ECE 309 Introduction to Thermodynamics and Heat Transfer

Tutorial # 6

Entropy

Problem 1 Air is compressed steadily by a 5-kW compressor from 100 kPa and 17oC to 600 kPa and 167oC at a rate of 1.6 kg/min. During this process, some heat transfer takes place between the compressor and the surrounding medium at 17oC. Determine (a) the rate of entropy change of air and (b) the rate of entropy generation during this process.

600 kPa 167oC 17oC

AIR COMPRESSOR 5 kW

100 kPa 17oC

Solution: Step 1: Write the given data from the problem statement o State 1: P1 = 100 kPa, T1 = 17 C o State 2: P2 = 600 kPa, T2 = 167 C o Surrounding temperature: Tsurr = 17 C Work input to the compressor: W& =- 5 kW (negative sign indicates work is done on the system) Mass flow rate: m& 1 = m& 2 = m& = 1.6kg / min = 0.02666kg / s

Step 2: Write what we are asked to solve for:

(a) rate of entropy change of air: !S&air = ?

(b) rate of entropy generation during the process: S& gen = ?

Step 3: State the assumption(s): (1) At the specified conditions air can be assumed as an ideal gas (2) Assuming steady state steady flow process (SSSF) (3) Changes in kinetic and potential energies to be negligible

1 Tutorial 6

Step 4: Write the Energy balance (First Law of Thermodynamics) and Entropy Balance (Second Law of Thermodynamics) equations for the system shown in Figure P1

Energy Balance (First Law of Thermodynamics):

Q& !W& = m& (h2 ! h1 ) (1.1) for ideal gas, we know that

dh = C p,av dT (1.2)

Q& = W& + m& C p,av (T2 !T1 ) (1.3)

Entropy Balance (Second Law of Thermodynamics):

Q& + S& gen = m& (s2 " s1 ) (1.4) ! T

Again for ideal gas, we know that

T2 P2 s2 ! s1 = C p,av ln ! R ln (1.5) T1 P1 where R is gas constant

Q& ' T P $ S& m C ln 2 R ln 2 ! + gen = & % p,av ( " (1.6) T & T1 P1 #

Step 5: Solve for the unknown quantities

From Eq. (1.3), calculate the heat transfer during the process

Q& = !5 kW + (0.02666 kg / s)(1.010 kJ /(kg " K ))(440 ! 290)K = !0.96101 kW (1.7)

Note: In the above calculation of heat transfer, Cp,av is found from Table A-2b at Tav= 365 K

(a) Rate of entropy change ( !S&air ) & T P # S& m s s m C ln 2 R ln 2 ( air = & ( 2 ' 1 )= & $ p,av ' ! (1.8) % T1 P1 " where s2 ! s1 is replaced by Eq. (1.5)

2 Tutorial 6

& 440 K 600 kPa # S& 0.02666 kg / s 1.010 kJ / kg K ln 0.287 kJ / kg K ln ) air = ( )$( ' ) ( ( ' ) ! (1.9) % 290 K 100 kPa "

"S&air = !0.00248 kW / K

(b) Rate of entropy generation during the process ( S& gen )

Using Eq. (1.6) and result (a), we can find the rate of entropy generation during the process

Q& ! 0.96101 kW S&gen = "S&air ! = !0.00248 kW / K ! (1.10) Tsurr 290 K

S& gen = 0.00083 kW / K

3 Tutorial 6

Problem 2 Steam enters a turbine at 30 bars and 400°C with a velocity of 160 m/s. Saturated vapor exits at 100oC with a velocity of 100 m/s. At steady state, the turbine develops work equal to 540 kJ per kg of steam flowing through the turbine. Heat transfer between the turbine and its surroundings occurs at an average outer surface temperature of 500K Determine the rate at which entropy is produced within the turbine per kilogram of steam flowing, in KJ/kgK. Neglect the change in potential energy between inlet and exit.

Solution: Step 1: Write the given data from the problem statement o State 1: P1 = 30 bars, T1 = 400 C, V1 = 160 m/s o State 2: Saturated vapor, T2 = 100 C,V2 = 100 m/s Figure P2 Boundary temperature: Tb = 500 K Work output from the turbine: W = 540 kJ/kg

Step 2: Write what we are asked to solve for:

Rate of entropy produced per kg of steam flowing within the turbine: S gen = ?

Step 3: State the assumption(s): (1) Assuming steady state steady flow process (SSSF) (2) Change in potential energy between the inlet and outlet is negligible (3) Heat transfer between the turbine and the surroundings occurs at a boundary

temperature Tb

Step 4: Write the Energy balance (First Law of Thermodynamics) and Entropy Balance (Second Law of Thermodynamics) equations for the system shown in Figure P2

Energy Balance (First Law of Thermodynamics):

& V 2 'V 2 # Q& W& m$h h 2 1 ! ' = & $ 2 ' 1 + ! (2.1) % 2 " 2 2 Q& W& V2 !V1 = + h2 ! h1 + (2.2) m& m& 2

2 2 Q& V2 !V1 Q = = W + h2 ! h1 + (2.3) m& 2

4 Tutorial 6

& # 2 2 2 $ ! & kJ # & kJ # &100 '160 #& m # 1N & 1kJ # Q 540$ ! 2676.1 3230.9 $ ! $ !$ !$ !$ ! (2.4) = $ ! + ( ' )$ ! + $ !$ 2 ! $ 3 ! % kg " % kg " % 2 "% s "$ kg m !%10 N m " $1 2 ! % s " Q = !22.6kJ kg

o {Note: From Table A-4, @ T1 = 400 C and P1 = 30 bars, h1 = 3230.9 kJ/kg and s1 = 6.9212 kJ/kg K o From Table A-2, @ T2 = 100 C and x2 = 1, h2 = 2676.1 kJ/kg and s2 = 7.3549 kJ/kg K }

Entropy Balance (Second Law of Thermodynamics):

Q& + S& gen = m& (s2 ! s1 ) (2.5) T b & Q& # $ ! S&gen % m& " = (s2 ' s1 )' (2.6) m& Tb

& kJ # (' 22.6)$ ! & kJ # $ kg ! S 7.3549 6.9212 % " gen = ( ' )$ ! ' (2.7) % kg K " 500 K

S gen = 0.4789kJ / kg K

5 Tutorial 6