Can some one please solve the following problem and work through it showing and
ID: 2995270 • Letter: C
Question
Can some one please solve the following problem and work through it showing and explaining every step clearly? The only way I can learn from the solution is if I can clearly follow each step. I will award points to the most thorough and easy to follow answer. Thanks!!!!!!
Your company uses 1.4 kg/s of high pressure steam for some industrial process. After being used for this process, the waste steam is saturated vapor at 50 bar. You currently just run the steam through a condenser, rejecting thermal energy to the atmosphere, before re-circulating it to the boiler as a sub-cooled liquid. A consultant has proposed (and offered to sell you) a proprietary power cycle that would produce work from the waste steam. She claims you can get 725 kW of power from her cycle while condensing your steam steam at constant pressure to a saturated liquid. In other words, the condensing steam serves as the high-temperature thermal reservoir for the power cycle. (Physically, this would be a heat exchanger with the steam condensing on one side and the working fluid of the power cycle on the other side.)
Do a preliminary analysis of this proposal, based on the 2nd Law. Be sure to include your assumptions, theory, equations, and comment on engineering feasibility, both theoretical and practicalyes or no is NOT good enough! (Dont worry about economics; only thermodynamics.) A sketch might help you picture the various energy flows.
Thanks!!!
Explanation / Answer
Given exit steam as saturated Vapor (Inlet to the condenser)
Properties of Steam at P1 = 50 bar are (from steam tables)
T1 = Tsat = 264 deg = 537 K
h1 = hg = 2794 kJ/kg
Exit of the condenser is in SAturated liquid state and Pressure is Constant
P2 = P1 = 50 bar
T2 = Tsat = 264 deg = 537 K
h2 = hf = 1154 kJ/kg
Given mass flow rate of steam, m = 1.4 kg/s
From First law of thermodynamics
Heat transfer,Q
Q = m*(h1-h2) = 1.4*(2794-1154) = 2296 kW
Q = 2296 kW
From the second law of Thermodynamics
Exergy, E = Q*(1-Tl/Th)
Tl is the lower temperature (or ambient temp = 300 K)
Th is the Higher Temperature,Th = 537 K
E = 2296*(1-300/537)
E = 1013 kW