CHE 301 HW 3 - Derivation of fundamental equations PDF

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Derivation of fundamental equations...

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QUESTION 1: Enthalpy and Entropy Calculations for Liquid Fluids

Derivation of how to calculate specific enthalpy of water: H = H (T, P) Take a partial derivative of above equation,

dH=

( ∂∂ HP ) dP+( ∂∂ HT ) dT T

P

( ∂∂TH ) =C

P

P

Therefore,

( ) dP

dH=C P dT + ∂ H ∂P

--------------- (1)

T

One of the three fundamental equations is shown below:

dH = TdS + VdP

dP dS dH +V =T dP dP dP

( Divided by dP )

dS dH +V =T dP dP

( ∂∂ HP ) =T ( ∂∂ SP) +V T

---------------- (2)

( partial derivative is taken)

T

Plug in equation (2) into equation (1),

[

dH=C P dT + V +T

From

( ∂∂ PS ) ]dP

---------------(3)

T

dG=−SdT +VdP , Stefan Maxwell equation yields:

( ∂∂ PS ) =− ( ∂∂TV ) T

P

volume expansivity ( β ) =

( )

1 ∂V ∗ V ∂T

P

=β∗V ( ∂V ∂T ) P

( ∂∂ SP ) =− β∗V

---------------(4)

T

Plug in equation (4) into equation (3),

dH=C P dT +[ V +T∗(− β∗V )] dP dH=C P dT + V ( 1−β∗T ) dP

Assumption #1:

---------------(5)

β ≈ 0 for liquids(water∈this problem)

dH =C P dT + VdP Integrate above equation: T2

P2

T1

P1

∆ H=∫ C P dT +∫ VdP

Assumption #2:

C P=constant for liquids(water∈this problem)

Assumption #3:

density (ρ)= constant for liquids(water ∈this problem)

∆ H=C P ∆ T +V ∆ P Now,

V=

1 ρ

1 ∆ H=C P ∆ T + ∆ P ρ

(

∆ H= 4.184

)

J ∗(55 ℃−35℃ )+ g∗℃

( )

1 ∗( 500 k Pa−300 kPa ) g 1 3 cm

kPa∗c m 3 ∗1m 3 g ∗1 J (100 cm )3 J ∆ H=83.68 +200 g 1 kPa∗m 3 J ∗1000 g g ∆ H=83.6802 1 kg ∆ H=83,680.2

J kg

∆ H that is calculated above is a measured value .

Actual values were obtained using the software called Athena, and they are shown below in the table at different temperature and pressure. Athena  Tools  Steam Properties

Specific Enthalpy (

J ) kg

At T= 35 ℃ , P = 300 kPa At T= 55 ℃ , P = 500 kPa

146723.8408 230425.7426

Change in specific enthalpy

230425.7426 - 146723.8408 = 83701.9018

( 83680.2−833701.9018 ) Relative Error =¿ 83701.9018

J kg

j kg

∨¿ 100

Relative Error=0.02593

Derivation of how to calculate specific entropy of water:

One of the three fundamental equations is shown below:

dH = TdS + VdP dH=C P d T +V (1−β∗T ) dP Set above both equations equal to each other: C P dT + V ( 1− β∗T )dP =TdS + VdP C P dT +VdP−β∗V ∗TdP=TdS + VdP TdS=C P dT −β∗V∗TdP Divide the above equation by “T”

dS=C P

dT −β∗VdP T

Assumption #1:

dS=C P

β ≈ 0 for liquids(water∈this problem)

dT T

Integrate above equation: T2

∆ S=∫ C P T1

dT T

Assumption #2:

∆ S=C P∗ln C P=4.184

C P=constant for liquids(water ∈this problem)

T2 T1

J J =4.184 g∗ K g∗℃

T 1 =35℃ +273.15=308.15 K T 2 =55℃ +273.15=328.15 K

∆ S=4.184

328.15 K J ∗ln 308.15 K g∗K

J ∗1000 g g∗K ∆ S=0.2632314 1 kg ∆ S=263.2314

J kg∗K

∆ S that is calculated above is a measured value .

Actual values were again obtained using the software called Athena, and they are shown below in the table at different temperature and pressure. Athena  Tools  Steam Properties Specific Entropy (

J ) kg∗K

At T= 35 ℃ , P = 300 kPa At T= 55 ℃ , P = 500 kPa

504.4251 766.9654

Change in specific entropy =

766.9654 - 504.4251 = 262.5403

(263.2314−262.5403 ) Relative Error=¿ 262.5403

J kg∗ K

j kg∗K

∨¿ 100

Relative Error=0.2633

QUESTION 2: Enthalpy and Entropy Calculations for Liquid Fluids Derivation of how to calculate specific enthalpy of water: Assumptions:

Ev =0

Density ( ρ ¿=constant Heat Capacity ( C P ¿=constant

Volume Expansivity β ≈ 0

Efficiency of the pump ( η) :

η=

Ideal Work Real work ID EAL

Ws η= Ws

Efficiency of the pump is 100%, which means η = 1.

1=

W sIDEAL Ws

W s =W IDEAL s To calculate the work:

∆

( Pρ )=−E +W where ,−E =loss due ¿ friction v

s

v

For Ideal Work ; E v =0 That is also my assumption

∆P IDEAL =W s ρ IDEAL

=

Ws

IDEAL

Ws

=

P2−P1 ρ 1500 kPa−350 kPa g 1.02 c m3 3

kPa∗c m 3 ∗1m g ∗1000 Pa ( 100 cm)3 ∗1000 g 1 kPa W sIDEAL=1127.451 1 kg

IDEAL

Ws

=1127.451

Pa∗m 3 kg

1 Pa∗m3=1 J Assuming there is a steady flow rate, kg ∗1 hr hr ∗1 hp J ∗250,000 3600 s IDEAL Ws =1127.451 J kg 735.50 s IDEAL

W s =W s

=106.452 hp

To find an increase in a temperature for 100% efficiency pump: Assumption: Heat (Q) = 0 ∆ H=Q+ W s IDEAL ∆ H =W s=W IDEAL (becauseW s =W s ) s

∆ H =106.452 hp Using equation (5), dH=C P dT + V ( 1−β∗T ) dP

Assumption #1:

---------------(5)

β ≈ 0 for liquids(water∈this problem)

dH =C P dT + VdP Integrate above equation: T2

P2

∆ H=∫ C P dT +∫ VdP T1

P1

Assumption #2:

C P=constant for liquids(water ∈this problem)

Assumption #3:

density (ρ)= constant for liquids(water ∈this problem)

∆ H=C P ∆ T +V ∆ P V=

Now,

1 ρ

1 ∆ H=C P ∆ T + ∆ P ρ

----------------(i)

( Pρ )=−E +W where ,−E =loss due ¿ friction

∆

v

s

For Ideal Work ; E v =0 That is also my assumption

∆P IDEAL =W s =W s -------------------(ii) ρ Substitute equation (ii) into (i) IDEAL

∆ H=C P ∆ T +W s ∆T= ∆T=

-----------(iii)

∆ H−W IDEAL s CP 106.452hp−106.452 hp kJ 2.8 kg∗ K

∆ T =0 K There is no temperature increase across the pump.

When the pump’s efficiency of the pump is 75%:

η=

Ideal Work Real work IDEAL

W η= s Ws

Efficiency of the pump is 75%, which means η = 0.75.

0.75=

106.452 hp Ws

v

W s =141.936 hp hp∗735.50 1 hp

∗3600 s

1 hr 250,000 kg

W s =141.936

W s =1503.273

J s

∗1 hr

J kg

To find am temperature increases when the pump efficiency is 75%: From equation (iii): IDEAL

∆ H=C P ∆ T +W s ∆ H =Q+ W s

And as I assumed, Q = 0: ∆ H =W s

Therefore, IDEAL

C P ∆ T +W s

=W s

IDEAL

W −W s ∆T= s Cp

J J −1127.451 kg kg kJ ∗1000 J kg∗K 2.8 1kj

1503.273 ∆T=

∆ T =0.1342 K There is a temperature increase of 0.1342 K across the pump when its efficiency is 75%.

SUMMARY: For the first problem, I made some of the assumptions to calculate mass specific enthalpy and entropy. After calculating measured values for mass specific enthalpy and entropy, actual values were

obtained from Athena Visual Studio TM . Following that, % error was calculated to see how effective the assumptions are. My relative error for specific enthalpy was 0.02593 % and for specific entropy, it

was 0.2633 % . These errors were so close to zero that means assumptions that I made were very good. If error were to be more than 5%, then I would have needed to rethink my assumptions. For the second problem, I again started with some assumptions and some were made as I go through my derivations. For the 100% efficiency pump, my shaft work was

106.452hp

and there was no change in temperature across the pump. However, for the 75% efficiency pump, my shaft work was 141.936 hp and there was change of 0.1342 K in temperature across the pump. As efficiency of the pump decreased, shaft work increased with small amount of change in temperature....