Answered

Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Explore in-depth answers to your questions from a knowledgeable community of experts across different fields. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

Water steam enters a turbine at a temperature of 400 o C and a pressure of 3 MPa. Water saturated vapor exhausts from the turbine at a pressure of 125 kPa. At steady state, the work output of the turbine is 530 kJ/kg. The surrounding air is at 20 o C. Neglect the changes in kinetic energy and potential energy. Determine (20 points) (a) the heat transfer from the turbine to the surroundings per unit mass flow rate, (b) the entropy generation during this process.

Sagot :

batolisis

Answer:

a) -505.229 kJ/Kg

b) -1.724 kJ/kg

Explanation:

T1 = 400°C

P1 = 3 MPa

P2 = 125 kPa

work output   = 530 kJ/kg

surrounding temperature = 20°C = 293 k

A) Calculate heat transfer from Turbine to surroundings

Q = h2 + w - h1

h ( enthalpy )

h1 = 3231.229 kj/kg

enthalpy at P2

h2 = hg = 2676 kj/kg

back to equation 1

Q = 2676 + 50 - 3231.229  = -505.229 kJ/Kg  ( i.e.  heat is lost )

b) Entropy generation

entropy generation = Δs ( surrounding )  + Δs(system)

                                =  - 505.229 / 293   + 0

                                = -1.724 kJ/kg  

We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.