An insulated centrifugal separator is to be used to separate the liquid and vapor portions of a saturated mixture flow. A saturated mixture of water with a quality of 0.30 and a temperature of 120oC enters the system with a mass flow rate of 4.50 kg/s. Saturated liquid water exits through one exhaust pipe at a temperature of 120oC at a rate of 3.00 kg/s, and 1.50 kg/s of saturated water vapor exit at 120oC through the other exhaust pipe. (a) Determine the amount of power that was added to the system to generate these outlet flows. (b) Using your separation chamber model, plot the power required to produce flows of saturated vapor at 120oC ranging from 1.40 kg/s to 4.0 kg/s.
State 1: Inlet stream; State 2: Saturated liquid exit stream; State 3: Saturated vapor exit.
Given: T1 = T2 = T3 = 120oC; x1 = 0.30; x2 = 0.0; x3 = 0.0; m?1=4.50kgs; m?2=3.00kgs; m?3=1.50kgs
Assume: Q?=0 (insulated). Given no other information regarding the mixing chamber, make the following common assumptions: ?KE=?PE=0. We are told there is power.
Also, assume the mixing chamber is a single-inlet, multiple-outlet, steady-state, steady-flow device.
What will be an ideal response?
Solution: The First Law for Open Systems reduces to W?=m?1h1?(m?2h2+m?3h3)
(a) For water: h1 = 1164.43 kJ/kg; h2 = 503.78 kJ/kg; h3 = 2705.93 kJ/kg
Solving: W?=????????????? ????????
(b) Keeping in mind the conservation of mass: m?1=m?2+m?3
Plotting the power for the range of mass flow rates for the saturated vapor (m?3):
Keep in mind, originally there is only (0.3)(4.5kg/s)= 1.35 kg/s of saturated vapor. To create a larger mass flow rate of vapor, there needs to be an input of energy – in this case work. So, to get more saturated vapor, one needs more power (a larger negative value).
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