Answer to Question #242591 in Mechanical Engineering for Marho1da

Question

Calculate the internal energy and enthalpy of 1kg of air occupying 0.05m ^ 3 at 20bar. If the internal energy is increased by 120kJ as the air is compressed to 50bar, calculate the new volume occupied by 1kg of air. (250.1kJ/kg. 350.1kJ/kg;0.0296m^ 3 )

Answer 1

$U = 120 kJ = n * INT{ Cv dT } = n * INT{ (Cp-R) dT } = n * INT{ Cp dT } - n * R*(T_2-T_1)$ . Here, $R = 8.314 J/mol-K$ , n = Mair / MWair and MWair ~ 29g/mol. Cp is a polynomial that you can look up in your book or on the NIST website: $Cp = A + (B/1000)*T + (C/1000^2)*T^2 + (D/1000^3)*T^3 + (E/1000^-2)/T^2$ , where T is in Kelvin and Cp is in J/mol-K. The values of the constants are: $A = 26.84107, B = 7.7816776, C = -1.8103208, D = 0.14594026, E = -0.01102637$. You can integrate this polynomial fairly easily.

But we do not know T1 or T2. Use the ideal gas equation of state to determine $T1: P1 V1 = n R T1. So, T1 = P1 * V1 / (n * R). I got T1 = 348.81 K = 75.66C.$

The problem is that T2 is the ONLY unknown in the equation. You could use SOLVER in EXCEL or a more powerful math package to determine a value of T2 that yields a ?U of 120 kJ. Once you know T2, use the ideal gas equation of state to determine $V_2: P_2 V_2 = n R T_2 or V_2 = n R T_2 / P_2.$

An approximate solution could be obtained if you assume the Cp of air is a constant, say about 1 kJ/kg-K. In this case, ?U = m Cp * (T2-T1). Solving for T2 yields: T2 = T1 + ?U/(m Cp). Plugging in values yields: T2 = 468.81 K = 195.66C.

Finally, V2 = 0.02688 m^3.