How much steam will be produced if water at 125°C is cooled to 100°C?
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How much steam will be produced if water at 113 kPa and 125°C is reduced to pressure of 101 kPa and 100°C?
The result can also be obtained by calculating the change in enthalpy and dividing it by the heat of vaporization of water
I checked the relevant data: the enthalpy of water at 125°C is 525.33 kJ/kg, while that of water at 100°C is 419.54 kJ/kg. The latent heat of vaporization of water – at which temperature should this value be taken? It’s approximately 2200 kJ/kg. The calculation is as follows: (525.33 – 419.54) / 2200 = 105.79 / 2200 = 0.048. In other words, when 1 ton of water at 125°C is cooled to 100°C, 0.048 * 1 ton = 48 kg of water vapor is produced. Is this calculation correct?
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Dizzy! At 113 kPa and 125°C, water is in the form of steam; even when the pressure is reduced, it remains steam! This question, ugh!
Our plant has a pressure vessel in which the pressure is maintained at 113 kPa, and the temperature at the bottom is 125°C. Continuous drainage occurs from the bottom of this vessel; the main component of this drainage is water. Downstream of the discharge outlet, there is a waste heat recovery system operating at normal pressure, with the temperature controlled at 100°C. We need to calculate how much steam can be generated and how much water at 100°C will remain
At 113 kPa and 125°C, water is in the form of steam; even when the pressure is reduced, it remains steam