portuguese
Newbie level 6
Hello! I'm studying for a Thermodynamics exam and I need to solve these problems in EES (Engineering Equation Solver). Below each question, there's my resolution. Can someone confirm if they are right?
1) The vapor quality of water, using u=2000 kJ/kg and s=6 kJ/kg×K (Steam_IAPWS tables).
x_1=Quality(Steam_IAPWS;s=6;u=2000)
2) The specific volume of water, using x=0.8 e h=2400 kJ/kg (Steam_IAPWS tables).
v_2=Volume(Steam_IAPWS;h=2400;x=0,8)
3) The enthalpy and temperature of R134a fluid, using v=0.8 m^3/kg e u=300 kJ/kg (R134a tables). In which phase is the fluid? Justify your answer using a calculation.
h_3=Enthalpy(R134a;v=0,8;u=300)
T_3=Temperature(R134a;v=0,8;u=300)
Phase: I know it's superheated steam, but how do I justify with a calculation?
4) Compare, calculating the percent error, the specific volume of air, real gas, with the specific volume of air, ideal gas, for T=-120 ºC and P=2000 kPa.
v_4=Volume(Air_ha;T=-120;P=2000)
v_5=Volume(Air;T=-120;P=2000)
error_1=((v_5-v_4)/v_4)*100
5) Compare, calculating the percent error, the specific volume of water, real gas, with the specific volume of water, ideal gas, for T=-120 ºC and P=2000 kPa.
v_6=Volume(Water;T=380;P=23000)
v_7=Volume(H2O;T=380;P=23000)
error_2=((v_7-v_6)/v_6)*100
Thanks!
1) The vapor quality of water, using u=2000 kJ/kg and s=6 kJ/kg×K (Steam_IAPWS tables).
x_1=Quality(Steam_IAPWS;s=6;u=2000)
2) The specific volume of water, using x=0.8 e h=2400 kJ/kg (Steam_IAPWS tables).
v_2=Volume(Steam_IAPWS;h=2400;x=0,8)
3) The enthalpy and temperature of R134a fluid, using v=0.8 m^3/kg e u=300 kJ/kg (R134a tables). In which phase is the fluid? Justify your answer using a calculation.
h_3=Enthalpy(R134a;v=0,8;u=300)
T_3=Temperature(R134a;v=0,8;u=300)
Phase: I know it's superheated steam, but how do I justify with a calculation?
4) Compare, calculating the percent error, the specific volume of air, real gas, with the specific volume of air, ideal gas, for T=-120 ºC and P=2000 kPa.
v_4=Volume(Air_ha;T=-120;P=2000)
v_5=Volume(Air;T=-120;P=2000)
error_1=((v_5-v_4)/v_4)*100
5) Compare, calculating the percent error, the specific volume of water, real gas, with the specific volume of water, ideal gas, for T=-120 ºC and P=2000 kPa.
v_6=Volume(Water;T=380;P=23000)
v_7=Volume(H2O;T=380;P=23000)
error_2=((v_7-v_6)/v_6)*100
Thanks!