Question

Nitrogen starts at -20 degrees Celsius and finishes at 200 degrees Celsius during an adiabatic process where the index is 1.24. The start pressure is 18bar what is the final pressure in pascal

Answer 1

Answer on Question #45010 – Physics – Molecular Physics | Thermodynamics

Question.

Nitrogen starts at -20 degrees Celsius and finishes at 200 degrees Celsius during an adiabatic process where the index is 1.24. The start pressure is 18 bar what is the final pressure in pascal.

$$T_1 = -20{}^\circ \text{C} = 253\,K$$ $$T_2 = 200{}^\circ \text{C} = 473\,K$$ $$P_1 = 18\,bar = 18 \cdot 10^5\,Pa$$ $$\gamma = 1.24$$ $$P_2 = ?$$

Solution.

The adiabatic equation:

$$PV^\gamma = const$$

But from the equation of ideal gas we know

$$PV = RT \rightarrow \frac{PV}{T} = R = const \rightarrow V = \frac{const}{P} T$$

Therefore,

$$P^{1-\gamma} T^\gamma = const$$

In our case,

$$P_1^{1-\gamma} T_1^\gamma = P_2^{1-\gamma} T_2^\gamma = const \rightarrow \frac{T_1}{T_2} \gamma^\gamma = \frac{P_2}{P_1} \gamma^{1-\gamma}$$

So,

$$P_2 = P_1 \cdot \frac{T_1}{T_2} \gamma^{\frac{\gamma}{1-\gamma}} = P_1 \cdot \frac{T_2}{T_1} \gamma^{\frac{\gamma}{\gamma-1}}$$

Calculate:

$$P_2 = 18 \cdot 10^5 \cdot \frac{473}{253} \gamma^{\frac{1.24}{0.24}} = 18 \cdot 10^5 \cdot 1.87^{5.167} = 18 \cdot 10^5 \cdot 25.387 = 457 \cdot 10^5 = 4.57 \cdot 10^7\,Pa = 45.7\,MPa$$

Answer.

$$P _ {2} = P _ {1} \frac {T _ {2}}{T _ {1}} ^ {\frac {Y}{Y - 1}} = 4.57 \cdot 10^{7} \mathrm{Pa} = 45.7 \mathrm{MPa}$$

http://www.AssignmentExpert.com/