Technological University (Toungoo)

ME-1013 Basic Engineering Thermodynamic

Sample Questions

1

1.During a test on a four-stroke cycle oil engine the following data and results

were obtained:

Mean height of indicator diagram 21mm

Indicator spring number, 27 kN/m2/mm

Swept volume of cylinder, 14 litres

Speed of engine ,6.6 rev/sec

Effective brake load, 77kg

Effective brake radius, 0.7m

Fuel consumption, 0.002kg/s

Calorific value of fuel, 44000 kJ/kg

Cooling water circulatin, 0.15 kg/s

Cooling water inlet temperature, 38 C°

Cooling water outlet temperature, 71 C°

Specific heat capacity of water, 4.18 kJ/kg

Energy to exhaust gases, 33.6 kJ/s

Determine the indicated and brake outputs and the mechanical efficiency. Draw up an

overall energy balance in kJ/s and as a percentage. (20 marks)

Solution,

Mean height of indicator diagram 21mm

Indicator spring number, 27 kN/m2/mm

Vst = 14 litres = 14 310−× m3/s

N = 6.6 rev/s

m = 77 kg

F =77×9.81 Nr = 0.7m

kg/s 0.002m f =°

CV = 44000 kJ/kg

mw = 0.15 kg/s

T1 = 38 C°

T2 = 71 C°

cpw = 4.18 kJ/kg

Qex = 33.6 kJ/s

ip =? bp=? ?ηmech =

Indicated mean effective pressure

2

Pm = mean height of diagram× indicator spring number

= 21×27

= 567 kN/m2

Indicated power ip = Pm LANn

= 567 ×14×10-3×3.3×1

= 26.19 kW

Brake power bp = TNπ2 ×××

TorqueT = F ×r

= 77×9.81×0.7 Nm

0.79.81776.6π2bp ×××××=

= 21927.12 Watt

= 21.927 kW

Mechanical efficiency = 83.72%100%0.837226.19

21.219ipbp

=×==

Energy balance

Energy from fuel, fQ° = CVm f ×°

= 0.002×44000 = 88 kJ/s

Energy to brake power = 21.927 kW = 21.927 kJ/s

Energy to coolant = ( )TCm PWw. Δ××

= 0.15×4.18×(71-38)

= 20.69 kJ/s

Energy to exhaust = 33.6 kJ/s

Energy lost to surroundings = Energy from fuel-(Energy to bp + Energy to bp

+ Energy to coolant + Energy to exhaust)

= 88 - (21.927 + 20.69 + 33.6)

=11.78 kJ/s

Energy balance kJ/s %

Energy from fuel 88 100

Energy to brake power 21.927 24.92

Energy to coolant 20.69 23.5

Energy to exhaust 33.6 39.2

Energy to surroundings 11.78 13.3

3

2.The diameter and stroke of single cylinder gas engine, working on the constant

volume cycle, are 200mm and 300mm, respectively, and the clearance volume is 2.78

litres.

When running at 300 rev/min,the number of firing cycle/min was 135,the indicated

mean effective pressure was 518 kN/m2 and the gas consumption 8.8 m3/hr.Calorific

value of the gas used =16350 kJ/m3.

Determine

(a) the air standard efficiency;

(b)the indicated power developed by the engine;

(c) the indicated thermal efficiency of the engine ;

Assume γ =1.4 (20marks)

Solution

D = 200 mm = 200×10-3 m

L = 300 mm = 300×10-3 m

Vcl = 2.78 L =2.78×10-3

m3

N = 135 rpm

/sm 8.8/3600m 3f =°

imep = 518 kN/m2

CV = 16350 kJ/m3

ip =?

?η?η

ith

air

==

(a)Stroke volume ,Vst = Ld4π 2 ×

9.421L

/sm109.242

)10(300)10(2004π

33

323

=×=

×××=

−

−−

V1 = ,Vst +Vcl

= 9.42 + 2.78

= 12. 2 L

V2 = Vcl = 2.78 L

3.478.22.12

VV

r2

1 ===

4

% 44.2 100%0.442

(4.3)1-1

r11η

1-1.4

1γair

=×=

=

−= −

(b)indicated power ip =PmLANn

= 518×300×10-3× ( ) 160

135102004π 23 ××× −

= 11 kW

(c) 27.5%100%0.27516350

36008.8

11CVm

ipηf

ith =×=×

=×°

=

3.During a trial on a six cylinder petrol engine, a Morse test carried out as the means of estimating the indicated power of the engine .When running at full load , all cylinders in, the brake power output was 52 kW.The measured brake power out puts, in kW ,When each cylinder was cut out in turn and the load reduced to bring the engine back to its original speed were as follows:

1 2 3 4 5 6

40.5 40.2 40.1 40.6 40.7 40.0

For this data, estimate

(a)the indicated power of the engine ;

(b)the mechanical efficiency of the engine . (20 marks)

Solution

(a)ip cylinder 1 = 52 - 40.5 = 11.5

ip cylinder 2 = 52 - 40.2 = 11.8

ip cylinder 3 = 52 - 40.1 = 11.9

ip cylinder 4 = 52 - 40.7 = 11.3

ip cylinder 6 = 52 - 40.0 = 12.0

Total ip = 69.9 kW

5

(b)Mechanical efficiency =ipbp

= %4.74%100744.069.952

=×=

4.In a test on a single -cylinder oil engine operating on the four-stage cycle and fitted

with a simple rope brake ,the following reading were taken:

Brake wheel diameter 600 mm

Rope diameter 25.4 mm

Speed 50 rpm

Dead weight on rope 20 kg

Spring balance reading 3.25 kg

Area of indicator diagram 410 mm2

Length of indicator diagram 64 mm

Spring constant 100 kN/m2/mm

Bore 120 mm

Stroke 150 mm

Brake specific fuel consumption0.30 kg/kW-hr of oil CV = 41700 kJ/kg.

Calculate the bp, ip , mechanical efficiency and indicated thermal efficiency

of the engine. (20marks)

Solution,

Four stage cycle,

n =1,

Dw = 600mm = 600 310−× mm

d = 25.4mm = 25.4 310−× mm

N = 450rpm,

M = 20kg

spring balamce reading ,ms =3.25 kg,

Area of indicator diagram = 410mm2,

Length of indicator diagram = 64mm

spring constant =100kN/m2/mm,

bore D = 120 mm =120×10-3 m

stroke L = 150 mm =150 310−× m

b.s. f.c.= 0.30 kg/kW-hr,

6

CV = 41700kJ/kg

bp =?, ip=?, ?ηmech = , ?ηith =

bp=60NT2π

T = F× r

F = (M-ms)×9.81 = (20-3.25)×9.81=164.3175N

R = 0.3127312.7mm2

25.46002

dD==

+=

+ m

T = F× r

=164.3175×0.3127m=51.382Nm

∴ bp 2.42kW2421.3W60

51.3824502π==

××=

ip = PmLANn

2

m

kN/m 640.625

10064410

nCalibratio SprigIndicator ofLength

Indicator of AreaP

=

×=

×=

( ) ( ) 2232 0.0113m101204πD

4πA =×== −

ip = PmLANn

=640.625×150×10−3×0.0113× ⇐=××

4.07kW1260

450

59.4%100%0.5944.072.24

ipbpηmech =×==

kg/s102.016736002.420.30bpb.s.f.cfm 4−×=×=×=°

41700102.0164.07

CVfm.ipη 4ith ××

=×°

= −

=0.483×100%

= 48.3%

5. During a trail on a four- cylinder, compression ignition oil engine,a Morse test was

carried out in order to estimate the indicated power of the engine .At full load; with all

7

cylinders working, the engine developed a brake power of 45 kW.The measured brake

power outputs, when each cylinder was cut in turn and the load reduced to bring the

engine back to the original speed ,were, as follows:

1 2 3 4

31 32 31.8 31.2(kW)

Form this data, estimate:

(a) the indicated power of the engine ;

(b) the mechanical efficiency of the engine;

Solution

bp = b = 45kW

b1 = 31 kW

b2 = 32 kW

b3 = 31.8 kW

b4 = 31.2 kW

(a) ip =? (b) ?ηmech =

(a) i1 = b-b1 = 54-31 =14kW

i2 = b-b2 = 45-32 =13kW

i3 = b-b3 = 45-31.8 =13.2kW

i4 = b-b4 = 45-31.2 =13.8kW

total ip = 54 kW

(b) ⇐=×=== 83.3%100%0.835445

ipbpηmech

6.During a test on a four –stroke, single cylinder gas engine, the following

observations were made:

Calorific Value of gas, 18850 kJ/m3

Gas consumption 4.95 m3/h

Speed 5 rev/s

Effective brake diameter 0.9 m

Dead weight on brake 400 N

Spring balance reading 40N

Jacket cooling water 204kg/h

Temperature rise of jacket C30°

8

Cooling water

Indicated men effiective 455kN/m2

pressure

Cylinder diameter 165 mm

Piston stroke 305 mm

Determine

(a) the mechanical efficiency

( b) the indicated thermal efficiency

(c)the brake thermal efficiency

(d)Draw up an energy balance for the engine in kJ/s. (20marks)

Solution

Four stroke gas engine, single cylinder , n = 1

CV=18850 kJ/m3,

N = 5rev/s,

D = 0.9m,

M= 400N,

ms = 40N

mw=204kg/hr

CT °=Δ 30

Pm = imep = 455 kN/m2,

D =165mm =165×10−3m

L = 305mm =305×10−3m

(a) ?ηmech =

(b) ?ηith =

(c) ?ηbth =

(d) Draw up energy balance

ip =PmLANn, ( )24

DA π=

( )7.418kW

12510165

4π10305455 233

=

××××××= −−

( ) ( ) 162Nm2

0.940400rmMrFT s =×−=×−=×=

9

⇐==×××== 5.089kW508.94W1620.5π2NT 2bp π

(a) 68.72%7.4015.086

ipbpηmech ===

(b) ⇐=×=××

=×°

= − 28.6%1000.28618850101.375

7.419CVfm

ipη 3ith

(c) ⇐=××

=×°

= − 19.6%18850101.375

5.086CVfm

bpη 3bth

(d) Draw up energy balance,

Energy from fuel, 25.92kW188503101.375CVfmQ f =×−×=×°=°

Energy to bp =5.08 kW

Cpw = 4.18 kJ/kgK

Energy to cooling water, ( ) 7.1kW304.183600204ΔTC mQ pww =××=°= w

Energy lost to surrounding, [ ] [ ] ⇐=+−=+−°= 13.73kW7.15.0825.2QwbpfQQ

Energy balance kJ/sec %

Energy from fuel 25.9 100

Energy to brake power 5.08 19.6

Energy to coolant 7.1 27.4

Energy to exhaust_ 0 0

Energy to surroundings 13.73 53

7.In a trial on a single-cylinder, four stroke cycle oil engine :

Stroke cycle oil engine:

250 mm bore by 450 mm stroke; the following results were recorded;

Duration of trial, 30 min

Total revolution, 7962;

Average dead load on brake, 940 N

Average spring balance reading, 110 N

Brake radius,1m

Average indicated mean effiective pressure,565kN/m2

Total fuel used,2.9 kg of calorific value,44000 kJ/kg

Total jacket water,200 kg;

Inlet temperature, C17°

Outlet temperature, C67° ;specific heat capacity of water 4.18 kJ/kg K

10

Calculate:

(a)the indicated power;

(b)the brake power;

(c)the mechanical efficiency;

(d)the brake thermal efficiency;

(e)the percentage energy loss to the jacket.

Solution

Four stroke oil engine

d = 250 mm = 250 ×10-3m

L= 450 mm = 450×10-3m

N = 7962 rev/30min

M×9.81= 940N

ms×9.81 = 110 N

r = 1 m

n2.9kg/30mim f =°

imep = 565 kN/m2

CV = 44000 kJ/kg

mw = 200 kg

T1 = C17°

T2 = C67°

cpw = 4.18 kJ/kg

(a)ip = ?

(b)bp = ?

(c) ?ηmech =

(d) ?ηbth =

(e)% of energy loss

(a) 223 0.049m)10(2504πA =×= rev/sec 2.2116

260)(30

7962

N =×=

ip = PmLANn

= 12116.2049.010450565 3 ××××× −

= 27.55 kW

11

(b)bp =2π NT

T = F r×

F =(940 - 110) = 830 N

T = F r× = 830 ×1= 830 Nm

bp =2π NT = 2π 830423.4 ×× = 23.054 kW

(c) 83.6%100%0.836827.55

23.054ipbpηmech =×===

(d)

( )32.5%100%0.325

44006030

2.923.054

CVmbpη

fbth =×=

××

=×

=

Energy from fuel, kW 70.88440006030

2.9CVmQ ff =××

=×°=

Energy to water )T(TcmQ 12pwww −××°=

= )1767(18.46030

200−××

×

= 23.22kW

% loss to jacket = %75.32%1003275.088.7022.23

=×=

8.A six cylinder, four –stroke cycle, marine oil engine has cylinder diameter of 610

mm and a piston stroke of 1250 mm.When the engine speed is 2 rev/s it uses 340 kg

of fuel oil of calorific value 44200 kJ/kg in one hour. The cooling water amounts to

19200 kg/h, entering at C15° and leaving at C63° .The torque transmitted at the

engine coupling is 108 kN.mand the indicated mean effective pressure is 775kN/m2.

Determine:

(a) the indicated power

(b) the brake power

(c) the percentage of the energy supplied/kg of fuel lost to the cooling

water;

(d) the brake thermal efficiency;

12

(e) the brake mean effective pressure;

(f) the mechanical efficiency;

(g) the fuel used/kW-h, on a brake power basis.

Solution

Six-cylinder, four stroke Engine

n = 6

D =610 mm

L = 1250 mm

N = 2 rev/s

m = 340 kg

T = 108 kN.m

CV= 44200 kJ/kg

mw =19200 kg/h

T1 = C15°

T2 = C63°

imep = 775 kN/ m2

(a) ip =?

(b) bp = ?

(c)%

(d) ?ηbth =

(e)bmep =?

(f) ?=mechη

(g)b.s.f.c = ?

(a) ip = PmLANn

( )kW 1697.25

62210610

4π101250775 233

=

××××××= −−

(b) NTπ2bp =

= 10822 ×××π

= 1357.168 kW

(c)Energy from fuel, CVmQ ff ×°=°

= 44200094.0 ×

= 4154.8 kJ/kg

13

Energy to coolant , ( )cpwww ΔTcmQ ××°=

= ( )156318.43600

19200−××

=1070.8 kW

% loss to cooling water = 1008.41544.1069×

= 25.73 %

(d)CVm

bpηf

bth ×°=

=44200094.048.1356

×

= 0.326 ×100%

= 32.6%

(f) 79.9%100%0.7991697.251357.168

ipbpηmech =×===

(e)

2

mech

mech

kN/m 619.717750.799i.m.e.pηb.m.e.p

i.m.e.pb.m.e.pη

=

×=×=

=

(g) hr0.249kg/kW3600106.92971356.48

0.094bp

mb.s.f.c 5f −=××==

°= −

9. In a test on a two-stroke, heavy-oil engine, the following observations were made;

Oil consumption, 4.05 kg/h

Calorific value of oil, 43000 kJ/kg;

Net brake load, 579N;

Mean brake diameter, 1m;

Mean effective pressure, 275 kN/m2

Cylinder diameter, 0.02m

Stroke, 0.250m;

Speed, 6 rev/s;

Calculate;

(a) the mechanical efficiency;

14

(b) the indicated thermal efficiency;

(c) the brake thermal efficiency;

(d) the quantity of jacket water required per minute if 30% of the

energy supplied by the fuel is absorbed by this water. Permissible

rise in temperature is 25 C° .

Solution

Two-Stroke Engine,

kg/s101.12536004.05

kg/hr 4.05m

3

f

−×==

=°

CV = 43000 kJ/kg

F =M-ms = 579N

Db = 1 m

Pm = 275 kN/m2

Dc = 0.20 m

L = 0.250 m

N = 6 rev/s

?(d)m?(c)η?(b)η

?(a)η

w

bth

ith

mech

=°===

289.5Nm

0.5579rFT

=×=

×=

10.913kW10913.89W

289.56π2NT2bp

==

×××== π

12.9525kW

16(0.20)4π0.250275

LANnPip

2

m

=

×××××=

=

(a) 0.842%100%0.842312.9510.908

ipbpηmech =×===

15

(b) 26.7%100%0.26743000101.125

12.95CVm

ipη 3f

ith =×=××

=×°

= −

(c) 22.54%100%0.225443000101.125

10.908CVm

bpη 3f

bth =×=××

=×°

= −

(d)Energy to coolant = 30% of Energy from fuel

( )kW 14.51

43000101.1250.3

CVm0.33

f

=×××=

×°×=−

( )

n8.328kg/mi600.1388

0.1388kJ/sm 254.18m14.1

ΔTcmcoolant Energy to

w

w

pww

=×=

=°××°=

××°=

10.A four cylinder four stroke cycle ,petrol engine 75 mm bore by 90 mm stroke

operates on the constant volume cycle and has a compression ratio of 6 to 1 ,the

efficiency ratio being 55%.Calculate the indicated thermal efficiency .Take γ =1.4

,When running at 40 rev/s the engine developed a brake mean effective pressure of

725 kN/m2and uses 9.2 kg of fuel/hr of calorific value 44000 kJ/kg.Calculate the

brake thermal efficiency ,the mechanical efficiency and the specific fuel consumption

is kg/kW-hr.

Solution

Four cylinder four stroke cycle

Constant volume cycle

D = 75 mm = 75 ×10-3 m

L = 90 mm = 0.09 m

r = 6

efficiency ratio = 55%

?ηith =

1.4γ =

Speed N = 40 rev/s

16

bmep = Pm = 725 kN/m2

kg/s36009.2kg/hr 9.2m f ==°

CV = 44000 kJ/kg

?b..s.f.c?η?η

bth

mech

===

0.51166

11r11η 11.41γair =−=−= −−

28.14%100%0.2814η 0.5116η0.55

ηηratio efficiency

ith

ith

air

ith

=×=

=

=

kJ/kg 31.64

4400036009.20.2814ip

=

××=

( )

72.88%10031.641

23.06125

100%ipbpη

kJ/kg 23.06125

42400.090.075

4π725

LANnPbp

mech

2

m

=×=

×=

=

××××=

=

20.51%44000

36009.223.06125

CVmbpη

fbth

=×

=

×°=

17

hrkg/kW 0.2907731.649.2

ipmi.s.f.c f −==°

=

hrkg/kW 0.3989323.06125

9.2bp

mb.s.f.c f −==°

=

11.A single cylinder four stroke oil engine has a bore of 18 cm and a stroke of 36

cm.The clearance volume is 590 cm3.During a test the fuel consumption was 3

kg/hr,the engine speed is 308 rpm,the indicator card area 420 mm2:the indicator card

spring rating 108 kN/m2/mm.If the fuel has calorific value of 42650 kJ/kg .Calculate

the efficiency relative to that of the constant volume air standard cycle .Asume

γ =1.4.

Solution

Single cylinder 4 stroke

d = 18 cm

L = 36 cm

Vc = 590 cm3

3kg/hrm f =°

N = 380 rpm

indicator card Area = 420 mm2

indicator card Length =63 mm

spring rating =108 kN/m2/mm

CV = 42650kJ/kg

relative efficiency =?

1.4γ =

18

( ) 32

s

9160cm36184π

LAV

=××=

×=

16.53590

5909160V

VVrc

sc =+

=+

=

( )67.43%100%0.6743

16.5311

r11η 0.41γair =×=−=−= −

2kN/m 72010863

420i.m.e.p =×=

( ) 20.89kW1602

3800.360.184π720

nNLAi.m.e.pip2 =×

××××=

××××=

58.75%100%0.587842650

36003

20.89CVm

ipηf

ith

=×=×

=

×°=

87.17%100%0.67430.5878

ηηratio Efficiency

air

ith =×==

19

12.Determine the specific liquid enthalpy specific enthalpy of evaporation and

specific enthalpy of dry saturated stream at 0.5 MN/m2.

Looking up stream tables,the various values will appear as follows,

P(MN/m2) Sat.temp C° hf Specific

enthalpy hfg

Kj/kg

hg

0.50 151.8 640.1 2107.4 2747.5

Thus

Specific liquid enthalpy = 640.1 kJ/kg

Specific enthalpy of evaporation = 2107.4 kJ/kg

Specific enthalpy of dry saturated stream = 2747.5 kj/kg

Note that ,

hg = hf +hfg

= 640.1 + 2107.4 = 2747.5 kj/kg

Note also that saturation temperature = 151.8 C°

13.Determine the saturation temperature ,specific liquid enthalpy,specific enthalpy of

evaporation and specific enthalpy enthalpy of dry saturated stream at pressure of 2.04

MN/m2.

p = 2.04 MN/m2

ts =?

hf s=?

hfg =?

hg =?

P Sat.temp Specific enthalpy kJ/kg

MN/m2 C° hf hfg hg

20

2.1 214.9 920.0 1878.2 2798.2

2.0 212.4 908.6 1888.6 2797.2

Difference +0.1 +2.5 +11.4 - 10.4 +1.0

Require +0.04 5.21.0

04.0× 4.11

1.004.0

× 4.101.0

04.0−× 0.1

1.004.0

×

= +1.0 = +4.56 = - 4.16 = +0. 4

Adding 2.04 213.4 913.6 1884.44 2797.6

Hence,at 2.04 MN/m2

Saturation temperature = 213.4 C°

Specific liquid enthalpy = 913.16 kJ/kg

Specific enthalpy of evaporation =1884.44kJ/kg

Specific enthalpy of dry saturated steam = 2797.6 kJ/kg

13. Determine the specific enthalpy of stream at 2MN/m2 and with a temperature of

C°275

p = 2N/m2

t = 275 C°

h =?

At 2 MN/m2, from tables,tf = 212.4 C°

The steam must, therefore, be superheated since its temperature is above tf .

Drgree of superheated = 275- 212.4 C° = 62.6 C°

The specific enthalpy can be looked up in steam tables under the heading superheated

states.

Looing up tables,

Specific enthalpy of steam at 2MN/m2 with a temperature of 275 C° =2965 kJ/kg.

Alternatively,

h = hg +cp(t-tf)

=2797.2 +2.0934×62.6 = 2797+132

=2929.2 kJ/kg.

It will be observed that this gives only an opproximation.

21

14.Determine the specific enthalpy of stream at 2.5MN/m2 and with a temperature of

320 C° .

Looking up steam tables shows that at 2.5 MN/m2 the saturation temperature is

223.9 C° .The steam is therefore superheated.

Degree of superheat = 320 – 223.9 = 96.1 K

The specific enthalpy could be estimated using the equation,

( )

kJ/kg 3001.92012800.996.12.09342800.9

ttchh fpg

=+=×+=

−+=

However,a more accurate value can be interpolated from tables giving specific

enthalpy against temperature.Looking up this table will show that neither the pressure

of 2.5 MN/m2 nor the temperature of 320 C° are given.Interpolation for both pressure

and temperature is therefore required.

A note is made of values of specific enthalpy on either side of the pressure and

temperature as follows.

Pressure MN/m2 2 4

Temp C° Spec. enthalpy kJ/kg

325 3083 3031

300 3025 2962

Pressure +25

Require +20

+58 +69

4.46582520

+=× 2.55692520

+=×

Adding 320 3071.4 3017.2

This gives values of specific enthalpy at the temperature of 320 C° .An interpolation is

now required for the pressure.

Pressure MN/m2 Specific enthalpy,kJ/kg

4 3017.2

2 3071.4

Difference +2 -54.2

+0.5

6.132.5425.0

−=−×

22

Adding 2.5 3057.8

Thus, the specific enthalpy of steam at a pressure of 2.5 MN/m2 and with a

temperature of 320 C° =3057 .8 kJ/kg

Note that the estimated value of 3001.9 kJ/kg was not very occurate.

16.Determine the specific enthalpy of wet stream at a pressure of 70 kN/m2 and

having a dryness fraction of 0.85.

x = 0.85

kJ/kg 2283.3h0.7barkJ/kg 376.8h0.7bar

fg

f

=⇒=⇒

h = hf + xhfg

kJ/kg 2321.81975376.8

2283.30.85376.8

=+=

×+=

17. Determine the specific volume of water at saturation temperature for a pressure of

4.0 MN/m2.

p = 4.0 MN/m2

vf =?

Looking up steam tables, values will appear as follows,

Pressure Sat.Temp.Specific Volume m3/kg

MN/m2 C° vf

4.0 250.3 0.001252

This shows that at a pressure of 4.0 MN/m2,saturation temperature is 250.3 C° and

the specific volume of water is 0.001252 m3/kg

18.1.5 kg of stream originally at a pressure of 1MN/m2 and temperature 225 C° i9s

expanded until the pressure becomes 0.28 MN/m2.The dryness fraction of the stream

is then 0.9.Determine the change of internal energy which occurs.

Solution

m = 1.5 kg

23

p1 =1 MN/m2

T1 = 225 C°

p2 = 0.28 MN/m2

x = 0.9

Δu = ?

At 1 MN/m2 and 225 C°

h1 = 2886 kJ /kg

v1 = 0.22m3/s

u1 = h1 – p1v1

2666kJ/kg2202886

0.22101012886 3

6

=−=

××−=

At 0.28 MN/m2 and dryness fraction 0.9,

kJ/kg 2504.491953.09551.4

2170.10.9551.4

xhhh fgf22

=+=

×+=

+=

v2 = xvg2

= /kgm 0.58140.6460.9 3=×

kJ/kg 2341.69162.82504.49

0.58141010.282504.49

vphu

3

62222

=−=

××−=

−=

Hence

u2 –u1 = 2341.69-2666

= -324.31kJ/kg

This is a loss.

For 1.5 kg.

Loss of internal energy = 234.31×1.5

= 486.47 kJ/kg

24

19.A closed vessel of 0.6 m3 capacity contains dry saturated stream at 350kk,n/m2

.The vessel is cooled until the pressure is reduced to 200 kn/m2.

Calculate

(a)the mass of stream in the vessel ,

(b)the final dryness of the stream,

(c)the amount of heat transferred during the cooling process

Solution

(a)At 350 kN/m2 ,vg =0.5241 m3/kg

kg 1.14480.5241

0.6in vessel steam of mass ==∴

(c) At 200 kN/m2, vg = 0.885 m3/kg

Since the volume remains constant,

Specific volume after cooling

Specific volume before cooling

steam of dryness final0.5920.8850.5241x ===∴

(d) For a non –flow process ,Q =Δu+W,and for a constant volume change

W= 0

Q =Δu

Hence in this cse the heat transferred is equal to the change of internal energy.

Now , h = u +pv

u = h- pv

at 350 kN/m2,dry saturated,

kJ/kg 2548.6183.427320.52413502732u1 =−=×−=

At 200 kN/m2, dryness 0.592

kJ/kg 18081303.3504.7

2201.60.592504.7h2

=+=

×+=

g1703.2kJ/k104.81808

0.52412001808u2

=−=

×−=∴

25

Change in specific internal energy

= u2-u1

=1703.2 -2548.6

= -845.4 kJ/kg , a loss

But there are 1.1448 kg of steam in the vessel.

Hence the amount of heat transferred during the cooling process

= - 845.4 ×1.1448

= - 967.8 kJa loss

20.Stream at 4mn/m2 and dry fraction 0.95 received heat at constant pressure until its

temperature becomes 350 C° .Determine the heat received by the stream/kg.

solution

At 4 MN/m2 and 0.95 dry,

g2714.7kJ/k1627.31087

1712.90.951087.4h1

=+=×+=

At 4 MN/m2 and temperature 350 C°

h2 = 3095 kJ/kg

Note that the steam in this case is super heated since saturation temperature at 4

MN/m2

= 250.3 C°

heat received = h2-h1

=3095-2714.7

=380.3 kJ/kg

21.A quantity of dry saturated stream occupies 0.2634 m3 at 1.5 MN/m2.Determine

the final condition of the stream if it is compressed until the volume is halved;

(a) if the compression is carried out in an isothermal manner;

(b) if the compression follows the law PV = constant.

In case (a) determine the heat rejected during the compression.

Solution

26

Extract from Steam Tables

Press Sat.Temp Spec.enthalpy ,kJ/kg Spec.Vol,m3/kg

MN/m2 tf C° hf hfg hg vg

1.5 198.3 844.7 1945.2 2789.9 0.1317

3.0 233.8 1008.4 1793.9 2902.3 0.0666

(a)vg at 1.5 MN?m2 = 0.1317 m3/kg

2kg0.13170.2634present steam ofQuantity ==∴

In this case the steam is operating in the evaporation region since the temperature

remains constant.

volumespecific final/kg0.0659m2

0.13172v 3g ===∴

Hence, final condition is at 1.5 MN/m2 with a dryness fraction of

For 2 kg

Enthalpy = kJ 3634.61817.32 =×

The loss of heat during this process will be the loss of enthalpy of evaporation,

changing from dry saturated steam to wet steam of dryness fraction 0.5 at constant

pressure.

Heat loss = 972.6kJ/kg1945.20.50.5fg =×=

For 2 kg

kJ 1945.22972.6lossHeat =×=

(b)If the compression is according to the law pv = constant then

p1v1 = p2v2

2

2

112 MN/m 3.021.5

vvpp =×==∴

Specific volume after compression = 0.0659 m3/kg.

At 3.0 MN/m2

vg = 0.0666 m3/kg

dryness fraction after compression

kJ/kg 1817.31945.20.5844.7enthalpy specific

0.50.13170.0659

=×+=

=

27

989.00666.00659.0

=

Specific enthalpy = 1008.4+0.989 9.1793×

=1008.4 +1774.2

= 2782.6 kJ/kg

For 2 kg,Enthalpy = 2 6.2782×

=5565.2 kJ

22A quantity of stream at a pressure of 2.1 MN/m2 and 0.9 dry occupies a

volume of 0.2562 m3.It is expaneded according to the law PV1.25 = constant to

a pressure of 0.7 MN?m2.Determine

(a)the mass of stream present

(b)the external work ,done

(c)the change of internal energy ,

(d)the heat exchange between the stream and surrounds,stating the direction of

transfer.

Extract from steam tables

Press Sat.tem Spec.enthalpy kJ/kg Spec.vol.

MN/m2 tf C° hf hfg hg m3/s

0.7 165 697.1 2064.9 2762.2 0.273

2.1 214.9 920.0 1878.2 2798.2 0.0949

(a) specific volume of steam at 2.1 MN/m2 and 0.9 dry.

/kg0.0854m0.09490.9vxv 3g111 =×==

3kg0.08540.2562present steam of mass ==∴

(b)For the expension, 1.2522

1.2511 vpvp =

28

1/1.251/1.25

2

112 0.7

2.10.0854ppvv ⎟

⎠⎞

⎜⎝⎛×=⎟⎟

⎠

⎞⎜⎜⎝

⎛×=∴

( )/kg0.2058m

2.410.085430.08543

1/1.25

=

×=×=

Steam is wet after expansion

0.7540.273

0.2058vvv

g2

2 ===

( )

140.8kJ/kg0.1408100.251

0.035210

11.250.20580.70.08542.110

1nvpvpdone work External

6

6

6

2211

=×=

×=

−×−××

=

−−

=

External work done for 3 kg = 140.8 422.4kJ3 =×

(b) u1 = h1-p1v1 = 920.0 + 0.9 2.1878×

= 920.0 + 1690.4

= 2610.4 kJ/kg

g2431.1kJ/k179.32610.4

0.08542.110102610.4u 3

6

1

=−=

××−=∴

u2 = h2-p2v2

2064.90.754697.1hxhh f22f22 ×+=−=

= 679.1+1556.9

= 2254.0 kJ/kg

0.20580.710102254.0u 3

6

2 ××−=∴

=2254.0-144.1= 2109.9 kJ/kg

loss a ,321.2kJ/kg2431.12109.9uuenergy internalin Change 12 −=−=−=∴

For 3 kg of steam,

loss of internal energy = 963.6kJ321.23 =×

(d) WUQ +Δ=

29

= -963.6 +422.4 =-541.2 kJ a loss to the surroundings.

23.Determime the volume occupied by 1kg of at a pressure of 0.85MN/m2 and

having a dryness fraction of 0.95.

(b)This is expanded adiabatically to a pressure of 0.17 MN/m2;the law of expansion

being PV1.13 = constant .Determine ,

(i) the final dryness fraction of the stream,

(ii)the change of internal energy of the stream during the expansion.

Extract from steam Table

Press Spec.vol.

MN/m2 m3/kg

0.17 1.031

0.85 0.2263

(a) v1 = x1vgf = 0.95×0.2263 = 0.2150 m3/kg

(b) (i) p1v11.13 = p2v2

1.13

1/1.13

2

112 p

pvv ⎥⎦

⎤⎢⎣

⎡=

= 13.1/1

17.085.02150.0 ⎥⎦

⎤⎢⎣⎡×

= ( ) 13.1/152150.0 ×

= 0.2150 /kg0.8923m4.15 3=×

At 0.17 MN/m2,vg2 = 1.031 m3/kg

0.8651.0310.8923

vvx

g2

22 ===

(ii) For an adiabatic expansion

u = - W,

( )1n

vpvpuu 221112 −

−−=−

30

( )

( )

energy. internal of loss akJ/kg, 239.20.239210

0.130.031110

10.130.15170.182810

11.130.89230.170.21500.85

1010

3

33

3

6

−=×−=

×−=−

−×−=

−×−×

×−=

24.A vessel of volume 0.03m3 contains dry saturated steam at 17 bar .Calculate the

mass of steam in the vessel and the enthalpy of this steam.

Solution

V1=0.03 m3 (dry saturated steam)

P1 =17 bar

m =?

h = ?

V1= vg = 0.1167

V1 = m×v1

0.03 =m×0.1167

m = 0.257 kg

h1=hg = 2793kJ/kg

H1 =mh1

=0.257×2795

=718.5kJ

25. Steam at 7 bar and C250° enters a pipeline and flows along it at constant

pressure. If the steam rejects heat steady to the surrounding, at what temperature will

droplets of water begin to form in the vapor?

Solution

P1 = 7 bar t2 = ?

t1=250 C° p2 = 7 bar

Q12 =?

t2 = t3 = 165 C°

h1 = hsuper (p1=7 bar, ts = 165<t1 = 250 C° (steam is superheat)

31

=2955 kJ/kg

h2 = hg2 (p2 = 7 bar)

= 2764 kJ/kg

Q12 = (h2 –h1) =2764-2955

=-191 kJ/kg

26. Determine the enthalpy, volume and density of 1 kg of steam at a pressure of 20

bars and a temperature of C.300°

Solution

h =?

v =?

ρ =?

m =1kg

p = 20 bar

t = 300 C°

p = 20 bar C,300212.4ts °<=⇒ steam is superheated.

h = 3025 kJ/kg

H=h×m

= 3025×1 =3025kJ

v = 0.1255 m3/kg

V = v×m

=0.1255×1=0.1255m3

3kg/m 7.970.1255

1v1ρ ===

27Steam at a pressure of 28 Kpa is passed into a conderser and it leaves as condensate

at a temperature of C59° .Cooling water circulates through the conderser at the rate of

45 kg/min.It enters at C15° and leaves at C30° .If the steam flow rate is 1.25

kg/min,determine the quality of the steam as it enters the conderser.

Solution

p1 =0.28 bar

32

t2 =59 C° , ?x kg/min, 1.25m 1f ==°

C30TC,15Tkg/min 45m

outin

cw

°=°==°

Steam leave at condenser

h2 = hf2 (p =0.28 bar ,t2 =59 C° )

=247.36 kJ/kg

Heat lost of steam =heat gain of water

( ) ( )( ) ( )

kg2508.34kJ/h 15304.18745247.36h1.25

TTcmhhm

1

1

inoutpwcw21s

=−×=−

−°=−°

( )

0.95 x2339x2832508.34

hxhh

1

1

fg1f11

=×+=

+=

28.Determine the enthalpy volume and density of 1 kg of steam at a pressure of 50 bar

and with a quality of 0.9.

Solution

h = ?

v = ?

ρ =?

m = 1kg

p = 50 bar

x = 0.9

( )

kJ/kg 30.12616390.91155

xhhh fgf

=×+=

+=

For m = 1 kg

H = 2630.1×1 =2630.1kg

/kgm 0.03540.039440.9

xvv3

g

=×=

=

33

3

3

kg/m 28.250.0354

1v1ρ

m 0.03540.03541vmV

===

=×=×=

29. 0.05 kg of steam at 10 bar is contained in a rigid vessel of volume 0.0076m3

.What is the temperature of the steam?If the vessel is cooled , at what temperature will

the steam be just dry saturated ?cooling is contained until the pressure in the vessel is

11 bar ,calculate the final quality of the steam ,and the heat rejected between the

initial and the final states.

Solution

m =0.05 kg

p1=15bar

V1 =0.0076 m3

t = ?

/kgm 0.1520.05

0.0076mVv

mvV

311

11

===

=

( )C250t/kg.152mv

dsuperheate is steamv0.131715barv

13

1

1g

°=⇒=∴

∴<=

P2 =15 bar ,(steam is dry saturated)

t2 = ?

t2 =ts = 198.3 C°

p3 = 11 bar ,x3 =? Q1-3 =?

v3 = v2 = v1 =0.152 m3/kg

0.8560.17740.152

vvx

g3

33

==

=

( ) ( )( )

kJ/kg 2697

C250t15bar,puucvmuuQ

11superheat1

1321

=

°====−=− Q

34

( )[ ]( )

kJ/kg 1547.4278025860.856780

11barpat uuxuu 3fg3f3

=−+=

=−+=

( )( )

kJ 57.480.0526971547.42

muuQ 1331

−=×−=

−=∴ −

30.A quality of steam 0.7 dry,occupied a volume of 0.197 m3 at a pressure of 109

MN/m2 .If the pressure is kept constant,determine the heat to be supplied to make the

steam just dry and the percentage of this heat which appears as external work.If

now,the volume is kept constant while heat is extracted until the pressure fails to

1.25MN/m2 ,find the dryness fraction and the heat transferred from the steam.

Solution

Wet steam

x1=0.7 p=c Dry saturated steam

V1 = 0.197 m3 p2 = 14 bar

p1 = 1.4 MN/m2 Q1-2 =?

=14 bar % heat = ?

( )

2790kJ/kghh2202kJ/kg

19600.7830

xhhh

g2

14barfgf1

===

×+=

+=

/kg0.09856m0.14080.7

xvvvvx

3

g1

g1

11

=

×=

=

=

kg 1.99880.09850.197

vVm

1

1 ===

( )( )

kJ/kg 1175.41.998822022790

mhhQ 1221

=−=

−=−

35

( )( )

( )kJ 118.21

1.99880.098560.1408101.4 mvvp

mvpvpw

312

112221

=×−×=

−=−=−

10.05%1175.4118.21

Qwheat of %

21

21 ===−

−

v2 = 0.1408 m3/kg v =c v3 =0.1408 m3/kg

p2 =14 bar x3 =?

Q2 -3 =? p3 = 1.25 MN/m2 =12.5 bar

0.890.15720.1408

vvx

vxv

g3

33

g333

===

=

u2 = ug2 =2593 kJ/kg

( )( )

kJ/kg 2398.8580525890.89805

uuxuu fg3f3

=−+=

−+=

0wcv =→=

( )( )

kJ 400.241.998825932392.76

muuQ 2332

−=×−=

−=−