Steam heat exchange problem

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Seeking advice from experts: I am currently trying to calculate the thermal efficiency. The main calculations involve: 1) Converting water at 0°C into steam (under conditions of 0.1 MPa). For the evaporation of water vapor, heat exchange is carried out using steam at pressures of 0.5 MPa, 0.6 MPa, and 0.7 MPa respectively. After heat exchange, the steam turns into saturated water at 0.5 MPa, 0.6 MPa, and 0.7 MPa respectively. Calculate the ratio of the amount of steam consumed to the amount of water evaporated. For example: 1.3 kilograms of steam are required to evaporate 1 kilogram of water (the steam is converted into saturated water at 0.1 MPa). 1.4 kilograms of steam are required to evaporate 1 kilogram of water (the steam is converted into saturated water at 0.5 MPa). To evaporate water at 25°C into steam (under conditions of 0.1 MPa), steam with pressures of 0.5 MPa, 0.6 MPa, and 0.7 MPa is used respectively. After heat exchange, the steam turns into saturated water at 0.5 MPa, 0.6 MPa, and 0.7 MPa respectively. Calculate the ratio of steam consumed to water evaporated. For example: 1.2 kilograms of steam are required to evaporate 1 kilogram of water (with the steam converting into saturated water at 0.1 MPa), and 1.3 kilograms of steam are needed to evaporate 1 kilogram of water (with the steam converting into saturated water at 0.5 MPa). Question 2 is identical to Question 1, except that the basic conditions for water evaporation are 0 degrees and 25 degrees respectively. Is it correct to use the following method for calculation? First, the steam at 0.5 MPa is converted into saturated water at 0.5 MPa; the energy released is calculated by subtracting their specific enthalpies: 2748.11 – 640.19 = 2107.92 ①. Next, water at 0°C and 0.1 MPa is converted into steam at 100°C and 0.1 MPa; the energy required for this conversion is calculated in the same way by subtracting their specific enthalpies: 2676 – 0.06 = 2675.94 ②. Then, dividing ② by ① gives the ratio of steam consumed to water evaporated. Similarly, the energy released when steam at 0.5 MPa is converted into saturated water at 0.1 MPa is calculated as 2748.11 – 417.51 = 2258.94 ③. Dividing ② by ③ yields another ratio relating the amount of steam consumed to the amount of water evaporated (in this case, when the steam is converted into water at 0.1 MPa). The same method is applied when the water temperature is 25°C. I’m asking whether this method is correct: Please ask experts for help. Is there a better method or software? Experts, please advise me. Moreover, the higher the pressure, the more steam is consumed in converting it to water at the same pressure, and the ratio of this steam consumption to the amount of water that evaporates actually increases, meaning that more energy is used. Is this phenomenon correct? I would appreciate it if an expert could explain this to me.