Question

Steam at 150 bars and 600°C passes through process equipment and emerges at 100 bars and 700°C. There is no flow of work into or out of the equipment, but heat is transferred. a. Compute the flow of heat into the process equipment per kg of steam.b. Compute the value of enthalpy at the inlet conditions.Consider steam at 1 bar and 600°C as an ideal gas. Express your answer as (Hin – Hig)/RTin

Answer 1

Answer:

a) q_in = 286.9 KJ/kg

b) H_ig = 1918.9413 KJ/kg ,  (Hin – Hig)/RTin = 4.43

Explanation:

Given:

State 1: Steam @

  - P_1 = 150 bars

  - T_1 = 600 C

State 1: Steam @

  - P_2 = 100 bars

  - T_1 = 700 C

- No work done W_net = 0

Find:

- q_in

- Compare h_1 from property table with h_1 with ideal gas steam @P = 1 bar and T = 600 C.

Solution:

State 1:

Use Table A-5

h_1 = 3583.1 KJ/kg

State 2:

Use Table A-5

h_2 = 3870 KJ/kg

- The Thermodynamic balance is as follows:

                                q_in - w = h_2 - h_1

                                q_in = h_2 - h_1

                                q_in = 3870 - 3583.1 = 286.9 KJ/kg

- Enthalpy H_ig of ideal gas @P = 1 bar and T_in = 600 C

                                c_p = 2.1981 KJ / kg.K         @ T_in = 600 + 273 = 873 K

                                H_ig = c_p*T_in

                                H_ig = 2.1981*873 = 1918.9413 KJ/kg

- Enthalpy H_in of steam @P = 1 bar and T_in = 600 C

  Using Table A-5

  H_in = 3705.6 KJ/kg

- Relative Enthalpy:

                               (Hin – Hig)/RTin = (3705.6 - 1918.9413) / 0.462*873

                               (Hin – Hig)/RTin = 4.43