Answer to Question #124313 in Molecular Physics | Thermodynamics for Oreoluwa

Question

12kg of a fluid per minutes goes through a reversible steady flow process. The properties of the fluid at inlet are p1=1.4bar, P1=25kg\m3 C1= 120m\s, U1=920kj\kg and at the exit p2=5.6kg\m3,C2=180m\s,P25kg\m3,U2=720kg\kg. During the passage the fluid rejects 60m. Determine the change in enthalpy and work done during the process

Answer 1

Solution.

$m=12kg/min=0.2kg/s;$

$P_1=1bar=10^5Pa;$

$\rho_1=25kg/m^3;$

$C_1=120m/s;$

$u_1=920kJ/kg=920\sdot10^3J/kg;$

$P_2=5.6bar=5.6\sdot10^5Pa;;$

$\rho_2=5kg/m^3;$

$C_2=180m/s;$

$u_2=720kJ/kg=720\sdot10^3J/kg;$

$Q=60kJ/s=60\sdot10^3J/s;$

$\Delta Z=60m;$

$\Delta h=\Delta u+\Delta(pv);$

$\Delta (pv)=\dfrac{P_2v_2-P_1v_1}{1}=\dfrac{P_2}{\rho_2}-\dfrac{P_1}{\rho_1};$

$\Delta h=(720-920)\sdot10^3+\dfrac{5.6\sdot10^5}{5}-\dfrac{10^5}{25}=-88.48\sdot10^3J/kg=-88.48kJ/kg;$

$Q=\Delta KE+\Delta PE+\Delta h+W;$

$\Delta KE=\dfrac{C_2^2-C_1^2}{2};$

$\Delta KE=\dfrac{180^2-120^2}{2}=9000J=9kJ$ or $=9kJ/kg;$

$\Delta PE=\Delta Zg;$

$\Delta PE=60\sdot9.81=588.6J =0.588kJ$ or $=0.588kJ/kg;$

$Q=\dfrac{60\sdot10^3}{0.2}=300kJ/kg;$

$-300=9+0.588-88.48+W;$

$W=-221.1kJ/kg=-221.1kW;$

Answer: $\Delta h=-88.48kJ/kg;$

$W=-221.1kW.$