Question

Steam produced in a boiler is frequently “wet”—that is, it is a mist composed of saturated water vapor and entrained liquid droplets. The quality of a wet steam is defined as the fraction of the mixture by mass that is vapor.

A wet steam at a pressure of 5.0 bar with a quality of 0.85 is isothermally “dried” by evaporating the entrained liquid. The flow rate of the dried steam is 52.5 m³/h.

(a) Use the steam tables to determine the temperature at which this operation occurs, the specific enthalpies of the wet and dry steams, and the total mass flow rate of the process stream.

(b) Calculate the heat input (kW) required for the evaporation process.

Answer 1

• Given : Pressure of Steam is 5 bar
• Quality x= 0.85
• Also given that flow rate of dried steam is 52.5 m³/h

• So first from the steam table the saturation temperature at Pressure P= 5.0 bar

T_sat = 151.831 ^oC                  -----------------------Answer

• Also from saturated steam table at 5 bar

Specific enthalpies of Saturated Liquid and Vapor are

h_f = 640.09 kJ/Kg

h_f = 2748.1 kJ/Kg

• Therefore specific enthalpy of wet steam with quality x= 0.85

h_wet = h_f + x* (h_g - h_f )

h_wet = 640.09 + 0.85* (2748.1 – 640.09 ) kJ/Kg

h_wet = 2431.8985 kJ/Kg

• Since dry steam is the saturated steam at 5 bar so its specific enthalpy is the specific enthalpy of saturated vapor.

h_dry = h_g = 2748.1 kJ/Kg

• Given the volumetric flow rate of dry steam = 52.5 m³/h

Now from saturated steam table at 5 bar , specific volume of saturated vapor

v_g = 0.37481 m³/Kg

Therefore, the mass flow rate of the process steam is

Mass flow rate:

(b) Now heat supplied to the steam

Q = m(h_dry – h_­wet )

Q = 140.07 *(2748.1 - 2431.89 ) kJ/h

Q = 44291. 53 kJ/h