# Performance of I.C. Engines

10.7.1 Indicated Power (IP)

The total power which is developed by combustion of fuel in the combustion chamber is called indicated power. It is given by

IP = Work done by the cycle x no of cycles per second.

IP = n Pm LANK / 60 (KW)

n =Number of cylinders

PM = Indicated mean effective pressure (KPa)

= KN/ M2

L = Length of stroke (metre)

A = Cross sectional area of piston (metre2)

K = 1/2. for 4 – stroke engine.

K = 1 for 2- stroke engine.

N =Number of revolution of crank shaft/minute (rpm) .

[Note that in previous system (MKS), -it was called IHP {indicated horse power) 1HP =

746 Watt.

IHP = n Pm LANK / 4500 , where Pm is in bars.

10.7.2 Brake Power (BP)

The power developed by an engine at the output shaft is called brake power (BP). It is given by

BP = 2π NT / 60 x 1000 KW

Where                                     N = Speed in rpm

T = Torque developed at shaft in Nm

10.7.3 Frictional Power (FP)

If BP is subtracted from IP, it is frictional power (FP).

FP = IP-BP.

10.7.4 Mechanical Efficiency

The ratio of B.P to I.P is the mechanical efficiency.

% ɳ mech = B.P / I.P x 100

10.7.5 Volumetric Efficiency

The ratio of actual volume (reduced to NTP) of the charge, drawn in during suction stroke to swept volume of the piston is called volumetric efficiency.

10.7.6 · Thermal Efficiency

The ratio of. IP to energy supplied per second by the fuel. It is of two types.

10.7.6.1 Indicated Thermal Efficiency

Indicated ɳ thermal = IP / m f x C = work done per cycle / heat added per cycle

Where                                     m f = mass of fuel supplied in Kg per second

C = lower calorific value of the fuel

10.7.6.2 Brake Thermal Efficiency

Brake ɳ thermal = BP / m f x C

10.7.7 Specific Output

It is the brake power per unit piston displaced volume.

Where                         A = π / 4 d2 in m2

and                  d = bore diameter in m

L = stroke length in m.

SOLVED EXAMPLES

Examples Based on Otto Cycle

Example 10.1. An engine working on ideal otto cycle has bore and stroke of the cylinder 17 cm and 30 cm respectively. The clearance volume is given 0.002025 m3  Calculate the air standard efficiency of the engine. Assume r= 1.4. p

Solution. Given

d = diameter of cylinder

=bore= 17 cm= 0.17m

l = stroke length= 30 cm = 0.30 m

Clearance volume = Vc = v2 = 0.002025 m3

Clearance volume = Vc = v2 = 0.002025 m3

γ – 1.4 , ɳ = ?

us = swept volume = π / 4 d2 l

= π / 4 (0.17)2 x 0.3 m3

= 6.81 x 10-3 m3

u1 = u c + us = 0.002025 = 4.363

ɳ cycle = 1 – 1 / r γ-1 = 1 – 1 / (4.363)1.4 -1 = 44.526 %. Ans

Example 10.2. Calculate the efficiency of a four stroke cycle gas engine, assumed to be working on the constant volume cycle, of stroke 45 cm, piston diameter 30 cm, and a clearance of 11.4 litres (Take r= 1.4)

Solution. Given        d = 0.3 m

l = 0.45 m

V2 = Vc = 11.4 x 10-3 m3

Vs = π / 4 (0.3)2 x 0.45

= 0.0318 m3

Total volume = V1 = VS + VC = 0.0318 + 11.4 x  10-3 = 0.0432 m3

r = V1 / V2 = 0.0432 / 11.4 x 10-3 = 3.789

ɳ cycle = 1 – 1 / r γ-1 = 1 – 1 / (3.789)1.4 -1 = 0.413 %. Ans

Example 10.3. A four-cylinder petrol engine has a total swept volume of 3000 cm3 and clearance volume of each cylinder is 70 cm3. If the pressure and the temperature at the beginning of compression are 1.03 bar and 25°C and the maximum cycle temperature is 1500°C, calculate (i) the air standard efficiency (ii) all pressure and temperature of the cycles. (iii) the mean effective pressure. [Take r= 1.4, Cv = 0.718 KJ/kg °K]

R = 0.287

Solution. (i)

Swept volume per cylinder = 3000 / 4 = 750 cm3.

compression ratio = r = us + u c / u c = 750 + 70 / 70 = 11.714 = u1 / u2 = u4 / u3

ɳ air standard  = 1 – 1 / γ-1 = 1 – 1 / (11.714)0.4 = 62.63 % Ans.

(ii)                                           P1 = 1.03 bar

T1 = 273 + 25 = 2980 K

T3 = 273 + 1500 = 17730 K

P2 = P1 (u1 / u2)γ = 1.03 (11.714)1.4 = 32.286 bar

P1 u1 / T1 = P2 u2 / T2

T2 = P2 / P1 . u2 / u1 T1 = 32.286 / 1.03 x 1 / 11.714 x 298 = 797.420 K

P2 u2 / T2 = P3u3 / T3 or  P2 / T2 = P3 / T3

P3 = T3 / T2 .P2 = 1773 / 797.42 x 32.286 = 71.785 bar

P4 = p3 (u3 /u4) γ = 71.785 ( 1/ u4) γ = 71.785 / (11.714)1.4

= 2.29 bar

P3 u3 / T3 = p4 u4 /T4  or T4 = T3 (P4/P3) = 2.29 / 71.785 x 11.714 x 1773

= 662.55 k

Heat supplied                  = q s = C v (T3-T 2) = 0.718 (1773 – 797.42)

= 700.47KJ/ kg cycle

Heat rejected            q R = C v (T4– T l) = 0.718 (662.55 – 298)

Q R = 261.75 KJ /kg cycle

Work done                     = q s– q R = 700.47 – 261.75

= 438.72KJ/kg.

(iii)                   [ check ɳ cycle = q s – q R  / q s = 438.72 / 700.47 = 62.63%

P1V1 = mRT1

102 x 1.03 (750 + 70)10-6 = m x 0.287×298

m = 9.87 x 10-4 kg = mass in each cylinder

W net= 438.72 x 9.87 x 10– 4

= 0.433 KJ

Pm = W net / V s = 0.433 kJ/ 750 x 10-6 m3

= 577.33 KN / m3

= 577.33 / 102 bar = 5.7733 bar.

Example 10.4. Show that for maximum work, the intermediate temperature in otto cycle is given by T2 = T4 = T1 T3 T1 and T3 are initial maximum temperatures in compression and expansion process respectively.

Solution.

Work done during the cycle  =w = qs – qR

for process 1-2  = cu ( T3 – T2) – Cu (T4 – T1)

T2 / T1 = ( u1 / u2γ-1 = rγ-1

Again for process (3- 4)

T3 / T4 = (u4 / u3) γ-1 = rγ-1

The following expression of w is a function of r when T3 & T1 are constant (fixed). The value of ‘γ  is constant

w = cu ( T3 – T1 r γ-1) – cu ( T3 / r γ-1 – T1)

= cu T3 – cu T1 r γ-1 – cu T3 r1-r + cv T1

T3 / T1 = ( 1-γ) r γ-2 / (1-γ) r – γ = r γ-2+γ = r2 (γ-1)

√ T3 / T1 = r γ-1 = T3 / T4 => (from (ii) we get)

T3 / T1 = T3 / T4

T4 = T1 T3

Example 10.5. An engine working on the Otto cycle is supplied with air at 0.1 MPa, 35°C compression ratio is B. Heat supplied is 2100 KJ/kg. Calculate the maximum pressure temperature of the cycle, the cycle efficiency and the mean effective pressure, (for air cp = 1 Cv = 0.718, and R = 0.287 KJ/kg 0k).

Solution.

P1 = 0.1 MPa = 100 I<N/m2

qs = 2100 KJ /kg= Cv (T3– T2)

T1 = 273 + 35 = 308°K

r = 8 = u1 / u2

p3 = p max = ? , T3 = T max = ?

ɳ = ? , pm ?

ɳ cycle = 1 – 1 / r γ-1 = 1 – 1 / 8 1.4 -1 = 56.47 %

γ = cp / cu = 1.005 / 0.718 = 1.4

p1 u1 / T1 = P2 u2 / T2 or  P1 / T1 = P2 / T2 x ( u2 / u1) = p2 / 8 T2

100 / 308 = P2 / 8 T2

P2 / T2 = 2.6

2100 = 0.718 (T3 – T2)

T3 – T2 = 2924.8 K

P2 = P1 (u1 / u2) γ = 100 (8) 1.4 = 1837.92 KN / m2

T2 = p2 / 2.6 = 1837.92 / 2.6 = 706.9 k

T max = T3 = 2924.8 + Y2 = 2924.8 + 706.9

= 3631.7 K

P3 u3 /T3 = P2 u2 /T2

P3 = ( P2 / T2) T3 = 2.6 x 3631.7

= 9442.42 KN / m2

= 9.44242 mpa

w net = qs x ɳ = 2100 x 0.5647

= 1185.87 kj /kg

Pm = w net / us = w net / u1 –u2

u1/ u2 = 8 and p1 u1 = RT1

u1 = RT1 / P1 = 0.287 x 308 / 100

= 0.884 m2 /kg

u2 = u1 / 8 = 0.884 / 8 = 0.11 m3 / kg

pm = w net / us = w net / u1 – u2 = 1185.87 / 0.884 – 0.11 = 1532.13 kpa

Example 10.6. In an otto ‘cycle air at 1 bar, 1JOC is compressed isentropically until the pressure is 15 bar. The heat is added at constant volume until the pressure rises to 40 bar. Calculate the air standard cycle efficiency and the mean effective pressure for the cycle. [Take cv = 0.717 K/1 kg°K, R = 0.287 K]/kg°K, r= 1.4)

Solution. Given

P1 = 1 bar= 105 N/m2

T1 = 273 + 17 = 290°K

P2 = 15 bar= 15 x 105 N/m2

P3 = 40 bar= 40 x 105 N/m2

= 6.919 = r

ɳ cycle = 1 – 1 / r γ-1 = 1 1 / ( 6.91901.4 -1 = 53.87 %

q s = cu (T3 –T2), P3 u3 / T3 = P2 u2 /T2

T3 = 2.67 x T2 = 2.67 X 628.7 = 1678.63 °K

qs = cu, (T3– T2) = 0.717 (1678.63- 628.7) = 752.8 K] /kg

W net = ɳ  x qs

= 0.5387 x  752.8 K] I kg = 405.53 K] I kg

Pm = w net / us

U1 / u2 = 6.919 , p1 u1 / T1 = R where m = 1 kg

or                                        u1 = RT1 / P1 = 0.287 x 290 / 100 = 0.120 m3 / kg

p m = 495,53 / 0.8323 – 0.12 =  569.3 KN /m2

= 5.693 bar

Example 10.7. An engine working on the otto cycle has a clearance of 17% of stroke volume and initial pressure of 0.95 bar and temperature 30°C. If the pressure at the end of constant volume heating is 28 bar. Find (i) Air standard efficiency (ii) the maximum temperature in the cycle (iii) Ideal mean effective pressure. Assume working fluid to be air.

Solution. y= 1.4,cp = 1.005 KJ/kg°K, cv = 0.714 KJ /kg°K

γ = 1.4 , c p = 1.005 kj / kg k , cu = 714 kj / kg k

u c = 0.17 v s

r = u1 / u2 = u c + us / u c = 1 + us / v c

r = 1 + 1 / 0.17 = 6.88

ɳ cycle = 1 – 1 / r γ-1 = 1 – 1 / (6.88)0.4

= 53.76%

P1 = 0.85 bar = 95 KN /m2

T1 = 273 + 30 = 303 k

P3 = 28 bar = 2800 KN /m2

T1 / T2 = (u2 / u1) γ-1 = ( 1/ 6.88) 0.4

T2 = t1 (6.88)0.4  = 303(6.88)0.4 = 655.35 k

P2 = p1 (u1 / u2)γ = 95 (6.88)1.4 = 1413.66 KN / m2

P2 u2 / T2 = P3 u3 / T3

T3 = P3 / T2.T2 = 2800 / 1413.66 x  (655.35) = 1298 K

Maximum temperature= 1298°K,v1 = RT1 / 1 = 0.287 x 303 / 95 m3 = 0.9153 m3/  kg

u2 = u1 / 6.88 = 0.9153 / 6.88 = 0.133 m3 / kg

Example 10.8. An engine working on ideal otto cycle has temperature and pressure at the beginning of compression as 25°C and 1.5 bar respectively. Find the compression ratio if r= 1.4 and thermal efficiency= 48%. Also find the temperature and pressure at the end of compression.

Solution. Given T1 = 273 + 25 = 298 °K, P1 = 1.5 bar, y = 1.4, i1 = 0.48

ɳ = 1 – 1 / γ-1

0.48 = 1 – 1/ γ1.4 -1 or 5.13 = u1 / u2

We have

T1 / T2 = (u2 / u1)γ-1 , or  T2 = T1 r γ-1

= 298 (5.13)0.4

= 573 k = 300 c

P1 u1 / T1 = p2 u2 / T2 p1 T2 / T1(u1 / u2) = p2

1.5 x 573 / 298 x 5.13 = p2

14.79 bar = p2

Example. 10.9. An engine working on the otto cycle has a compression ratio of 8.5:1. The temperature and pressure at the beginning of compression are 93°C and 0.93 bar respectively. The maximum pressure in the cycle is 38 bar. Determine the pressure and temperature at the salient points of the cycle and the air standard efficiency.

Solution. Given T1 = 273 + 93 = 366°K, P1 = 0.93 bar

r = 8.5 = u1 /u2 , p3 = 38

= 18.606 bar

= 8611.5 k

At the beginning of expansion

Process 2 – 3               p3 u3 / T3 = P2 u2 /T2

T3 = T2 .P3 / P2 = 861 x 38 / 18.606 = 1759.5 k

Process 3 – 4

P4 = p3 (u3 / u4) γ = p3 ( u2 / u1) γ = 38 / 8.5 1.4

P4 = 1.89 bar

Process 4-1                 p4 u4 / T4 =p1 u1 / T1

T4 = T1 ( P4 / P1) = 366 (1.89 / 0.93) = 743.81 K

ɳ air stand = 1 – 1 / r γ-1 = 1 1 / ( 8.5) 1.4 -1 = 57.51 %

Example 10.10. The compression ratio of an otto cycle is 8. At the beginning of compression the pressure and temperature are 1 bar and 300°K respectively. The heat transfer to the air per cycle is 1900 KJ/kg of air. Calculate (a) the pressure and temperature at the end of each process of the cycle (b) thermal efficiency and (c) the mean effective pressure. Take c p = 1.005 KJ/kg°K1 C v = 0.718 KJ/kg o K, R = 0.287 KJ/kg0K.

Solution . (a)              p1 = 1 bar , T1 = 300 K , γ  = c p / c u = 1.005 / 0.718 = 1.4

P1u1 = RT1  or u1 = RT1 / P1 = (0.287 x 103) / 1x 105 = 0.861 m3

Process 1 – 2               T2 / T1 = r γ-1 => T2 = T1 r γ-1  = 300 (8)0.4 = 689.22 k

u1 / u2 = r or u2 = u1 / r = 0.861 / 8 = 0.1076 m3 / kg

Heat added per kg during constant volume. Process 2-3 is given by (q5) = “v (T3 – T2)

1900 = 0.718 (T3 – 689.22)

T3 = 3335.46°K

Process 2 – 3               V3 = V2 P3 = p2 x T3 / T2 = 18.38 x 3335.46 / 689.22 = 88.95 bar

Process 3-4                 T4 = T3 / r γ-1 = 3335.46 / 8 0.4  = 1451.84 k

Process 3-4                 p4 = p3 / r γ = 88.95 / 8 1.4 = 4.839 bar

(b)                               ɳ = 1 – 1 / r γ-1 = 1 – 1 / 8 0.4 = 56.47 %

Heat rejected per kg of air during constant volume process (4 – 1) is given by

q R= C11 (T .. -T1)

= 0.718 (1451.84 – 300)

= 827.02 KJ !kg.

Net work done            = q, – q R = (1900- 827.02)

= 1072.98 KJ/kg

(c) Mean effective pressure

Pm = met work done / stroke volume = 1072.98 / u1 – u2

= 1072.98 / 0.861 – 0.1076 kn / m2

= 1424.18 KN/m2

= 14.2418 bar. Ans.

Example 10.11. An engine working on otto cycle has a volume of 0.5 m3, pressure 1 bar and temperature 27°C at the beginning of compression stroke. The pressure at the end of compression is 10 bar. The heat added during constant volume process is 200 Kf. Calculate (a) percentage clearance (b) efficiency and (c) mean effective pressure.

Solution.                   P1 = 1 bar, p2 = 10 bar, v1 = 0.5 m3, T1 = 273 + 27 = 300°K

or                      (u1 /u2) γ = p2 / p1

or                                 r γ = p2 / p1  or  r = ( p2 / p1) 1 /γ = (10 / 1) 1/ 1.4

r = 5.18

T2 = T1 r γ-1 = 300 (5.18)0.4 = 579.23 k

P1u1 / T1 = P2 u2 / T2  or  V2 = P1 V1 / T1 x  T2 / P2

V2 = 1 x 0.5 x 579.23 / 300 x 100 = 0.0965 m3

Heat supplied              = Qs = m cu (T3 –T2)

Where                         m = P1 V1 / RT1 = 1 x 105 x 0.5 / 0.287 x 103 x 300

= 0.58 kg

(1)                               Qs = 200 kj

200 = 0.58 x 0.718 x ( T3 – 579.23)

T4 / T3) = (u3 / u4) r-1 = ( u2/ u4) r-1 = ( 1/ r) r-1

T4 = T3 ( 1/ r) r-1 = 1059.5 ( 1/ 5.18) 1.4-1

= 548.74 k

(a)  percentage clearance = V2 / V1 – V2 x 100

= 0.0965  x 100 / 0.5 – 0965 = 23.915%

ɳ =1 – 1 / r γ-1 = 1 – 1/ (5.18)0.4 = 48.21

Heat supplied              QR = m cu (T4 – T1)

= 0.58 x 0.718 ( 548.74 – 300)

= 103.58 kj

Net work done                        = Qs – QR = 200 – 103.58 = 96.42 KJ

(c) mean effective pressure    = w net / v s

= 96.42 x 103 / (V1 – V2) Nm / m3

= 96.42 x 103 / 0.5 – 0.0965 N.m2

= 2389591.1 N.m2

= 2.389591 bar

Example 10.12. The upper and lower temperature limits for an otto cycle are 1400°K and 300°K. Find the maximum theoretical power developed by the engine working on this cycle when the rate of flow of air is 0.33 kg/min.

Solution. The condition for maximum work is given by

T2 = t4 = t1 t3

= 300 x1400 = 648 k

Work done per kg                       = cu [ T3 – T2]- CU [ T4 – T1]

= 0.718 ( 1400 – 648) – 0.718 (648 – 300)

w net =290.072 kj / kg

Maximum power developed

= 290.072 kj / kg x 0.33 kg / 60 sec

= 1.594 kw.

Example 10.13. The bore and stroke of an engine working on the otto cycle are 17 em and 30 cm respectively. The clearance volume is 0.001025 m3  Calculate the air standard efficiency.

Solution.                V1 – V2 = swept volume = V s = π/4 d2 l = π/ 4 x 172 x 30 = 6809 cc

V2 = clearance volume= V, = 0.001025 x 106 cc = 1025 cc

compression ratio             = r V1 / V2 – VC + VS / VC = 1025 + 6809 / 1025

= 7.64.

Air standard efficiency          ɳ = 1- 1 / r γ-1 = 1 – 1 / 7.64 1.4 -1

ɳ = 55.7%

Example 10.14.Jn an ideal constant volume cycle the pressure and temperature at the beginning of compression are 97 KN/n/ and 40°C, respectively. The volume ratio of compression is 7:1. The heat supplied during the cycle is 1200 KJ/kg of working fluid. Determine (a) the maximum temperature attained in the cycle (b) the thermal efficiency of the cycle and (c) the work done during the cycle/kg of working fluid.

Assume                       r = 1.4, cu = 0.718 KJ/kg°K

Solution.                (a) T2 = T1 r γ-1 = (273 + 40)71.4 – 1 = 681.7 °K

Q s = Cu (T 3-T 2)

1200 = 0.718(T3-681.7),  T3=2353K = T max· Ans.

ɳ = 1 – 1 / r γ-1 = 1 – 1 /71.4 -1 = 54.1 %

(c) Work done/ kg I cycle =heat supplied/kg/ cycle x ɳ

1200 x 0.541 = 649 KJ.

Alternatively:

Work done                  = q s – q R

= cu (T3 – T2) – cu (T4 – T1)

Now                     T3 = T4ry

or                               = T3 / r γ-1 = 2353 / 7 1.4-1 = 1080 k

work done                = 1200- 0.718 (1080 – 313)

= 649 KJ. Ans.

Example 10.15. In an ideal otto cycle the compression ratio is 8. The initial pressure and

temperature of the air are 1 bar and 100°C. The maximum pressure in the cycle is 50 bar. For 1 kg of air flow, calculate the values of the pressure, volume, and temperature at the four salient points of the cycle. What is the ratio of heat  supplied to the heat rejected?

for air R = 0.287 KJ/kg°K, r= 1.4.

Solution.                          P1 = 1 bar

P1 = 1 bar

T1 = 273 + 100 = 373°K

P1 V1 = mRT1

V1 = 1 x 0.287 x 373 / 1 x 100 = 1.0705 m3 = u1 as m = 1 kg

Process 1-2

or                                             p2 = p1 ( u1 / u2) γ = 1 (8) 1.4 = 18.38 bar

u2 = u1 / 8 = 1.0705 / 8 = 0.1338 m3

p1 u1 /T1 = P2u2 / T2

T2 = P2 / P1. u2 / u1.T1 = 18.38 x 0.1338 x 373 / 1 x 1.0705 = 856.8

Process 2-3                 u3 = u2 = 0.1338 m3 , p3 = 50 bar

P3 u3 / T3  = p2 u2 / T2 ,or T3 = P3 / P2 T2

T2 = P2 /P2 . T3 = 18.38 / 50 x T3

T3 = P3 / P2 T2 = 50 / 18.38 x 856.8  = 2336.4 k

Process 3 – 4

P4 = p3 ( u3 / u4) γ = 50 ( 1/8) 1.4 = 2.72 bar

P4 u4 /T4 = P1 u1 / T1

T4 = p4 / p1 . t1 = 2.72 / 1 x 373 = 1014.6 k

cu = R / γ-1 = 0.287 / 14 -1 = 0.718 KJ / KG K

q s = cu (T3 – T2) = 0.718 (2336.4 – 856.8)

= 1062.4 KJ / Kg

Q R = u c (T4 – T1) = 0.718 ( 1014.6 – 373) = 460 KJ / kg

or                                 q s / q R = heat supplied / heat rejected = 1062.4 / 460.7 = 2.306

Points              Pressure bar                Volume m3                 Temp °C

1                      1.0                               1.0705                        100

2                      18.4                            0.1338                         583.8

3                      50                               0.1338                         2063.4

4                      2.72                             1.0705                         741.6

Example 10.16. A Jour stroke engine having swept volume of 0.13 m3 , operates on otto cycle. the compression ratio is 6 and tile conditions at the beginning of compression are 1 kg f/cm and 60°C. The heat supplied is 36 Kcal/Kg/cycle. Calculate the work done, efficiency. Take c1, = 0.237 Kcal/kg K, cv = 0.169 Kcal/kg K. r= 1.4, (air)

Solution.     Vs = 0.13 m3, T1 = 60°C = 60 + 273 = 333°K

r = 6, P1 = 1 kg f/cm2 , q5 = 36 kcal/cycle

r = u1 / u2 = vc + us / vc

6 = uc + 0.13 / uc , or uc = 0.026 m3

u1 = uc + us = 0.156 m3

u2 = uc = 0.026 m3

process 1-2     T2 = T1 r γ-1 = 333 (6)0.4 = 681.88 k

qs = vc ( T3 –T2)

36 = 0.169(T3 – 681.88)

T3 = 894.89 K

Process 3-4     T4 = T3 / r γ-1 = 894.89 / 60.4 = 437 k

Q R = cu (T4 – T1) = 0.169 ( 437 – 333) = 17.5 Kcal / kg

= qs – q R = 36 – 17.5 = 18.5 kcal / kg

= 18.5 x 4.184 kj / kg

= 77.4 kj / kg

ɳ = 1 – q R/ qs , – 17.5 / 36 = 51.38%

Examples Based on Diesel Cycle

Example 10.18. 1 kg of air is taken through a diesel cycle. Initially the air is at 15°C and 1 bar. The P compression ratio is 15 and the heat added is 1850 K]. Calculate the ideal cycle efficiency and mean effective pressure.

Solution .               u1 = RT1 / P1 = 0.287 x (15 + 273) / 100

= 0.81 m3 / kg [ 1 bar = 100 kn / m2

T2 = T1 r γ-1 = 288 (15) 1.4-1 = 851 k

u2 = u1 / r = 0.81 / 15 = 0.054 m3 / kg

2-3 , heat added = qs = cp (T3 – T2)

or                                  1850 = 1 x 1.005 ( t3 – 851)

process                        2-3 p2 u2 / T2 = P2 u2 / T3 , [ P2 = P3

u2 / T2 = u3 / T3

T3 / T2 = u3 / u2 = β

β = T3 / T2 = 2692 / 851 = 3.1633

u4 / u3 = u1 / u2 x u2 / u3 = r /β = 15 / 3.1633 = 4.742

T4 = T3 (u3 / u4)γ-1 = 2692 ( 1 / 4.742)0.4 = 1444 k

Head rejected              q R  = cu ( T4 – T1) = 1 x 0.718 ( 1444 – 288)

= 1850 – 830 = 1020 kj / kg

ɳ = work done / q s = 1020 / 1850 = 55.1%

Mean effective pressure sure = pm

Pm = net work done / u1 –u2 = 1020 / 0.81 – 0.054 KN / M2

= 1020 / (0.81 – 0.054) 100 bar

= 13.4 bar.

Example 10.19. In a diesel cycle the maximum pressure reached is 36 bar. The compression ratio is 13. The  temperature at the beginning of compression is 83°C and the temperature at the beginning of expansion is 1800°C. Determine (a) the temperature at the end of compression (b) the temperature at the end of expansion (c) the heat supplied and heat rejected per kg of working fluid, and (d) the thermal efficiency of the cycle. Take c P = 0.986 KJ/kg°K

Cv = 0.702 KJ/kg0 K.

Solution. Given T1 = 83°C = 83 + 273 = 356°K, r = 13,

T 3 = 1800°C = 1800 + 273 = 2073°K,