right)^{0.286}=210\ K=-63 C$$

ADDENDUM

Here are some calculations and analysis for your consideration based on ideal gas:

Initial pressure = 4500 psig = 307.1 atm.

Final pressure = 120 psig = 9.16 atm.

$$T_f=T_i\left(\frac{P_f}{P_i}\right)^{2/7}=210\ K=-63\ C$$ Calculation of initial moles $n_i$ in tank: $$n_i=\frac{500}{28}=17.857$$ Calculation of tank volume:$$V_T=\frac{n_iRT_i}{P_i}=\frac{(17.857)(0.08205)(573)}{307.1}=2.735\ liters$$ Calculation of final number of moles in tank: $$n_f=\frac{P_fV_T}{RT_f}=\frac{(2.735)(9.1633)}{(0.08205)(210)}=1.454\ moles=40.7\ grams$$ Application of the first law of thermodynamics to get the final temperature in the balloon:

Treating the mass of gas in the tank and the balloon as our closed system, and applying the first law of thermodynamics, we have: $$\Delta U=-\int_0^{V_f}{P_{ext}dV}$$where V is the volume of the balloon, and the change in internal energy of the gas is given by: $$\Delta U=n_fC_v(T_f-T_i)+(n_f-n_i)C_v(T-T_i)=(1.454)(2.5R)(210-573)+16.4(2.5R)(T-573)$$

We don't know the external pressure applied by the balloon rubber membrane to the gas because we don't know how the rubber behaves in terms of its stress strain behavior. But, we do know that the final pressure in the balloon is 20 psig. So, we rewrite the work term in our energy balance equation in a little different way:$$\int_0^{V_f}{P_{ext}dV}=\overline{P_{ext}}V_f=\frac{\overline{P_{ext}}}{P_{ext,f}}P_{ext,f}V_f$$where $$\overline{P_{ext}}=\frac{1}{V_f}\int_0^{V_f}{P_{ext}dV}$$We know that the ratio of the average external pressure to the final external pressure is going to be less than 1 because the pressure in the balloon is going to be increasing monotonically with time. So we can re-express the work as $$\int_0^{V_f}{P_{ext}dV}=kP_{ext,f}V_f$$with (0 < k < 1). From the ideal gas law, we also know that $$P_{ext,f}V_f=(n_i-n_f)RT$$So, combining previous equations, we have: $$n_fC_v(T_f-T_i)+(n_f-n_i)C_v(T-T_i)=-k(n_i-n_f)RT$$or equivalently: $$(1.454)(2.5)(210-573)+16.4(2.5)(T-573)=-k(16.4)T$$ Solving this for values over the allowable range of k, we obtain:

T = 603 K = 325 C for k = 0

T = 432 K = 159 C for k = 1

T = 506 K = 234 C for k = 0.5

Questions?

$\endgroup$
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Feb 23 '19 at 15:16
  • $\begingroup$ @ Chester, I've got questions for you, check out the chat. $\endgroup$ – TechDroid Feb 24 '19 at 10:00
  • $\begingroup$ I checked out the chat, but I don't see any questions there. Why don't you ask them in comments right below here? $\endgroup$ – Chet Miller Feb 24 '19 at 12:33
  • $\begingroup$ So in your newly added analysis, the final values for Ts over the k interval (0,1) are the final temperature for what? And what condition does k represent in a practical sense. If they are the final temperature in the balloon, then where goes the adiabatic cooling. $\endgroup$ – TechDroid Feb 24 '19 at 13:15
  • $\begingroup$ For we have come this far, I think it's time I pour out my thoughts and understanding of the situation so far. Since the adiabatic cooling happens in the tank, I believe as temperature in the tank drops, temperature of gas leaving it also drops. Now let assume we track every 10 g of nitrogen leaving the tank, so definitely every 10 g is cooler than the last, and the last 10 is as cool as roughly ~-50°C. So averaging out the total change in temperature from 300°C to -50°C, we should be at a temperature low enough. $\endgroup$ – TechDroid Feb 24 '19 at 13:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.