Answered

Engage with knowledgeable experts and get accurate answers on IDNLearn.com. Our platform offers reliable and detailed answers, ensuring you have the information you need.

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 :

Batolisisidn

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  

Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Find precise solutions at IDNLearn.com. Thank you for trusting us with your queries, and we hope to see you again.