CHETHE2Group 2

Barongan, Nikko M.

Mulingtapang, Justinn Donn S.

Opoc, Dave Patrick G.

Submitted To:

Dr. Joseph Auresenia

7.2

a. ( ∂T∂ P )

S

b. ( ∂T∂V )

U

a. ( ∂T∂ P )

S

Consider S=f ( P ,T )

dS=( ∂S∂P )

T

dP+( ∂ S∂T )

P

dT=( ∂S∂ P )

T

dP+CvdTT

From Euler’s chain relation

( ∂S∂ P )

T( ∂ P

∂T )S( ∂T

∂S )P

=−1

( ∂S∂ P )

T

=−( ∂T∂ P )

S( ∂S∂T )

P

( ∂S∂ P )

T

=−( ∂T∂ P )

S

CpT

( ∂T∂ P )

S

=−TCp ( ∂S

∂P )T

To express ( ∂S∂ P )

Tin term of more fundamental terms, that is, in terms of P,V,

or T, one must use a Maxwell relation. For this case consider the definition of Gibbs Free Energy.

dG=VdP−SdT

Meanwhile, it can be expressed as a function of P and T.

dG=( ∂G∂P )

T

dP+( ∂G∂T )

P

dT

Consider this expression

By inspection,

V=( ∂G∂ P )

T

;S=( ∂G∂T )

P

Differentiating both equations with respect to the other variable (as in exact differential equations)

( ∂V∂T )

P

=( ∂2G∂V ∂ P )∧( ∂S

∂ P )T

=−( ∂2G∂ P ∂V )

Since( ∂2G∂V ∂ P )=( ∂2G

∂ P ∂V ), we arrive at:

( ∂V∂T )

P

=−( ∂S∂P )

T

∨−( ∂V∂T )

P

=( ∂ S∂ P )

T

Going back to

( ∂T∂ P )

S

=−TCp ( ∂S

∂P )T

The rightmost term will be replaced as follows:

( ∂T∂ P )

S

= TCp ( ∂V

∂T )P

Note that V= ZRTP

, where Z is the compressibility factor and is a function of

the equation of state.

( ∂T∂ P )

S

=T

Cp ( ∂( ZRTP )

∂T )P

R and P are constants but note that T is not taken off from the partial derivative as it is a variable.

( ∂T∂ P )

S

= RTCp P ( ∂ (ZT )

∂T )P

Differentiate the rightmost term by product rule.

( ∂T∂ P )

S

=RT

Cp P (Z( ∂T∂T )

P

+T ( ∂ Z∂T )

P)

( ∂T∂ P )

S

=RT

Cp P (Z+T ( ∂Z∂T )

P)

( ∂T∂ P )

S

= ZRTCp P

+ RT2

Cp P ( ∂Z∂T )

P

In Example 7.5 from the book, the Joule-Thomson coefficient μ is equal to

RT 2

Cp P ( ∂Z∂T )

P

.

Therefore:

( ∂T∂ P )

S

= ZRTCp P

+μ

Or simply,

( ∂T∂ P )

S

= VCp

+μ

b .( ∂T∂V )

U

Consider U = f(T,V)

Consider this expression

dU=( ∂U∂T )

V

dT+( ∂U∂V )

T

dV =CvdT+( ∂U∂V )

T

dV

From Euler’s chain relation,

( ∂U∂V )

T( ∂V∂T )

U( ∂T∂U )

V

=−1

( ∂U∂V )

T

=−( ∂T∂V )

U( ∂U

∂T )V

( ∂U∂V )

T

=−( ∂T∂V )

U

Cv

( ∂T∂V )

U

=−1Cv ( ∂U

∂V )T

But ( ∂U∂V )

T

=T ( ∂P∂T )

V

−P

Note that P= ZRTV

, where Z is the compressibility factor and is a function of

the equation of state.

( ∂U∂V )

T

=T ( ∂(ZRTV )

∂T )V

−ZRTV

R and V are constants but note that T is not taken off from the partial derivative as it is a variable.

( ∂U∂V )

T

=RTV ( ∂(ZT )

∂T )V

−ZRTV

Differentiation by product rule yields

( ∂U∂V )

T

=RTV (Z ( ∂T

∂T )V

+T ( ∂Z∂T )

V)− ZRT

V

( ∂U∂V )

T

=RTV (Z+T ( ∂Z

∂T )V)−ZRT

V

( ∂U∂V )

T

=ZRTV

+ RT 2

V ( ∂Z∂T )

V

−ZRTV

( ∂U∂V )

T

=RT 2

V ( ∂ Z∂T )

V

Going back to

( ∂T∂V )

U

=−1Cv ( ∂U

∂V )T

Substitution of ( ∂U∂V )

T

¿ RT2

V ( ∂Z∂T )

V

will lead to:

( ∂T∂V )

U

=−1Cv

RT2

V ( ∂Z∂T )

V

And finally,

( ∂T∂V )

U

=−R T 2

Cv V ( ∂Z∂T )

V

7.10

Given: U1 = 230 ft/s U2= 2000ft/sP1= 130 psiaT1= 420oF P2= 35 psia

Req’d: a) state of the steam at the nozzle exit b) SG

Sol’n:

ΔH= -78.8 Btu/lbm

From table F.4 at 130 psi and 420oF:H1= 1233.6 Btu/lbmS1= 1.6310 Btu/lbm R

H2= H1 + ΔH = 1233.6 + (-78.8)= 1154.8 Btu/lbm

From table F.4 at 35 psia:Hl = 228.03 Btu/lbm Sl= 0.3809 Btu/lbm RHv= 1167.1 Btu/lbm Sv= 1.6872 Btu/lbm R

Let x= vapor fraction

1154.8= x(1167.1) + (1-x)(228.03) ; x= 0.987 Ans.

S2= 1.6872(0.987) + 0.3809(1-0.987) = 1.67 Btu/lbm R

SG= S2-S1= 1.67-1.6310= 0.039 Btu/lbm R Ans.

7.18Given:

T1= 500oCP1= 2400 kPa

P= 3500 kJ/s

P2= 20 kPaSaturated vapor

Req’d: a) b) ηT

Sol’n:

a) At 2400 kPa and 500oC, H1= 3462.9 KJ/kg, S1= 7.3439 kJ/kg K

For saturated vapor at 20 kPa, H2= 2609.9 kJ/kg

= Power / ΔHa = (3500kJ/s)/(I 2609.9- 3462.9 I)kJ/kg

= 4.103 kg/s Ans.

b) At 20 kPa:

Hl = 251.453 kJ/kg Sl= 0.8321 kJ/kg KHv= 2609.9 kJ/kg Sv= 7.9094 kJ/kg K

For isentropic work: S1 = S2’ = 7.3439 kJ/kg KLet x= vapor fraction

7.3439= x(7.9094) + (1-x)(0.8321) ; x= 0.92

H2’= 0.92(2609.9) + (0.08)(251.453) = 2421.224 kJ/kg

Ws = ΔHs = 2421.224 – 3462.9 = -1041.676 kJ/kg

TURBINE

Wa = ΔHa = 2609.9 – 3462.9 = -853 kJ/kg

ηT = Wa / Ws = (-853)/(-1041.676) = 0.819 Ans.

7.26

Given:

η=0.065+0.080 ln|W|

ṅ=175mol s−1

C p=72

R=72

(8.314 ) Jmol K

=72

(8.314 x10−3 ) kJmol K

Required:

a.) Wa

b.) ηturbine

c.) SG

Solution:

ΔE = W + Q

but Q = 0 for adiabatic expansion. Hence,

ΔE = W

CvdT = -PdV

T 1=550K

P1=6 ¿̄P2=1.2 ¿̄

CvdT = -(RT/V)dV

∫ (-Cv/R)(dT/T) = ∫ dV/V

(-Cv/R) ln (T2/T1) = ln (V2/V1)

(T2/T1)-Cv/R = (V2/V1)

(T2/T1)-Cv/R =

R T 2

P2RT 1

P1

(T2/T1)-Cv/R = (T2/T1)(P1/P2)

(T2/T1)-Cv/R – 1 = (P1/P2)

(T2/T1)Cv/R + 1 = (P2/P1)

(T2/T1)(Cv+R)/R = (P2/P1)

(T2/T1)Cp/R = (P2/P1)

(T2/T1) = (P2/P1)R/Cp

T2 = T1(P2/P1)R/Cp

T2 = 550(1.2/6)8.314/(3.5*8.314)

T2 = 347.26 K (Final temperature at reversible work)

Wa = η*ΔH = η*ṅ ∫CpdT = η*ṅ*Cp*(T2-T1)

W a=¿

The actual work is solved by means of successive substitutions starting at the initial guess of, say, -600 kJ/s. After doing so,

Wa = -594.72 kJ/s = -594.72 kW

The actual work delivered is 594.72 kilowatts.

η = 0.065 + 0.08 ln |-594.72| = 0.576

The turbine efficiency is 0.576.

At actual work = -594.72 kW, the final temperature at irreversible work is computed.

Wa = ΔH = ṅ ∫CpdT = ṅ*Cp*(T2-T1)

−594.72 kJs

=(175mols )( 72 (8.314 x10−3 ) kJ

mol K )(T 2−550K)

T 2=433.21K

This temperature is needed for the computation of the entropy generation.

dS=C pdTT

−( ∂V∂T )dP=C p

dTT

−RdPP

ΔS=ṅC p lnT2T1

−ṅR lnP2P1

ΔS=(175 mols )(72 8.314 J

mol K )(ln 433.21K550K )−(175 mol

s )(8.314 Jmol K )¿

ΔS=SG=1126.14J

K−s=1126.14W

K=1.1261 kW

K

The rate of entropy generation is 1.1261 kW/K.

7.34Given:

Sat. steamP1= 125 kPa P2= 700 kPa

= 2.5 kg/s

Adiabatically, ηc = 0.78Req’d: Power, H2, S2

Sol’n: From steam table:H1

sat.v = 2685.2 kJ/kgS1

sat.v = 7.2847 kJ/kg K For isentropic work: S1= S2’ = 7.2847 kJ/kg K

Interpolation at 700 kPa for H2’ using S2’= 7.2847:H: 3017.7 H2’ 3059.8S: 7.2250 7.2847 7.2997

(H2’-3017.7)(7.2997-7.2847) = (3059.8-H2’)(7.2847-7.2250)

H2’= 3051.346 kJ/kgWs = ΔHs = 3051.346-2685.2= 366.146 kJ/kgWa = ΔH = Ws / ηc = 366.146/0.78 = 469.418 kJ/kg

H2= H1 + ΔH = 2685.2 + 469.418 = 3154.618 kJ/kg Ans.

Interpolation at 700 kPa for S2 using H2= 3154.618:H: 3112.1 3154.618 3164.3S: 7.3890 S2 7.4745

(3154.618-3112.1)(7.4745-S2) = (3164.3-3154.618)(S2-7.3890)

S2= 7.4586 KJ/kg K Ans.

P = ΔH = (2.5)(469.418) = 1173.545 kW Ans.

COMPRESSOR

7.42Given:

P1= 1atm T1= 308.15 K η = 0.65P2= 50 atm T2= 473.15 K Q= 0.5 m3/s

Req’d: a) number of stagesb) mechanical-power requirement per stagec) Heat duty for each intercoolerd) Water as coolant for intercoolers. Enters at 25oC and leaves at 45oC.

Calculate the cooling-water rate per intercooler. Assume air is ideal gas with Cp= (7/2)R

Sol’n: Vm = (RT1)/P1 = (0.08205)(308.15)/(1) = 25.2837 L/mol = 0.0252837 m3/molṅ = Q/Vm = 0.5/0.0252837 = 19.7756 mol/s

From eq. 7.23, solving for T2’:

T2’ = (T2-T1) η + T1 = (165)(0.65) + 308.15 = 415.4 K

From eq. 7.18 with the addition of term N as the number of stage:

T2’ = T1(P2/P1)R/N Cp ; solving for N:

a) N= (R/Cp)(ln P2/P1)/(ln T2’/T1) = (2/7)(ln 50)/(ln 415.4/308.15) = 3.743 ~ 4 stages Ans.

b) Calculate pressure ratio (r) using equation, r = (P2/P1)1/N = (50/1)0.25 = 2.659

Power = (ṅ)(Cp)(T1)(rR/Cp -1)/ η = (19.7756)(3.5)(8.314)(308.15)(2.6592/7-1)/0.65Power = 87940 W = 87.94 kW Ans.

c) Since the gas (ideal) leaving the intercooler and the gas entering the compressor is at the same temperature (308.15 K), there is no enthalpy for the compressor/interchanger system hence from the 1st law of thermodynamics, q= -W = -87.94 kW Ans.

d) At 25oC, Hsat. liq = 104.8 kJ/kg, at 45oC, Hsat. liq = 188.4 kJ/kg

Energy balance on the interchanger: ΔHH20 = (188.4-104.8) = 83.6 kJ/kg

=( IqI) / ΔHH20 = (I-87.94I) / 83.6 = 1.052 kg/s Ans.