chap5 open system

1DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

CHAPTER 5MASS AND ENERGY

ANALYSIS OFCONTROL VOLUMES

(OPEN SYSTEM)

OBJECTIVES

DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

At the end of the course, students should be able to :

Develop the conservation of mass principle

Apply the conservation of mass principle to various systems for steady flow control volumes

Apply the first law of thermodynamics as the statement of the conservation of energy principle to control volumes.

solve energy balance problems for common steady flow devices such as nozzles, compressors, turbines, throttling valves, mixers and heat exchangers.

CHAP 5 : MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES

2MASS AND VOLUME FLOW RATES


The amount of mass flowing through a cross section per unit time is called the mass flow rate and is denoted by m&

The differential mass flow rate of fluid flowing across a small area element dA is expressed as

dAVm n =&

Density(kg/m3) Velocity normal

To dA (m/s)

Small areaelement

The mass flow rate through the entire cross-sectional area of a pipe or duct is

kg/sA V dAVmm avgA nA === &&

The volume flow rate, AVAVdAVV avgn === A&

VVm or VVVm ===== &&& A


CONSERVATION OF MASS PRINCIPLE


The conservation of mass principle for a control volume can be expressed as

The net mass transfer to or from a control volume during a time interval t is equal to the net change in the total mass within the volume during t

=

t duringCV the within mass in changeNet

t duringCV the leaving mass Total

t duringCV the entering mass Total

kg/s dt

dmmm

kg mmm

cv

cvin

=

=

outin

out

&&

For steady flow processes

kg/s dt

dmmm cv= outin

&&

(kg/s) VAVA or mm ==outinoutin&&

0


For steady, incompressible flow, the density is assumed constant, thus/sm VV or VAVA 3 ==

outinoutin

&& For single stream/flow 221121 AVAVVV == &&

3EXAMPLE 5-1



EXAMPLE 5-2



A 1.2 m high, 0.9 m diameter cylindrical water tank whose top is open to the atmosphere is initially filled with water. Now the discharge plug near the bottom of the tank is pulled out and a water jet whose diameter is 1.3 cm streams out. The average velocity of the jet is given by , where h is the height of water in the tank measured from the center of the hole and g is the gravitational acceleration. Determine how long it will take for the water level in the tank to drop to 0.6 m from the bottom.

2ghV =

Assumptions : Water is incompressible substance, the distance between the bottom of the tank and the center of the hole is negligible.

Analysis : Water level drops, variable control volume H unsteady-flow

dtdmmm cvoutin = && ( ) 2ghAVAm jetoutout ==& hAVm tankcv ==

( )dtdh

4D2 ktan == 2gh

4D

hAd2ghA2jet

tankjet

2ghdh

DDdt 2

jet

2tank=

Integrating from t = 0 at which h = ho to t = t at h = h22

jet

tank2oh

h2jet

2tankt

0 DD

g/2hh

t h

dh2gD

Ddt 2o

==

Substituting, min 11.6 s 6940.013

0.99.807/2

0.61.2t 2

==

=

4DESIGNED AND PREPARED BY : MOHD KAMAL ARIFFIN/2003

FLOW WORK OR FLOW ENERGY

Work done on the element of mass,W = Fdl = pA dl = pdV pF

Cross-sectional area of piston, A

1 2

An element of mass, m

pV +mgz + Vm + U = E energy, total The 221

Force, F (F = PA) is needed to push a mass into the system

The total Work done by the surrounding,

W = pV = Flow Work

Internal energy Flow energyKinetic

energy Potential energy

Pipe


Substituting H = U + pV,

mgz + Vm + H = E energy, total The 221

EXAMPLE 5-2



Steam is leaving a 4 L pressure cooker whose operating pressure is 150 kPa. It is observed that the amount of liquid in the cooker has decreased by 0.6 L in 40 min after the steady operating conditions are established, and the cross sectional area of the exit opening is 8 mm2. Determine (a) the mass flow rate of the steam and the exit velocity (b) the total and flow energy of the steam per unit mass (c) the rate at which energy leaves the cooker by steam.

Analysis : The liquid has the properties of sat liquid and the and the vapor has the properties of sat vapor.

( )( ) m/s 34.310 x 8

1.159410 x 2.37A

mA

mV

kg/s 10 x 2.37 kg/min 0.01421000 x 40

0.0010530.6tV

tmm

6

4g

g

4-ff

====

=====

&

(b) Neglecting KE and PE, thus

Assumption : Steady flow, KE and PE are negligible, saturations conditions exist within the cooker all the time.

(a) From Table A-5, the saturation properties of water at 150 kPa are f = 0.001053 m3/kg, g = 1.1594 m3/kg, ug = 2519.2 kJ/kg and hg = 2693.1 kJ/kg.

Total energy, Etotal = u + p + ke + pe = h = 2693.1 kJ/kgeflow = p = h u = 2693.1 2519.2 = 173.9 kJ/kg

(c) The energy rate, kW 0.638 kJ/s 0.638 )(2693.1)10 x (2.37EmE -4 ==== &&

5ENERGY ANALYSIS FOR STEADY FLOW SYSTEMS


1) Properties (pressure, temperature, density, mass flow rate and etc) are constant at any point on the system and its boundary (inlet or outlet), average values are used

2) Inlet mass flow rate = outlet mass flow rate

5) The heat and work interactions between system and its surroundings do not change with time

1 2

T = 30 oC

T = 30 oC

T = 30 oC

minV1m& 2m&

1 2

0mmmmmm

s12

12s ===&&&&&&

1m& 2m&

VAm =&

2

22

1

11 VAVA =

Vm&

& = 2

2

1

1 VV

&& =

VAm =& 222111 VAVA =

Vm && = 2211 VV && =


3) Boundary work is zero (volume of the control volume is constant)4) The total energy of CV remains constant, thus the change in the total energy of the CV

is zero (ECV = 0)

ENERGY ANALYSIS FOR STEADY FLOW SYSTEMS


=

CV of energy

total the in Change

CV the leaving energy Total

CV the entering energy Total

(Qin+ Win + Ein) (Qout + Wout Eout) = Esystem

Energy enteringThe CV

Energy leavingthe CV

Change in thetotal energy

Ein = U + KE + PE + pV U + KE + PE Eout = U + KE + PE + pV

In general, we can write


dtdEEE

EEE

systemoutin

systemoutin

==

&&

6ENERGY ANALYSIS FOR STEADY FLOW SYSTEMS


Inlet

Outlet

1

2

p11u1gz1

Q

W

SystemUs, KE, PEz1

z2p22u2gz2

)gz + 2

V+u (m = )gz+ 2

V+ p +(um-W - )gz +2

V+ p +(um +Q S2S

SS2

22

22221

21

1111 &&&&&

2V21

2V21

Energy enteringThe system

Energy leavingThe system

Total energyIn the system

Rate ofHeat transfer

Power

Inlet Mass flowrate, kg/s

Outlet massFlow rate, kg/s

Mass flow rateOf the system(m/s) velocityV

GravitationalAcceleration


ENERGY ANALYSIS FOR STEADY FLOW SYSTEMS


)gz + 2

V+u (m = )gz+ 2

V+ p +(um-W - )gz +2

V+ p +(um +Q S2

SSS2

22

22221

21

1111 &&&&&

For steady flow, mass flow rate of the system is zero, 0mmm 12s == &&&

0 = )gz+ 2

V+ p +(um-W - )gz +2

V+ p +(um +Q 22

222221

21

1111 &&&&

0 = )gz+ 2

V+ (m-W - )gz +2

V+ (m +Q 22

2221

21

11 hh &&&&

)gz +2

V+ (m- )gz+ 2

V+ (m W-Q 12

1112

22

22 hh &&&& =

mmm 21 &&& ==

kW )gz +2

V+ (- )gz+ 2

V+ (m W-Q 12

112

22

2

= hh&&&mmm 21 &&& ==

kW )gz-(gz+ 2

VV)- (m W-Q 122

12

212

+= hh&&&

kJ )gz-(gz+ 2

VV)- (m W-Q 122

12

212

+= hh

kJ/kg )gz-(gz+ 2

VV)- ( w-q 122

12

212

+= hh

Inlet mass flow rate = Outlet mass flow rate,


STEADY STATE FLOW EQUATION

7kJ/kg )gz-(gz+ 2

VV)h- h( w-q 1221

22

12

+=

NOZZLES AND DIFFUSERS


Nozzles and diffusers are commonly utilized in jet engines, rockets, spacecraft and even garden hoses.

21

12

( ) 2

VV+h-h =021

22

12

2V-V= h-h

22

21

12

12 VV >D2 < D1 p2 < p1

D2 > D1 12 VV p1

Energy balance for a diffuser or nozzle is

EXAMPLE 5-4



Assumptions : Air is ideal gas, PE = 0, Q 0, KEout , W = 0Analysis :

(a) Using ideal gas equation at inlet,

/kgm 1.01580

283 x 0.287p

RT 31

11 ===

( )( ) kg/s 78.80.42001.015

1VA1VAm 111

111 ==== &

(b) Energy balance for diffuser

( )2

VVTTchh22

21

12p12==

K 302.9 2832000x1.005

0-200T2c

VVT22

1p

22

21

2 =+=+=

Air at 10oC and 80 kPa enters the diffusers of a jet engine steadily with a velocity of 200 m/s. The inlet area of the diffuser is 0.4 m2. The air leaves the diffuser with a velocity that is very small compared with the inlet velocity. Determine

(a) the mass flow rate of the air(b) the temperature of the air leaving the diffuser

8EXAMPLE 5-5



Steam at 1.8 MPa and 400oC steadily enters a nozzle whose inlet area is 0.02 m2. The mass flow rate of steam through the nozzle is 5 kg/s. Steam leaves the nozzle at 1.4 MPa with a velocity of 275 m/s. Heat losses from the nozzle per unit mass of the steam are estimates to be 2.8 kJ/kg. Determine (a) the inlet velocity and (b) the exit temperature of the steam

Assumptions : Steady flow, no work interactions, PE = 0Analysis : For steady flow, mmm 21 &&& ==

(a) At 1.8 MPa, Ts = 207.11oC, T1 > TsH shsp1 = 1.8 MPaT1 = 400oC

1 = 0.16849 m3/kgh1 = 3251.6 kJ/kg

Table A-6

( )( ) m/s 42.1V 0.02V0.16849

15 V AV1m 1111 ===&

(b) kJ/kg )gz-(gz+ 2VV)h- (h w-q 12

21

22

12

+= 2

VV)h- (h q21

22

12

+=0 0

kJ/kg 3211.92000

42.1275-2.8)3251.6 2

VVqhh222

122

12 =

+=

+= (

p2 = 1.4 MPah2 = 3211.9 kj/kg

T2 = 378.6oC (Table A-6)

P1 = 1.8 MPaT1 = 400oCA1 = 0.002 m2

P2 = 1.4 MPas/m275V2 =

2.8 kJ/kg

5 kg/s

Steam enters a nozzle at 40 bar, 400 oC and with a velocity of 10 m/s steadily. The steam exits at 14 bar and with a velocity of 665 m/s. The flow rate of the steam is 2 kg/s. Heat transfer and change in potential energy can be neglected. Determine the cross sectional area of the outlet opening in m2.

MY EXAMPLE 5-1


2

22 V

m = A &

At 40 bar, Ts = 250.3 oC, T>TS , shs

2V-V= h-h

22

21

12

kJ/kg 2992.5 = 2000

665-10+3213.6=h22

2

At 14 bar, hg = 2790.0 kJ/kg, h2 > hg shs,

( )/kgm 0.1743 =

0.16350.16350.18232927.5-3040.42927.5 -2992.5 =

3

+

2h 2927.2 0.1635

2992.5 23040.4 0.1823

2m 4-10 x 5.24 665

2(0.1743) = A Area, =

40 bar 400 oC h1 = 3213.6 kJ/kg

2

22

1

11 VAVA = m =& 21

p1 = 40 barT1 = 200 oC

m/s 14V1 =

kg/s 2m2 =&

p2 = 14 barm/s 665V2 =


9COMPRESSORS AND PUMPS


( )

12

21

22

12 z-zg+ 2VV+)h-(hm =W-Q &&&

)h - (hm = W 21&&

0 0 0

1

2

W

2222 z,V,T,p

1111 z,V,T,p

Compressors, pumps and fans are devices used to increase the pressure of a fluid

Works is supplied to these devices from an external source through a rotating shaft.

Compressor is capable of compressing a gas to very high pressures

Pumps handle liquids instead of gases and mainly used to increase the flow rate of a fluid

Fans increases the pressure of a gas slightly and is mainly used to mobilize a gas.

Potential energy and heat transfer are usually negligible

The process occur in these devices is a compression process.


EXAMPLE 5.6


Air at 100 kPa and 280 K is compressed steadily to 600 kPa and 400 K. The mass flow rate of the air is 0.02 kg/s and a heat loss of 16 kJ/kg occurs during the process. Assuming the changes in kinetic and potential energies are negligible, determine the necessary power input to the compressor.


( )

12

21

22

12 z-zg+ 2VV+)h-(hm =W-Q &&&

( ) ( )( )kW2.73

280))-1.005(400-0.02(-16 TTc-qm)h - (hqmh - (hm-Q = W 12p1212

==

== &&&&& )

0 0

10

( )

12

21

22

12 z-zg+ 2VV+)h-(hm =W-Q &&&

( ) 2

VV+h -h m-Q=W2

12

212

&&&

Air at 1 bar, 290 K and with a velocity 6 m/s enters a compressor steadily through an inlet pipe which has cross-sectional area of 0.1 m2. The conditions of the air at outlet are 7 bar, suhu 450 K and velocity 2 m/s. During the process heat is transferred from the compressor to the surrounding air at the rate of 180 kJ/min. Assuming the air is ideal gas, determine the input power of the compressor in kW. [Rair = 0.287 kJ/kgK and cP = 1.005 kJ/kgK]

MY EXAMPLE 5-2


0

1

11

VAm =&

( )kW 119.4- =

2000

26 + 450 -290005.1 0.72+60

180-= W 22

&

1

111

RTpVA= 1

2

W

m/s 2VC 450T

bar 7p

2

o

2

2

===

m/s 6VC 290T

bar 1p

1

o

1

1

===

Q=-180 kJ/min

( )( ) kg/s 72.0290287.0)101)(6)(1.0( 2 == x

Assumptions : Steady flow, PE = 0, ideal gas


A water pump is used to pump water in a pipe at the rate of 10 kg/s. The inlet conditions are 1.0 bar, 25 oC and velocity 3 m/s while the outlet conditions are 1.5 bar, 30 oC and velocity 15 m/s. The outlet pipe is located 15 m above the inlet pipe. Determine the power needed by the pump in kW. The gravitational acceleration is 9.81 m/s2.

MY EXAMPLE 5.3


p1 =1.0 barT1 = 25 oC

p2 = 1.5 barT2 = 30 oC

W

( ) )zz(g2

VV + +h - h m= W 2122

21

21

+

&&

Since T1 < TS at 1 bar and T2 < Ts at 1.5 bar, thus the condition of the inlet and the outlet water is compressed liquid. From Table A-4,

( )

+

100015)9.81(

2000153 + +125.74 - 104.83 10= W

22&

( ) )zz(g2

VV + +h - h m= W-Q 2122

21

21

+

&&&

h1 = hf at 25 oC = 104.83 kJ/kg h2 = hf at 30 oC = 125.74 kJ/kg

0

s/m 15V2 =

s/m 15V1 =

15 m

= -211.55 kW

Pump

Assumptions : Steady flow and heat transfer is negligible


11

TURBINE


Turbine is the device that drives the electric generator in steam, gas or hydro-electric power plant.

As the high velocity fluid passes through the turbine, work is done against the blades, which are attached to the shaft. As a result, the shaft rotates and the turbine produces work

Potential energy is negligible Heat transfer from turbines is usually negligible,

normally they are well insulated The process occurs in turbine is an expansion process

which amount of work is produced

1

2

W

( )

12

21

22

12 z-zg+ 2VV+)h-(hm =W-Q &&&

)h - (hm = W 21&&

0 0 0

2222 z,V,T,p

1111 z,V,T,p

Energy balance for a turbine is


EXAMPLE 5-7


( )

12

21

22

12 z-zg+ 2VV+)h-(hm =W-Q &&&

0


The power output of an adiabatic steam turbine is 5 MW and the inlet and the exit conditions of the steam are as shown in the Figure 5-28.a) Compare the magnitudes of h, ke and peb) Determine the work done per unit mass of the steam flowing through the turbinec) Calculate the mass flow of the steamAssumptions : Steady flow process, no heat transfer (adiabatic)a) At the inlet

P1 = 2 MPaT1 = 400oC

Shsh1 = 3248.4 kJ/kg (Table A-6)

At exit (mixture), h2 = hf + x2hfg = 225.94 + 0.9(2372.3) = 2361.01 kJ/kg

( ) ( ) kJ/kg 0.041069.81zzgpekJ/kg 14.95

2x100050180

2VVke

kJ/kg 887.393284.42361.01hhh

12

2221

22

12

======

===

b)

kg/s 5.73872.485000

wWm === &&c) Mass flow rate,

( ) kJ/kg 872.480.0414.95887.39W =+=&

12

Gas hasil pembakaran bahan api memasuki turbin loji kuasa turbin gas pada 7 bar dan 400 oC dan keluar pada 1 bar dan 30 oC. Kadar alir jisim gas adalah 2 kg/s dan garis pusatsalur masuk turbin adalah 200 mm. Semasa proses pengembangan berlaku, sebanyak150 kJ/min haba tersingkir dari turbin ke sekitaran. Gas hasil pembakaran boleh dianggapsebagai gas unggul dengan cp = 1.15 kJ/kgK dan R = 0.285 kJ/kgK. Dengan mengabaikanperubahan tenaga kinetik dan tenaga upaya, tentukan kuasa yang dihasilkan dan halaju gas semasa memasuki turbin

MY EXAMPLE 5-4


[ ] ( )[ ] kW 848.5 60

15030-400 1.15 2 +)T-(Tcm W 21p =

+=+= Q&&

1

11VA=m&

( ) 223-21 m 0.03144

200x10 4D === A

( ) kg/m 0.27410 x 7

273400 0.285p

RT 32

1

11 =+==

m/s 17.45===0.0314

0.274 x 2A V

1

11

m&

( )

12

21

22

12 z-zg+ 2VV+)h-(hm =W-Q &&&

0 0Q)h-(hcm =W 21p &&& +

1

11 A

V m&=


Throttling valves are any kind of flow-restricting devices that cause a significant pressure drop in the fluid and any device used to control the flow rate of a fluid.

Examples : adjustable valves, capillary tube and orifice tube Commonly used in refrigeration and air-conditioning applications Change in PE and KE and heat transfer are negligible and no work done

THROTTLING VALVES/DEVICES


1 2

h1 = h2

kJ/kg )gz-(gz+ 2

VV)h- h( w-q 1221

22

12

+=

00 00


13

Refrigerant-134a enters the capilary tube of a refrigerator as saturated liquid at 0.8 MPaand is throttle to a pressure of 0.12 MPa. Determine the quality of the refrigerant at the final state and the temperature drop during this process.

THROTTLING VALVES/DEVICES


h1 = h2 kJ/kg )gz-(gz+ 2VV)h- (h w-q 12

21

22

12

+=0


00

0

Assumptions : Steady flow, heat transfer, PE and KE are negligible, Analysis

MIXING CHAMBERS


The function of mixing chamber is to mix two or more streams of fluid

Examples : T-elbow or Y-elbow of a pipe, engine carburetor, the mixing process of fresh air and return air in a/c application.

PE, KE and heat transfer are negligible and do not involve any kind of work.

The total inlet mass flow rate is equal to the total outlet mass flow rate

mixingchamber

1

2

3

+++

++

++= 2

22

221

21

113

23

33 gz2Vhmgz

2Vhmgz

2VhmWQ &&&&& 332211 hmhmhm &&& =+

321 mmm &&& =+

outoutinin hm = hm &&

1m&

2m&3m&

0

outin m = m &&

0 0 0 0 0 0 0


Lets consider a mixing chamber with two inlets and one outlet. For conservation of mass, we can write

We can write the energy balance for this system as

In general,

14

EXAMPLE 5-9



Consider an ordirnary shower where hot water at 60oC is mixed with cold water at 10oC. If it is desired that a steady stream of warm water at 45oC be supplied, determine the ratio of the mass flow rates of the hot to cold water. Assume the losses from the mixing chamber to be negligible and the mixing to take place at a pressure of 150 kPa.

Assumptions : Steady flow process, KE, PE and heat losses are negligible. W = 0Analysis

Mass balance, 321 mmm &&& =+Energy balance, outoutinin hm = hm && 332211 hmhmhm &&& =+

( ) 3112211 hmmhmhm &&&& +=+Combining both equationsDividing by 2m&

32

11

2

22

2

11 hm

mmm

hmm

hm

+=+ &

&&&

&&

&

( ) 32132

12

2

11 h1yh yh h1mmh

mhm +=+

+=+ &

&&

&

Ts at 150 kPa is 111.35oC, thus, the water in all streams exist as a compressed liquid,h1 = hf @ 60oC = 251.18 kJ/kg, h2 = hf @ 10oC = 42.022 kJ/kg, h3 = hf @ 45oC = 188.44 kJ/kg

2.33188.44251.1842.022188.44

hhhhy

mm

31

23

2

1 ==

==&&

T1=60oC

T2=10oC T3=45oC

In a air conditioning duct system, a mixing process occurs between stream of outside fresh air stream of return air. 80 l/s of outside air at 30 oC is mixed with 240 l/s return air at the temperature of 26 oC at constant pressure of 1.1 bar. Determine the flow rate and the temperature of the mixed air. Assume air as an ideal gas with R = 0.287 kJ/kgK and cp = 1.005 kJ/kgK.

MY EXAMPLE 5-5


( )/kgm 0.7906

10 x 1.1273 30 x 0.287

pRT

3

21

11

=

+== ( )/kgm 0.7801 10 x 1.1

273 26 x 0.287p

RT

3

22

22

=

+==

kg/s 0.1012 0.7906 x 1000

80Vm1

11

===

&&

kg/s 0.3077 0.7801 x 1000

240Vm2

22

===

&&

kg/s 0.40890.30770.1012mmm 213 =+=+= &&&

p3

2p21p13 cm

TcmTcmT &

&& +=( ) ( ) K 300

0.4089273260.3077273300.1012 =+++=

332211 hmhmhm &&& =+ 3p32p21p1 TcmTcmTcm &&& =+

3

2211

mTmTm

&&& +=

Outside air80 l/s30 oC

Return air240 l/s26 oC

1

3

2

Mixed air


15

HEAT EXCHANGER


HE is a devices where two moving fluid streams exchange heat without mixing.

Widely used in various industries such as condenser in power plant, radiator in automobile engines, evaporator in a/c system, oil cooler and etc.

The heat transfer between two fluids occurs at constant pressure

Involve no work interactions (W = 0), PE, KE and heat transfer to the surrounding are negligible

Fluid A inlet

Fluid Boutlet Fluid Binlet

Fluid Aoutlet


Q = 0

Fluid A hotter than B

HEAT EXCHANGER


1 (Fluid A)

h (hm = Q 12AA )&&

( ) z-g(z2

V-V + h-h m= W-Q 1221

22

12AAA

+

)&&&


Q = 0a) Considering the whole HE as a system, the energy

balance can be written as( ) ( )( ) ( ) gzVhmgzVhm

gzVhmgzVhm= W-Q

3232

1331

212

111

4242

1442

222

122

+++++++++

&&&&&&

2 (Fluid A)

4 (Fluid B)

3 (Fluid B)Neglecting the Q, PE and KE and for steady flow, ( )4321 mm and mm &&&& ==

( ) ( ) ( ) ( )( ) ( )34B12A

33114422

hhmhhm0 hmhmhmhm

==+

&&&&&& -

b) Considering the fluid B as a system, the energy balance can be written as 2

1

Neglecting the PE and KE,

16

Refrigerant 134a is to cooled by water in a condenser. The refrigerant enters the condenser with a mass flow rate of 6 kg/min at 1 MPa and 70oC and leaves at 35oC. The cooling water enters at 300 kPa and 15oC and leaves at 25oC. Neglecting any pressure drops, determine the mass flow rate of the cooling water required and the heat transfer from the refrigerant to the water.

EXAMPLE 5-10 page 241



Assumptions : Steady flow process, KE and PE are negligible, heat losses from the system are negligible, W = 0Analysis : For each stream, R43W21 mmm and mmm &&&&&& ====

Energy balance, /dtdEEE systemoutin = &&( ) ( )34R12W44223311 hhmhhm hmhmhmhm =+=+ &&&&&&

( )( )

( ) kg/min 29.162.982104.83

303.85100.876hh

hhmm12

34RW =

== &&

Ts at 300 kPa = 133.52oC, thus the conditions of both water streams are compressed liquid,

h1 hf @ 15oC = 62.982 kJ/kg and h2 hf @ 25oC = 104.83 kJ/kg (Table A-4)

At 1 MPa and 70oC, R134a exist as superheated vapor, h3 = 303.85 kJ/kg (Table A-13)

At 1 MPa and 35oC, R134a exist as compressed liquid, h4 hf @ 35oC = 100.87 kJ/kg (A-11)

( ) ( ) kJ/min 121862.982104.8329.1hhmQ 12wW ===&



PIPE AND DUCT

Pipes are usually used to transport liquids, while ducts are usually used to transport gases or air, particularly in air conditioning system.

Under normal operating conditions, the amount of heat gained or lost by the fluid may be very significant, especially if the pipe/duct is long. Normally they are insulated to prevent any heat gain or loss.

If the pipe or duct involves a heating section, a fan or a pump, the work interactions should be considered.

The velocities in pipe and duct flow are relatively low, thus the KE changes are usually insignificant.

17



PIPE AND DUCT

In an air duct system, air is heated as it flows over a 15 kW resistance wires. Air enters the heating section at 100 kPa and 17oC with a volume flow rate of 150 m3/min. If heat is lost from the air in the duct to the surroundings at a rate of 200 W, determine the exit temperature of air.

Assumptions : Steady flow, air is an ideal gas, PE and KE are negligibleAnalysis : Heating section is taken as the system, heat is lost from the system and electrical work is supplied to the system

Energy balance, outin EE && =

( )12poutine,2out1ine, TTcmQW hmQhmW =+=+ &&&&&&&( )( )

kg/s 3.060 x 0.832

150Vm

/kgm 0.832100

2900.287p

RT

1

1

3

1

1

===

===

&

&

1

( ) C21.9C294.92901.00530.215T

cmQW

T oo1p

outine,2 ==+=+= &

&&



ENERGY ANALYSIS OF UNSTEADY-FLOW PROCESSES

_ Unsteady-flow or transient-flow processes are the processes which involve changes in mass and energy contents within the control volume with time as well as the energy interactions across the boundary

_ Example of unsteady-flow processes : the charging of rigid vessels from supply lines, discharging of fluid from a pressurized vessel, driving a gas turbine with pressurized air stored in a large container, inflating tires or balloons and cooking with an ordinary pressure cooker.

_ The differences between unsteady-flow and steady flow : Unsteady-flow processes start and end over some time

period, changes occur over some time interval, t. The mass within the system boundaries does not remain

constant during a process. Steady-flow systems are fixed in space, size and shape,

while the unsteady-flow systems are not and may involve moving boundaries, thus boundary work.

_ Most unsteady-flow can be represented by the uniform-flow process the fluid properties do not change with time or a cross section of an inlet or exit.

Charging of a rigid tank from a supply lines

Shape and size of a control volume may change

18



ENERGY ANALYSIS OF UNSTEADY-FLOW PROCESSES

The mass balance for any system undergoing any process can be expressed as

min mout = msystem or min mout = (m2 m1)CV

The general energy balance : Ein Eout = EsystemThe energy balance for a uniform-flow can be expressed as

( ) ( )( ) ( )( )system1122

outoutout

ininin

pekeumpekeum

pekehmWQpekehmWQ

++++=

++++

++++

If the ke and pe are negligible, then

( )system1122inout

umummhmhWQ += If no mass enters or leaves the CV during a process (mi = me = 0 and m1 = m2 = m) , this equation reduces to the energy balance for closed system



EXAMPLE 5-12

A rigid tank, insulated tank that is initially evacuated is connected through a valve to a supply line that carries steam at 1 MPa and 300oC. Now the valve is opened and steam is allowed to flow slowly into the tank until the pressure reaches 1 MPa at which point the valve is closed. Determine the final temperature of the steam in the tank.

Assumptions : uniform flow process, ke and pe are negligible for both streams and system, no work interactions involve, no heat transfer (insulated)

Analysis : the CV initially evacuated, thus m1 = 0 and m1u1 = 0The mass balance : min mout = msystemH mi = m2 m1 = m2The energy balance : Ein Eout = EsystemSince W = 0, Q = 0, ke = pe = 0, m1 = 0, thus mihi = m2u2Combining both equation, u2 = hi

pi = 1 MPaTi = 300oC

hi = 3051.6 kJ/kg = u2 (Table A-6)

p2 = 1 MPau2 = 3051.6 kJ/kg

T2 = 456.1 (Table A-6)

19


THE END


chap5 open system

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