2-3. the First Law of Thermodynamics

8/3/2019 2-3. the First Law of Thermodynamics

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Chapter 2The First Law of

Thermodynamics

Engi-3434Chemical Engineering Thermodynamics

Dr. Charles Xu @ Chemical Engineering, Lakehead University

Example 2.7Calculate the internal-energy and enthalpy changes that occur when air ischanged from an initial state of 4(ºC) and 10(atm), where its molar volume

is 0.002278m3(g mole )-1, to a final state of 60(°C) and 1 (atm). Assume forair that P V / T is constant and that C v = 20.93 and C P = 29.31J(g mole)-

1(°C)-1.

Take 1 g-mol air contained in an imaginary piston/cylinder as the system:

Because changes in properties (state functions) U

and H are independent of the path. We can assumetwo reversible steps in which 1 (g mole) of air is (a)

cooled at constant volume to the final pressure,and (b) heated at constant pressure to the finaltemperature.

T 1 = 4 + 273.15 = 277.15 K T 2 = 60 + 273.15 = 333.15 K

3

22

3

2

22

1

11 0274.015.333

)1(

15.277

)002278.0)(10(mV

K

V atm

K

matm

T

V P

T

V P=⇒=⇒=

Since PV/T = Const, the final volume (V2) is

Solution:

P1, V1, T1

P2, V2, T2

P’, V’, T’

V

P

a

b



Solution to Example 2.7 (cont’d)

K T T

matm

K

matm

T

V P

T

V P715.27'

'

)002278.0)(1(

15.277

)002278.0)(10(

'

' 331

1

11 =⇒=⇒=

(a) The first step: cooling at const V = V1 = 0.002278 m3 until the final pressure (P2=1atm) is reached. The temperature at the end of this step is thus

J K K T nC H nQ P 3.8952)715.2715.333()Kmol

Jmol)(29.31-g(1 =−

⋅=Δ=Δ=

(b) The second step: heating at const P = P2 = 1 atm, until the final volume V2=0.0274m3 and temperature T2 = 333.15K is reached.

Q =nΔU = nCVΔT = (1 g-mol)(20.93J/mol/K)(27.715K-277.15K) = −5220.7J

For the entire process (2 steps): ΔU = (−5220.7J) + (6406.8J) = 1186.1J

ΔH = Δ U + Δ(PV) = Δ U + (ΔP)(V) = – 5220.7J + (1×101325Pa –10×101325Pa)(0.002278m3) = −7298J

ΔU = Δ H - Δ(PV) = Δ H - PΔV= 8952.3J – (1.0×101325Pa)(0.0274m3 − 0.002278 m3) = 6406.8J

ΔH = (−7298J) + (8952.3J) = 1654.3J

2.12 Mass and Heat Balance for OpenSystem

• Measures of Flow

(kg/s)m&

Cold media in

Coldmediaout

• Relations between Measures of Flow

⎯ Mass Flowrate:

⎯ Molar Flowrate: (mol/s)n&

⎯ Volumetric Flowrate: /s)(m3

q

⎯ Velocity: (m/s)u

(kg/s)m&

n M m && ⋅= where M is molar mass (kg/mol)

ρ ρ uAqm =⋅= & where A is the cross-sectional area (m2) and ρ is specific density (kg/m3)

Auq = ρ ρ uAqn =⋅= & where A is the cross-sectional area (m2) and ρ is molar density (mol/m3)



2.12 Mass and Heat Balance for OpenSystem (Cont’d)

• Mass Balance for Open Systems

0)( =Δ+ fsCV m

dt

dm&

Consider a control volume (CV) with flow

streams (fs ), the mass balance can be written as

0)( =Δ+ fsCV m

dt

dm&

0)( =Δ+ fs

CV uAdt

dm ρ

0)()( =Δ=Δ fs fs uAm ρ &

0=dt

dmCV

or

⎯ Continuity equation of the system

At steady state, , thus

222111Const Au Aum ρ ρ ===&

When there is but a single entrance and a single exit,

(2.26)

or

2

22

1

11ConstV

Au

V

Aum ===& where V is specific volume (m3 /kg): V=1/ ρ

“Δ” denotes as “out – in”

2.12 Mass and Heat Balance for Open System (Cont’d)

rate)work ()2

1()

2

1(

)(

,

2

,

2 ++⎥⎦

⎤⎢⎣

⎡ ++−⎥⎦

⎤⎢⎣

⎡ ++= Qm zguU m zguU dt

mU d

out fsin fs

cv &&&

[ ] [ ]

W Qm zguPV U m zguPV U

W mPV mPV Qm zguU m zguU dt

mU d

out fsin fs

out fsin fs

out fsin fs

cv

&&&&

&&&&&&

++⎥⎦

⎤⎢⎣

⎡ +++−⎥⎦

⎤⎢⎣

⎡ +++=

+−++⎥⎦

⎤⎢⎣

⎡ ++−⎥⎦

⎤⎢⎣

⎡ ++=

,

2

,

2

,,

,

2

,

2

)2

1()

2

1(

)()()2

1()

2

1(

)(

• General Energy Balance for Open Systems

Consider a control volume (CV) with flow streams (fs ), the energy balance is

where U is the specific internal energy (J/kg), m is the mass the control volume.

The work associate with the CV mayinclude several forms: the PV work tomove the flowing streams through

entrance and exits, other types of workW (such as shaft work Ws, stirringwork, expansion and contraction of theCV, etc.). Thus

[ ] [ ] W mPV mPV out fsin fs

&&& +−= ,, )()(rate)work (

The energy balance is now:




W Qm zgu H dt

mU d

fs

cv &&& +=⎥⎦

⎤⎢⎣

⎡ ++Δ+ )2

1(

)( 2

W Qm zguPV U m zguPV U dt

mU d

in fsout fs

cv &&&& +=⎥⎦

⎤⎢⎣

⎡ +++−⎥⎦

⎤⎢⎣

⎡ ++++,

2

,

2)

2

1()

2

1(

)(

0)2

1( 2 ≈⎥⎦

⎤⎢⎣

⎡ +Δ fs

m zgu &

• General Energy Balance for Open Systems

Note that H = U + PV and denote “Δ” as “out – in”

For some applications, kinetic- and potential-energy changes in the flowingstreams are negligible, i.e.,

(2.28)

( ) W Qm H

dt

mU d fs

cv &&& +=Δ+)(

The energy balance equation (2.28) is simplified to

(2.29)

Example 2.8Show the first law of thermodynamics for a close system can be derived from thegeneral heat balance equation (2.29).

Solution:



Example 2.9

12.0 −= kgsm&

An insulated, electrically heated tank for hot water contains 190 kg of liquid water at 60°C whena power outage occurs. If a stream of cold water at 10°C flows in and a withdrawn stream flowsout the tank at a constant rate of . How long will it take for the temperature of the

water in the tank to drop from 60 to 35°C? Assume negligible heat losses from the tank and forliquid water, Cv = CP = C, independent of T and P, and assume a perfect mixing of the tank

water.Solution:


m

W

m

Q zg

u H s

&

&

&

&

+=Δ+Δ

+Δ2

2

sW Q zgu

H +=Δ+Δ

+Δ2

2

• Energy Balance for Steady-State Flow Systems

For steady-state systems, the accumulation term in Eq. (2.28) . Insuch system the mass within the control volume is constant, and no changes occurwith time in the properties within the whole system, and the work W = Ws

s

fs

W Qm zgu H &&& +=⎥⎦

⎤⎢⎣

⎡ ++Δ )2

1( 2

Const m =&

(2.30)

0)(

=dt

mU d cv

The energy balance equation (2.30) is simplified to

(2.32)

When there is but one entrance and one exit,

All terms in (2.32) are in units of J/kg-fluid

sW Q zgmum H m &&&&& +=Δ+Δ+Δ 2

2

1(2.31)

Or,



Example 2.10Air at 1 bar and 25°C enters a compressor at low velocity, discharges at 3 bar, andenters a well-insulated nozzle in which it expands to a final velocity of 600 ms-1 atthe initial conditions of pressure and temperature. If the work of compression is 240kJ per kilogram of air, how much heat must be removed during compression?

Solution (To be given at tutorial #3):

2-3. the First Law of Thermodynamics

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