Section XI – Second Law Applied to Open SystemsPage 1

SECTION XI – SECOND LAW APPLIED TO OPEN SYSTEM

6-54 Steam enters an adiabatic turbine at 5 MPa and 400C and leaves at a pressure of 200 kPa. Determine the maximum amount of work that can be delivered by this turbine.

Answer: 678.4 kJ/kg

Assumptions

 Flow is steady,

i.e.

 Flow is uniform at

the inlet and exit of turbine

ep and ekh

Given: Turbine is adiabatic

Conservation of energy applied to the turbine:

TI = 400C > Tsat = 263.99C corresponding to PI = 5 MPa  steam is superheated at the inlet  hi = 3195.7 kJ/kg, si = 6.6439 kJ/kg K

For maximum power output, flow through turbine must be isentropic (i.e. no loses since turbine is adiabatic)  si = se = 6.6459 kJ/kg K

sef < se < seg steam is a saturated liquid-vapour mixture at the exit

Exit quality:

Exit enthalpy:

6.61Steam expands in a turbine steadily at a rate of 25,000 kg/h, entering at 8 MPa and 450C and leaving at 50 kPa as saturated vapor. If the power generated by the turbine is 4 MW, determine the rate of entropy generation for this process. Assume the surrounding medium is at 25C. Answer: 8.38 kW/K

Assumptions:

Flow is uniform at exit and inlet of turbine

epekh

Given:

Flow is steady 

Conservation of energy applied to the turbine:

Inlet conditions: PI = 8 MPa, TI = 450C

TI = 450C > Tsat = 295.06C corresponding to PI = 8 MPa  steam is superheated at the inlet.

 hi = 3272.0 kJ/kg, si = 6.5551 kJ/kg K

Exit conditons: Pe = 50 KPa, saturated vapour

Second Law: dStot 0

6.65A 0.1 m3 rigid tank initially contains refrigerant-12 at 1 MPa and 100 percent quality. The tank is connected by a valve to a supply line that carries refrigerant-12 at 1.4 MPa and 30C. The valve is now opened, allowing the refrigerant to enter the tank, and it is closed when it is observed that the tank contains only saturated liquid at 1.2 MPa. Determine (a) the mass of the refrigerant that entered the tank, (b) the amount of heat transfer with the surroundings at 50C, and (c) the entropy generated during this process.

Answers: (a) 115.67 kg, (b) 1564 kJ, (c) 0.0907 kJ/K.

Assumptions:

Pressure drop through valve is

negligible

Flow is uniform at the inlet

R-12 is at uniform state in tank

ep and ek < h at the inlet

Given:

Tank is rigid  d = 0  W12 = 0

Conservation of mass

(a)

u1 = u1g = 186.32 kJ/kg, u2 = u2f = 83.22 kJ/kg

s1 = s1g = 0.6820 kJ/kg K, s2 = s2f = 0.3015 kJ/kg K

P1 = 1.4 MPa, TI = 30C < Tsat = 56.09C corresponding to PI = 4 MPa  R-12 is subcooled (or compressed) at the inlet.

Difference between hi(i) and hi(ii) is less than 1%  hi(i) will be used here

Conservation of energy:

(b)

(d=0)

Second Law:

(c)

5.66Fifty kilojoules of heat is transferred from a thermal-energy reservoir at 500 K to the environment at 15C. Calculate the change in entropy of the reservoir and the change in entropy of the environment. Show that the total entropy change satisfies the second law. If the amount of heat transfer remains unchanged, but the reservoir temperature is lower, what effect does this have on the total entropy change? Explain.

Entropy change of thermal-energy reservoir

Entropy change of the environment

Total change of entropy: (S)tot = (S)R + (S)E

 second law not violated

As TR TE, (S)tot 0  process reversible