Initial state (1): P1 = 7 bar, dry saturated steam (x1 = 1)

Here are the solutions for Question 1, involving the reversible adiabatic expansion of steam. Given: Initial state (1): P_1 = 7 bar, dry saturated steam (x_1 = 1) Final state (2): P_2 = 0.75 bar Process: Reversible and perfectly thermally insulated (adiabatic), meaning it is an isentropic process (s_1 = s_2). All calculations are per kilogram of steam. Step 1: Determine properties at the initial state (State 1). From saturated steam tables at P_1 = 7 bar and x_1 = 1: u_1 = u_g = 2572.5 kJ/kg h_1 = h_g = 2763.5 kJ/kg s_1 = s_g = 6.7085 kJ/(kg K) v_1 = v_g = 0.2729 m^3/kg Step 2: Determine properties at the final state (State 2). Since the process is isentropic, s_2 = s_1 = 6.7085 kJ/(kg K). From saturated steam tables at P_2 = 0.75 bar: s_f = 1.2130 kJ/(kg K) s_g = 7.4560 kJ/(kg K) Since s_f < s_2 < s_g, the steam at state 2 is a wet mixture. Calculate the dryness fraction x_2: s_2 = s_f + x_2 s_fg x_2 = (s_2 - s_f)/(s_g - s_f) = (6.7085 - 1.2130)/(7.4560 - 1.2130) = (5.4955)/(6.2430) = 0.88025 Now, calculate $u_2, h_2