Question

Steam at 150 bars and 600 ^oC passes through process equipment and emerges at 100 bars and 700 ^oC. There is no flow of work into or out of the equipment, but heat is transferred. Using the steam tables, compute the flow of heat into the process equipment per kg of steam

b.) Compute the value of enthalpy at the inlet conditions, Hin, relative to an ideal gas at the same temperature, Hig. Consider steam at 1 bar and 600 ^oC as an ideal gas. Express your answer as (Hin – Hig)/RTin

Answer 1

Inlet conditions of steam: Pressure, P = 150 bar Temperature. T1 = 600°C Outlet conditions of steam: Pressure.Pa = 100 bar Temperature. T2 = 700°C The open system energy balance equation is: Q = AH+W. Here there is no flow work. Then, Q = AH+W! Q=AH To get the flow of heat into the process, get the values of specific enthalpies at inlet and outlet conditions from steam tables. At inlet conditions: At P = 150 bar and T1 = 600°C the specific enthalpy value is H* = 3583.16 kg At outlet conditions: At P2 = 100 bar and T2 = 700°C the specific enthalpy value is H** = 3870 kad kg Then, flow of heat into the process will be, Q=AH = H - Hin ke Q= 287

(6) The steam at the inlet conditions is ideal gas. The enthalpy value at the inlet conditions is H" = 3583.1 The enthalpy at the ideal conditions of 1 bar and 600°C is H' = 3706. | 60 Hin- Now calculate. RTS Hi-H: 3583.1--3706.31 8.314 mo. K (600 +273) K (3583.1-37063)-xmol-wt (8.314x873) med (35831–3706.3)~15 min (8.314x873) (3583.1-3706.3)x18 (8.314x873) Hin-H RTin -=-0.3055