A distillation unit is to be used to generate pure water vapor. 2.0 kg/s of liquid water at 25oC enters the system, and 0.2 kg/s of saturated water vapor leaves through one port and 1.8 kg/s of saturated liquid leaves through a second port (along with impurities in the water which can be neglected in the analysis). The pressure throughout the whole system is 100 kPa. There is no work done in the process, and all energy is added through heat transfer. (a) Determine the rate of heat transfer needed to produce this flow of saturated water vapor. (b) Using your separation chamber model, plot the heat transfer rates needed to produce saturated water vapor flow rates between 0.1 kg/s and 1.0 kg/s (along with corresponding changes to the saturated liquid water exit flow rates).

Question

State 1: Inlet stream; State 2: Saturated vapor exit stream; State 3: Saturated liquid exit.
Given: P1 = P2 = P3 = 100 kPa; T1 = 20oC; x2 = 1.0; x3 = 0.0; m?1=2.0kgs; m?2=0.2kgs; m?3=1.80kgs

Assume: W?=0 (given). Given no other information regarding the mixing chamber, make the following common assumptions: ?KE=?PE=0. We are told there is heat transfer.
Also, assume the mixing chamber is a single-inlet, multiple-outlet, steady-state, steady-flow device.
What will be an ideal response?

Answer 1

Solution: The First Law for Open Systems reduces to Q?=(m?2h2+m?3h3)?m?1h1

For the entering liquid water, consider it to be a slightly compressed liquid (h1 = hf @T1)

For water: h1 = 83.92 kJ/kg; h2 = 2674.95 kJ/kg; h3 = 417.44 kJ/kg

(a) Solving: ?????= 1,120 kW

(b) Keeping in mind the conservation of mass: m?1=m?2+m?3

Plotting the heat transfer rate for the range of mass flow rates for the saturated vapor (m?2):

Much of the heat transfer is bringing the mixture up to the boiling temperature. Then, more heat is needed to boil some of the liquid water into vapor.