Superheated steam at 280 degrees is cooled to become saturated steam at 160 degrees – how can one calculate the additional amount of steam produced?
Thread Content
Dear seniors, with a flow rate of 50 t/h, superheated steam at 280 degrees is cooled to become saturated steam at 160 degrees. How can we calculate how much more saturated steam is produced? Can it be calculated as one ton of cooling water per ton of steam? Is there a formula for calculation?
Establish equations for energy balance and mass balance, and solve them together
Just calculate it based on enthalpy value. The pressure of saturated steam at 160 degrees is 0.618 MPa (a); assuming that the pressure remains constant during the temperature reduction process, and thus the pressure of the superheated steam is also considered to be the same, the enthalpy value of superheated steam at 280 degrees can be determined. The enthalpy value of the cooling water is also required. The energy balance equation is: the enthalpy of the 50t superheater plus the enthalpy of the n tons of desuperheated water equals the enthalpy of (50+n) tons of saturated steam at 0.618 MPa(a) and 160 degrees Celsius. The only unknown is n; once that is determined, the solution is straightforward. The key lies in determining the state parameters of the superheated steam and the desuperheating water, so as to find their enthalpy values.
The more cooling water that is injected, the more tons of saturated steam will be produced