Find the internal energy of 1 kg of superheated steam at a pressure of

1.	Find the internal energy of 1 kg of superheated steam at a pressure of 10 bar and 280ºC. If this steam is expanded to a pressure of 1.6 bar and 0.8 dry, determine the change in internal energy. Assume specific heat of superheated steam as 2.1 kJ/kg-K.
Answer» Given that State 1 : 10 bar 280°C State 2 : 1.6 bar, 0.8 dry Specific heat of superheated steam = 2.1 kJ/kg K Internal energy at state 1 is: u₁ = u_g+ m.c.(T₁ – T_sat) = (h_g - pv_g) + m.c. (T₁ - T_sat) = (2776.2 – 1000 × 0. 19429) + 2.1(280 – 179.88) = 2792.16 kJ/kg Internal energy at state 2; u₂ = uf₂ + xu_f g₂ = (h_f – Pv_f )₂ + x [h_fg – P (v_fg)]₂ = (h_f – Pv_f) + x (h_g – h_f) – P (v_g – v_f )] = (h_f – Pv_f) + x ((h_g – Pvg) – (h_f – P_vf)) = [475.38 + 160 (0.0010547)] + [(2696.2 – 160 * (1.0911)) – (475.38 – 160 (0.0010547))] = 475.21 + 0.8 (2521.62 – 475.21) = 2112.34 kJ/kg Change in internal energy = 211234 - 2792.16 = - 679.82 kJ/kg -ve sign shows the reduction in internal energy

Find the internal energy of 1 kg of superheated steam at a pressure of 10 bar and 280ºC. If this steam is expanded to a pressure of 1.6 bar and 0.8 dry, determine the change in internal energy. Assume specific heat of superheated steam as 2.1 kJ/kg-K.

Answer»

Given that

State 1 : 10 bar 280°C

State 2 : 1.6 bar, 0.8 dry

Specific heat of superheated steam = 2.1 kJ/kg K

Internal energy at state 1 is:

u₁ = u_g+ m.c.(T₁ – T_sat) = (h_g - pv_g) + m.c. (T₁ - T_sat)

= (2776.2 – 1000 × 0. 19429) + 2.1(280 – 179.88) = 2792.16 kJ/kg

Internal energy at state 2;

u₂ = uf₂ + xu_f g₂

= (h_f – Pv_f )₂ + x [h_fg – P (v_fg)]₂

= (h_f – Pv_f) + x (h_g – h_f) – P (v_g – v_f )]

= (h_f – Pv_f) + x ((h_g – Pvg) – (h_f – P_vf))

= [475.38 + 160 (0.0010547)] + [(2696.2 – 160 * (1.0911))

– (475.38 – 160 (0.0010547))]

= 475.21 + 0.8 (2521.62 – 475.21)

= 2112.34 kJ/kg

Change in internal energy = 211234 - 2792.16 = - 679.82 kJ/kg

-ve sign shows the reduction in internal energy