tutorial sheet 02 answers 2014

8/12/2019 Tutorial Sheet 02 Answers 2014

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EG-210 Tutor ial Sheet No. 2 (2013) Answers

The Second Law of Thermodynamics and Cyclic Processes

1) (a) Steam enters a horizontal pipe operating at steady state with a specific enthalpy of 3000

kJ/kg and a mass flow rate of 0.5 kg/s. At the exit, the specific enthalpy is 1700 kJ/kg. If

there is no significant change in kinetic energy from inlet to exit, determine the rate of heat

transfer between the pipe and its surroundings, in kW.

Answer

Diagram of system:

Starting with the general energy balance for open systems:

)()()(02

2

2

121

2121 vvzzghhmWQ S

No work is done (WS= 0) and ignoring potential and kinetic energy,

)(0 21 hhmQ

)( 12 hhmQ

kJ/kg650)30001700(5.0 Q

[5 marks]

(b) Steam enters a well-insulated turbine operating at steady state at 4 MPa with a specific

enthalpy of 3015.4 kJ/kg and a velocity of 10 m/s. The steam expands to the turbine exit

where the pressure is 0.07 MPa, specific enthalpy is 2431.7 kJ/kg, and the velocity is 90 m/s.

The mass flow rate is 11.95 kg/s. Neglecting potential energy effects, determine the power

developed by the turbine, in kW.

Answer


h1= 3000 kJ/kgm= 0.5 kg/s

h2= 1700 kJ/kgm= 0.5 kg/s


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)()()(0 222

121

2121 vvzzghhmWQ S

We are told the turbine is well insulated, therefore Q= 0 and we also assume PE change is

zero, giving:

)()( 222

121

21 vvhhmWS

We know m= 11.95 kg/s, then;

kW69271000/)9010()7.24314.3015(22

21 mWS

[5 marks]

(c) Steam enters a turbine operating at steady state with a mass flow of 10 kg/min, a specific

enthalpy of 3100 kJ/kg, and a velocity of 30 m/s. At the exit, the specific enthalpy is 2300

kJ/kg and the velocity is 45 m/s. The elevation of the inlet is 3 m higher than at the exit. Heat

transfer from the turbine to its surroundings occurs at a rate of 1.1 kJ per kg of steam flowing.Letg =9.81 m/s2. Determine the power developed by the turbine, in kW.

Answer


)()()(0 222

121

2121 vvzzghhmWQ S

Re-arranging gives:

)()()(2

2

2

121

2121 vvzzghhm

Q

m

WS

Therefore:

kJ/kg37.7981000/)4530()1000/381.9()23003100(1.122

21

m

WS

Now m= 10 kg/min = 0.1667 kg/s, so:

WS= 0.1667 798.37 = 133.09 kJ/s = 133.09 kW[5 marks]


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(d)Air enters a compressor operating at steady state at 1 atm with a specific enthalpy of 290

kJ/kg and exits at a higher pressure with a specific enthalpy of 1023 kJ/kg. The mass flow

rate is 0.1 kg/s. If the compressor power input is 77 kW, determine the rate of heat transfer

between the compressor and its surroundings, in kW. Neglect kinetic and potential energy

effects and assume the ideal gas model.

Answer


)()()(02

2

2

121

2121 vvzzghhmWQ S

Ignoring potential and kinetic energy, and re-arranging gives;

)( 12 hhmWQ S

kW7.3)2901023(1.077 Q [5 marks]

(e) A pump delivers water through a hose terminated by a nozzle. The exit of the nozzle has a

diameter of 2.5 cm and is located 4 m above the pump inlet pipe, which has a diameter of 5.0

cm. The pressure is equal to 1 bar at both the inlet and the exit, and the temperature is

constant at 20 C. The magnitude of the power input required by the pump is 8.6 kW and the

acceleration due to gravity, g = 9.81 m/s2. The specific volume of water at 20 C and 1 bar is

1.0018 10-3m3/kg. Determine the mass flow rate delivered by the pump in kg/s.

Answer


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)()()(0 222

121

2121 vvzzghhmWQ S

Assuming that heat transfer is negligible (Q= 0) and as the input and output water conditions

are the same (h1= h2), then;

)()(0 22212121 vvzzgmWS

Now assuming that the water is incompressible the velocity of water can be calculated using:

Avolumespecificmv /

So

mmv 5102.0)4/05.0/(100018.1 231

mmv 04085.2)4/025.0/(100018.123

2

1000/))04085.2()5102.0((1000/)4(81.9)6.8(022

21 mmm

mm 03924.0109524.16.80 33

Solving for m gives;

m= 15.98 kg/s

[5 marks]


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2) (i) The data listed below are claimed for a power cycle operating between hot and cold

reservoirs at 1000 K and 300 K, respectively. For each case, determine whether the cycle

operates reversibly, operates irreversibly, or is impossible.

(a) Qh= 600 kJ, Weng= 300 kJ, Qc= 300 kJ

(b) Qh= 400 kJ, Weng= 280 kJ, Qc= 120 kJ

(c) Qh= 700 kJ, Weng= 300 kJ, Qc= 500 kJ(d) Qh= 800 kJ, Weng= 600 kJ, Qc= 200 kJ

Answer

For each process first check the first law balance: Weng= Qh- Qcand if OK then check that:

For a reversible process:c

c

h

h

T

Q

T

Q

For an irreversible process:c

c

h

h

T

Q

T

Q

For an impossible process:c

c

h

h

T

Q

T

Q

a) First Law: Weng= Qh- Qc 300 = 600300 First law OK so,

6.01000

600

h

h

T

Q 1

300

300

c

c

T

Q therefore

c

c

h

h

T

Q

T

Q so process is irreversible

b) First Law: Weng= Qh- Qc 280 = 400120 First law OK so,

4.01000

400

h

h

T

Q 4.0

300

120

c

c

T

Q therefore

c

c

h

h

T

Q

T

Q so process is reversible

c) First Law: Weng= Qh- Qc 300 700 500 First law violated so process impossible

d) First Law: Weng= Qh- Qc 600 = 800200 First law OK so,

8.01000800

h

h

TQ 66.0

300200

c

c

TQ therefore

c

c

h

h

TQ

TQ so process is impossible

[4 marks]

(ii) A reversible power cycle operating between hot and cold reservoirs at 1000 K and 300 K,

respectively, receives 100 kJ by heat transfer from the hot reservoir for each cycle of

operation. Determine the net work developed in 10 cycles of operation, in kJ.

Answer

First calculate the efficiency of the heat engine using;


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7.01000

30011

h

c

T

T

The efficiency can also be calculated using:

h

eng

Q

W

inputheat

outputnet work

So,

Weng= Qh= 0.7 100 = 70 kJ

For 10 cycles power generated is:

Net work = no cycles Weng= 10 70 = 700 kJ [3 marks]

(iii) In a heat-treating process, a 1-kg metal part, initially at 1075 K, is quenched in a tank

containing 100 kg of water, initially at 295 K. There is negligible heat transfer between the

contents of the tank and their surroundings. The metal part and water can be modeled as

incompressible with specific heats 0.5 kJ/kg K and 4.2 kJ/kg K, respectively. Determine (a)

the final equilibrium temperature after quenching, in K, and (b) the amount of entropy

produced within the tank, in kJ/K.

Answer

First write the equation describing the energy change of the metal in cooling from 1075 K toT.

Qmet= mCpT= 1 0.5 (1075T) = 0.5 (1075T)

Next, write the equation describing the energy change of the water in being heated from 295

C to T.

Qwat= mCpT= 100 4.2 (T295) = 420 (T295)

Applying first law heat balance:

Qwat= Qmet

420 (T295) = 0.5 (1075T)

T= 295.93 K

Calculate the entropy change of the copper,

K/kJ645.0

1075

93.295ln5.01ln

1

2

T

TmCS pmet


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Calculate the entropy change of the water,

K/kJ322.1295

93.295ln2.4100ln

1

2

T

TmCS pwat

Calculate the overall entropy change,

K/kJ677.0322.1645.0 watmetT SSS [6 marks]

(iv) Propane at 0.1 MPa, 20 C enters an insulated compressor operating at steady state and

exits at 0.4 MPa, 90 C Neglecting kinetic and potential energy effects, determine (a) the

power required by the compressor, in kJ per kg of propane flowing. (b) the rate of entropy

production within the compressor, in kJ/K per kg of propane flowing. Data supplied: at 0.1

MPa, 20 C hpropane= 517.6 kJ/kg and spropane= 2.194 kJ/kg K, at 0.4 MPa, 90 C hpropane=

639.2 kJ/kg andspropane= 2.311 kJ/kg K,

Answer


)()()(0 222

121

2121 vvzzghhmWQ S

As the compressor is insulated Q= 0 and neglecting PE and KE, gives;

21 hh

m

WS

kg/kJ6.1212.6396.517 m

WS

Starting with the entropy balance:

genjjii

k

k SsmsmT

Q 0

As the mass flow is constant and Qis zero then;


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genSssm )(0 21

KkJ/kg117.0194.2311.212 ssm

Sgen

[3 marks]

(v) The pressurevolume diagram of a Carnot power cycle executed by an ideal gas with

constant specific heat ratio is shown below. Show that V4V2= V1V3

Answer

As step 12 occurs at a constant temperature which is higher than step 3 to 4 then,

T1= T2= Th and T3= T4= Tc

For an isothermal change:

1

2ln

V

VRTWQ therefore;

For step 1 to 2:

1

2

12 ln

V

VRTW h

For step 3 to 4:

3

4

34 ln

V

VRTW c

The cycle efficiency is given by:

h

c

in T

T

Q

WW

1

inenergy

net work 3412


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h

c

h

ch

T

T

V

VRT

V

VRT

V

VRT

1

ln

lnln

1

2

3

4

1

2

h

c

h

c

T

T

V

VRT

V

VRT

1

ln

ln

1

1

2

4

3

h

c

h

c

T

T

VVT

V

VT

1

2

4

3

ln

ln

Therefore,

1

2

4

3lnln

V

V

V

V

1

2

4

3

V

V

V

V

V3V1= V4V2

[6 marks]


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3) A heat engine cylinder contains 0.1 kg of air. The engine is assumed to operate on a Carnot

cycle comprising the following four processes:

Step A-B: Isothermal expansion at 1000 K during which the volume of the air isdoubled.

Step B-C: Adiabatic expansion until the temperature of the air reaches 350 K and thepressure 1 bar.

Step C-D: Isothermal compression of the air at 350 K. Step D-A: Adiabatic compression until the air temperature reaches 1000 K and the

cycle is complete.

Assuming that air behaves as an ideal gas and given the following data (specific heat capacity

of air in a constant volume process CV= 0.7175 kJ kg-1K-1; Gas constant for air R = 0.2867

kJ kg-1K-1), then

a) Sketch the cycle on a P-V diagram.b) What is the maximum pressure achieved in the cycle?c) Show that the volume of air is halved during the isothermal compression.d) Calculate the work done during each of the adiabatic processes.e) Calculate the heat and thus work effects for each of the isothermal processes.f) Determine the net work done per cycle.g) Confirm the cycle efficiency matches the prediction given by = 1(Tc/Th)

[25 marks]

Answer:

a)

[4 marks]

b) Step B-C is an adiabatic expansion from TB=1000 K to TC= 350 K,PC= 1 bar, therefore;

1

B

C

B

C

P

P

T

T


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Re-arranging

1

C

BCB

T

TPP

Now

CP= CV+R= 0.7175 + 0.2867 = 1.0042 kJ kg-1K-1

So,

1.39960.7175

1.0042

V

P

C

C

Resulting in:

bar53.39350

10001

13996.1

3996.1

BP

As air behaves as an ideal gas:

PAVA=PBVB

Also, VB= 2VA, so;

PAVA= 39.53 2 VA

PA= 79.06 bar

[6 marks]

c) For adiabatic compression from TD= 350 K to TA= 1000 K,PA= 79.06 bar, then

1

A

D

A

D

P

P

T

T

13996.1

06.79

1000

350

DP

bar00.21000

35006.79

13996.1

3996.1

DP

For an ideal gas:

PDVD=PCVC

VD=PCVC/PD= 1 VC/ 2 = 0.5VC[5 marks]


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d) Step B-C: Adiabatic expansion;

WBC= m CV(TBTC) = 0.1 0.7175 (1000350) = 46.64 kJ

Step D-A: Adiabatic compression;

WDA= m CV(TDTA) = 0.1 0.7175 (3501000) = -46.64 kJ

[3 marks]

e) Step A-B: Isothermal expansion at 1000 K

kJ87.1953.39

06.79ln10000.28670.1ln

B

AhABAB

P

PmRTWQ

Step C-D: Isothermal compression at 350 K

kJ96.62

1ln0350.28670.1ln

D

CcCDCD

P

PmRTWQ

[3 marks]

f) Net work;

Wnet= WAB+ WBC+ WCD+ WDA= 19.87 + 46.646.9646.64 = 12.91 kJ

[2 marks]

g) Efficiency is given by

65.087.19

91.12

AB

net

in

net

Q

W

Q

W

Check;

65.01000

35011

h

c

T

T

[2 marks]


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4)At the Bear Point refinery complex on the St Lawrence Seaway between Canada and the

USA, the process waste gas is being considered as a thermal energy source for raising steam

to generate electrical power in an energy park associated with the refinery. The preliminary

design of a Rankine Cycle steam turbine power plant is being considered to make use of this

thermal energy. In the proposed design, steam is to be supplied from a boiler to the turbine

inlet at a pressure of 15 MPa and temperature of 600 C whilst the exhaust pressure from theturbine is 10 kPa. The exhaust mixture is passed through a condenser and then pumped back

to the boiler. The power plant is assumed to be a closed system.

a) Draw a block flow diagram of the proposed power plant and a general Temperature-

Entropy diagram of the cycle, labelling the appropriate points.

[6 marks]

b) Calculate the operational data for the steam/water after each step of the cycle, namely:

temperature, pressure, quality, specific enthalpy, specific entropy. State any assumptions you

make.

[12 marks]

c) Calculate the thermal efficiency of the cycle.

[2 marks]

d) Determine the flow rate of water/steam required to produce 60 MW.

[2 marks]

e) What methods could be employed to improve the thermal efficiency of the cycle?

[3 marks]

Answer

a) Diagram:

Boiler

Turbine

Compressor

(pump)

Heat

P1= 10 kPa 1

P2= 15 MPa 2 3 P3= 15 MPa

T3= 600C

4 P4= 10 kPa

Qc

Qh

WoutWin


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[6 marks]

b) Assumptions:

Pump and turbine are isentropic P2= P3= 15 MPa T3= 600C P4= P1= 10 kPa The working fluid is completely condensed to a liquid in the condenser Kinetic and potential energy changes are zero

Known Data:

State T (C) P (kPa) h (kJ/kg) s (kJ/kg K) x

1 ? 10 ? ? ?

2 ? 15000 ? ? ?

3 600 15000 ? ? ?

4 ? 10 ? ? ?

At State 3:

P3= 15 MPa = 150 bar and T3= 600 C. From the Steam Tables we can see that this situation

corresponds to superheated steam (i.e. the water is all vapour and the quality of the steam, x,

is equal to 1) so:

hg3= h3= 3581 kJ/kg sg3= s3= 6.677 kJ/kg K

At State 4:

The turbine is isentropic so the water/steam mixture at State 4 has an entropy of:

s4= s3= 6.677 kJ/kg K

T

s

1

2

3

4

P = 10 kPa

P= 15 MPa

T3= 600oC


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The steam has been condensed in the turbine to a mixture of water and steam at a pressure of

P4= 10 kPa = 0.1 bar and a temperature of T4= 45.8 C (this temperature is from the given

pressure and Steam Tables). At these conditions:

sf4= 0.649 kJ/kg K sfg4= 7.5 kJ/kg K sg4= 8.149 kJ/kg K

hf4= 192 kJ/kg hfg4= 2392 kJ/kg hg4= 2584 kJ/kg

If we let the quality of steam (fraction of water as steam) = x, then:

smix= x sg+ (1x) sf

Now:

s4= x sg4+ (1x) sf4

6.677 = 8.149 x + 0.649 (1x)

x = (6.6770.649) / (8.1490.649) = 0.804

Also

hmix= x hg+ (1x) hf

so,

h4= x hg4+ (1x) hf4

h4= (0.804 2584) + [(10.804) (192)]

h4= 2115.2 kJ/kg

At State 1:

The pressure is P1= 10 kPa = 0.1 bar and the temperature T1= 45.8 C, therefore from Steam

Tables:

hf1= 192 kJ/kg hfg1= 2392 kJ/kg hg1= 2584 kJ/kg

sf1= 0.649 kJ/kg K sfg1= 7.5 kJ/kg K sg1= 8.149 kJ/kg K

Here we need to make an assumption that steam/water mixture from State 4 is condensed

down to just water at State 1 (so quality of steam, x, at State 1 = 0). This is a common

assumption in Rankine cycles as it makes pumping of the fluid easier. Therefore:

h1= 192 kJ/kg s1= 0.649 kJ/kg K

We are also given the specific volume of liquid water at 10 kPa = 0.00101 kg/m3in the

question. Therefore:


1 45.8 10 192 0.649 0

2 ? 15000 ? ? ?

3 600 15000 3581 6.677 1

4 45.8 10 2115.2 6.677 0.804


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At State 2:

The pump work is given by:

1212in hh)PP(W v

kJ/kg15.1J/kg15139.9)1000015000000(00101.0Win

Therefore the enthalpy at pump outlet:

h2= Win+ h1

h2= 15.1 + 192 = 207.1 kJ/kg

This process is also isentropic so:

s2= s1= 0.649 kJ/kg

From the steam tables, saturated liquid water at P = 15 MPa has T S= 342.1 C, hf= 1610 kJ /kg, sf = 3.685 kJ / kg K, which are greater than conditions at position [2] so water is

subcooled, giving:

T2= 45.8 C P2= 15 MPa h2= 207.1 kJ / kg s2= 0.649 kJ / kg K x = 0

Giving:


1 45.8 10 192 0.649 0

2 45.8 15000 207.1 0.649 0

3 600 15000 3581 6.677 14 45.8 10 2115.2 6.677 0.804

[12 marks]

c) We now have all the information we need to calculate the cycle efficiency.

Heat in to Boil er:

Qh= h3h2= 3581207.1 = 3373.9 kJ/kg

Heat removed by condenser:

Qc= h4h1= 2115.2192 = 1923.2 kJ/kg

Turbine work:

Wout= h3h4= 35812115.2 = 1465.8 kJ/kg

Pump work (alr eady calculated):

Win= 15.1 kJ/kg

Cycle effi ciency:

h

inout

h

net

Q

WW

Q

W


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42998.09.3373

1.158.1465

Q

W

h

net

Therefore the cycle efficiency is:

= 43 %

[2 marks]

d) Net work output = 60 MW:

Net work output = mass flow rate (Wout- Win)

= mass flow rate ((h3- h4) - (h2- h1))

60 MW = mass flow rate ((h3- h4) - (h2- h1))

mass flow rate = 60000/((35812115.2) - (207.1 - 192)) = 41.36 kg/s

[2 marks]

e) Thermal efficiency can be improved by:

(a) Lowering the condensing pressure (lower condensing temperature, lower Tc)

(b) Superheating the steam to higher temperature

(c) Increasing the boiler pressure (increase boiler temperature, increase Th)

[3 marks]


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5) Refrigerant 134a is the working fluid in an ideal vapour-compression refrigeration cycle

that communicates thermally with a cold region at -4C (minus four) and a warm region at

24C. Saturated vapour enters the compressor at -4C and saturated liquid leaves the

condenser at 24C. The mass flow rate of the refrigerant is 0.2 kg/s.

a) Draw a block flow diagram of the proposed refrigeration plant and explain how therefrigeration cycle works.

[5 marks]

b) Draw a Temperature-Entropy diagram of the cycle, labelling the appropriate points.

[2 marks]

c) Calculate the operational data for the refrigerant after each step of the cycle, namely:

temperature, pressure, quality, specific enthalpy and specific entropy. State any assumptions

you make.

[10 marks]

d) Calculate the compressor power, in kW.

[2 marks]

e) Calculate the refrigeration capacity, in kW.

[2 marks]

f) Determine the coefficient of performance of the ideal vapour-compression refrigeration

cycle.

[2 marks]

g) Determine the coefficient of performance of an ideal Carnot refrigeration cycle operating

at the same conditions.

[2 marks]

Answer:

a)Diagram:


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19

Process 12: Isentropic compression of the refrigerant from state 1 (usually from a saturated

vapour) to the condenser pressure at state 2 (usually a superheated vapour).

Process 23:Heat transfer from the refrigerant as it flows at constant pressure through the

condenser. The refrigerant exits at state 3 (usually as a saturated liquid).

Process 34:Throttling process from state 3 (usually as a saturated liquid) to a two-phase

liquidvapour mixture at state 4. This occurs at constant enthalpy.Process 41: Heat transfer to the refrigerant as it flows at constant pressure through the

evaporator to complete the cycle.

[5 marks]

b)

[2 marks]

c) Assumptions:

The compressor is isentropic. The condenser and evaporator operate at constant pressure. The working fluid is completely condensed to a liquid in the condenser The compressor and expansion valve operate adiabatically. Kinetic and potential energy changes are negligible. T3= 24C T1= T4= -4C

Known Data:

State T (C) P (bar) h (kJ/kg) s (kJ/kg K) x1 -4 ? ? ? 1

2 ? ? ? ? 1

3 24 ? ? ? 0

4 -4 ? ? ? ?

At State 1:

We have a saturated vapour at T1= -4 C. To get the other properties at this condition is a

straightforward look-up in the Saturated Refrigerant 134a Temperature table:

h1= hg @ -4C= 244.90 kJ/kg


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20

s1=sg @ -4C= 0.9213 kJ/kg K

P1=P@ -4C= 2.5274 bar

At State 2:

At state 2 we have a superheated vapour. The pressure of this vapour is the pressureaccording to the temperature of the liquid coming from the condenser i.e. the pressure when T

= 24C. From the Saturated Refrigerant 134a Temperature table this pressure is:

P2=P@ 24C= 6.4566 bar

The compressor is isentropic so the refrigerant at State 2 has an entropy of:

s2=s1= 0.9213 kJ/kg K

In order to calculate the enthalpy and temperature at state 2 we now need to use the tables for

the properties of the Superheated Refrigerant 134a. However, we have a problem as there isno exact table for a pressure of 6.4566 bar, hence we need to use the tables at 6 and 7 bar and

perform a double interpolation as follows.

First we need to generate a new T-h-s table for superheated vapour at a pressure of 6.4566

bar. We will do this by interpolating h and s data at each temperature data point i.e. Tsat,

30C, 40C, 50C etc. as far as is necessary.

We already know that when atP= 6.4566 bar, Tsat= 24C, so now we can interpolate hands

as follows:

)()(

)(64566.6

67

6@7@

[email protected]@ barbar

barbar

TbarTbar

TbarTbar PPPP

hhhh satsat

satsat

From tables:

hg @ 7 bar Tsat= 261.85 kJ/kg

hg @ 6 bar Tsat= 259.19 kJ/kg

so

kg/kJ40.260)64566.6()67(

)19.25985.261(19.2594566.6@

satTbar

h

Doing the same for the enthalpy;

)()(

)(64566.6

67

6@7@


barbar

TbarTbar

TbarTbar PPPP

ssss satsat

satsat

From tables:

sg @ 7 bar Tsat= 0.9080 kJ/kg


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21

sg @ 6 bar Tsat= 0.9097 kJ/kg

so

Kkg/kJ9089.0)64566.6()67(

)9097.09080.0(9097.0

4566.6@

satTbars

Repeat the calculations for the higher temperatures until the value of [email protected] barexceedss2. So

the next calculation is at T= 30C.

)()(

)(64566.6

67

C306@C307@


barbar

barbar

barbar PPPP

hhhh

From tables:

hg @ 7 bar 30C= 265.37 kJ/kg

hg @ 6 bar 30C= 267.89 kJ/kg

so

kg/kJ74.266)64566.6()67(

)89.26737.265(89.267C304566.6@

barh

Doing the same for the enthalpy;

)()(

)(

64566.667

C306@C307@

[email protected]@ barbarbarbar

barbar

barbar PPPP

ss

ss

From tables:

sg @ 7 bar 30C= 0.9197 kJ/kg

sg @ 6 bar 30C= 0.9388 kJ/kg

so

Kkg/kJ9301.0)64566.6(

)67(

)9388.09197.0(9388.0

304566.6@

bars

Hence we have already exceeded s2and our superheated refrigerant table at P= 6.4566 bar

reads:

Enthalpy

kJ kg-1

Entropy

kJ kg-1

K-1

T

C

hg sg

24 260.40 0.9089

30 266.74 0.9301


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22

We now move onto the second part of the interpolation to calculate T2and h2at s2= 0.9213

kJ/kg K. Using the table generated we get:

)()(

)(9089.09213.0

9089.09301.0

[email protected]@

9089.0@@2 2ss

ss

TTTT

ss

ss

C5.27)9089.09213.0()9089.09301.0(

)2430(24

2@2

sT

and

)()(

)(9089.09213.0

9089.09301.0

[email protected]@

9089.0@@2 2ss

ss

hhhh

ss

ss

kg/kJ11.264)9089.09213.0()9089.09301.0()40.26074.266(40.260

2@2

sh

Summarizing

h2= 264.11 kJ/kg

s2= 0.9213 kJ/kg K

P2= 6.4566 bar

T2= 27.5 C

At State 3:

We have a saturated liquid at T3= 24 C. Again this becomes a straightforward read off from

the Saturated Refrigerant 134a Temperature table:

h3= hf @ 24C= 82.90 kJ/kg

s3= sf @ 24C= 0.3113 kJ/kg K

P3=P@ 24C= 6.4566 bar

At State 4:

The throttle valve is isenthalpic, hence:

h4= h3= 82.90 kJ/kg

T4= T1= -4C

P4=P@-4C= 2.5274 bar

The refrigerant is a two-phase liquidvapour mixture at a temperature of -4C. At these

conditions:

hf4=hf @ -4C= 44.75 kJ/kg hg4= hg @ -4C= 244.90 kJ/kg


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23

If we let the quality of refrigerant (fraction of refrigerant as vapour) =x, then:

hmix=xhg+ (1x) hf

Now:

h4=xhg4+ (1x) hf4

82.90 = 244.90x+ 44.75 (1x)

x= (82.9044.75) / (244.9044.75) = 0.1906

We can now calculate the entropy using:

sf4=sf @ -4C= 0.1777 kJ/kg K sg4=sg @ -4C= 0.9213 kJ/kg K

and smix=xsg+ (1x)sf

s4=xsg4+ (1x)sf4

s4= (0.1906 0.9213) + [(10.1906) (0.1777)]

s4= 0.3194 kJ/kg

Results:

StateT(

C)

P(bar) h(kJ/kg) s(kJ/kg K) x

1 -4 2.5274 244.90 0.9213 1

2 27.5 6.4566 264.11 0.9213 1

3 24 6.4566 82.90 0.3113 0

4 -4 2.5274 82.90 0.3194 0.1906

[10 marks]

d) Calculate the compressor power

Win= m(h2h1) = 0.2 (264.11244.90) = 3.842 kW

[2 marks]

e) The refrigeration capacity

Qc= m(h1h4) = 0.2 (244.9082.90) = 32.4 kW

[2 marks]

f) The coefficient of performance

43.8842.3

4.32

in

c

W

QCOP

[2 marks]


24/24

g) Carnot COP

61.9)]4(273[)24273(

)]4(273[

CH

C

TT

TCOP

[2 marks]

tutorial sheet 02 answers 2014

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