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): $P_1 = 7 \text{ bar}$, dry saturated steam ($x_1 = 1$) Final state (2): $P_2 = 0.75 \text{ bar}$ 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 $P_1 = 7 \text{ bar}$ and $x_1 = 1$: $u_1 = u_g = 2572.5 \text{ kJ/kg}$ $h_1 = h_g = 2763.5 \text{ kJ/kg}$ $s_1 = s_g = 6.7085 \text{ kJ/(kg K)}$ $v_1 = v_g = 0.2729 m^3/\text{kg}$

Step 2: Determine properties at the final state (State 2). Since the process is isentropic, $s_2 = s_1 = 6.7085 \text{ kJ/(kg K)}$. From saturated steam tables at $P_2 = 0.75 \text{ bar}$: $s_f = 1.2130 \text{ kJ/(kg K)}$ $s_g = 7.4560 \text{ kJ/(kg K)}$ Since $s_f < s_2 < s_g$, the steam at state 2 is a wet mixture. Calculate the dryness fraction $x_2$: $s_2 = s_f + x_2 s_{fg}$ $x_2 = \frac{s_2 - s_f}{s_g - s_f} = \frac{6.7085 - 1.2130}{7.4560 - 1.2130} = \frac{5.4955}{6.2430} = 0.88025$ Now, calculate $u_2, h_2