Hyperloop: Compressor work

First we have to determine the constants $A$ and $\alpha$ . We will do this using the information given: $P_1=99 \text{ Pa}, P_2=2100 \text{ Pa}, T_1=293 \text{ K}, T_2=857 \text{ K}$ .

Using the ideal gas law $PV = nRT$ , we will find the corresponding volumes $V_1$ and $V_2$ .

The number of moles of air compressed per second is $n= \frac{490 \text{ g}}{29 \text{ g/mol}} = 16.897 \text{ mol}$ Thus we have $V_1=\frac{nRT_1}{P_1}=415.6 \text{ m}^3$ $V_2=\frac{nRT_2}{P_2}=57.3\text{ m}^3$

Now, since we are given that $P=AV^{-\alpha}$ , we know that $P_1=AV_1^{-\alpha}$ and $P_2=AV_2^{-\alpha}$ . Dividing the first equation by the second equation gives us $\frac{P_1}{P_2}=\left(\frac{V_1}{V_2}\right)^{-\alpha}=\left(\frac{V_2}{V_1}\right)^{\alpha}$ .

Substituting the numbers, we will get $\frac{99}{2100}=\left(\frac{57.3}{415.6}\right)^{\alpha}$ , which yields $\alpha = \frac{\log 0.04714}{\log 0.1379}=1.54$

Substituting this value of $\alpha$ into $P_1=AV_1^{-\alpha}$ gives us $99 = A(415.6)^{-1.54}$ . Hence $A=1.0676 \times 10^6$ .

Now the work done on the gas per second can now be calculated easily: $W= -\int_{415.6}^{57.3} P \, \mathrm{d} V = -\int_{415.6}^{57.3} AV^{-\alpha}\, \mathrm{d} V =\left. \frac{AV^{-0.54}}{0.54}\right|_{415.6}^{57.3}$ Putting in the numbers gives us $W = \frac{1.0676\times 10^{-6}}{0.54} (57.3^{-0.54} - 415.6^{-0.54}) = 145.9 \mathrm{ kJ}$

In each second, the input energy is $276 \mathrm{ kJ}$ . Thus the efficiency is $\eta = \frac{145.9}{276} = 0.53$

First, we determine $\alpha$ .

Since $P=AV^{-\alpha}$ and $\frac{PV}{T}=const$ , so $TP^{\frac{1-\alpha}{\alpha}}=const$ or $T_1P^{\frac{1-\alpha}{\alpha}}_1=T_2P^{\frac{1-\alpha}{\alpha}}_2$ .

Hence, $\frac{1-\alpha}{\alpha}=\frac{ln{\frac{T_1}{T_2}}}{ln{\frac{P_2}{P_1}}}$

$\alpha \approx 1.54$

Work done on the gas in time t is: $A=\int -PdV=\int\limits_{V_1}^{V_2} -P_1 V^{\alpha}_1 V^{-\alpha} dV$

$A=\frac{P_1 V^{\alpha}_1 (V_1^{1-\alpha} - V_2^{1-\alpha})}{1-\alpha}=\frac{P_1 V_1- P_2 V_2}{1-\alpha}=\frac{mR(T_2-T_1)}{\mu(\alpha-1)}$

$A=\frac{ptR(T_2-T_1)}{\mu(\alpha-1)}$ with p=0.49kg/s.

Therefore, $\eta=\frac{A}{P_0t}=\frac{pR(T_2-T_1)}{\mu P_0(\alpha-1)}=0.53$

First we find constants A and $\alpha$ :

$V_{1}=\frac{RT_{1}}{P_{1}}=24.61m^{3} ; V_{2}=\frac{RT_{2}}{P_{2}}=3.393m^{3}$ (note that this is the volume per mole)

Since $P=AV^{-\alpha} \Rightarrow \alpha = -log_{\frac{24.61}{3.393}} \frac{99}{2100} = 1.54 \rightarrow A=13738.7$

Then we find work done per mole: $W_{per mole} =$ integrals of $AV^{-\alpha}dV$ from $V_{1}$ to $V_{2} = 8641.54 (J)$

$\Rightarrow W_(s) = 8641.54 \times \frac{0.49}{0.029} = 146 (kW)$

$\Rightarrow \eta = 0.53$