T&HE: LESSON - 37 NUMERICAL PROBLEMS ON STEAM ENGINE INCLUDING PROBLEMS ON CALCULATIONS OF CYLINDER DIMENSIONS

LESSON - 37 NUMERICAL PROBLEMS ON STEAM ENGINE INCLUDING PROBLEMS ON CALCULATIONS OF CYLINDER DIMENSIONS

Problem 37.1: In a double acting steam engine, the cylinder diameter is 60 cm and stroke is 70 cm. The mean piston speed is 20 m/min. The intake pressure is 10 bar and exhaust pressure is 1.3 bar. The cut-off occours at 25% of the stroke. Neglect clearance and compression. Find the i.p. of the engine for a diagram factor of 0.78. Neglect the piston rod area.

Solution:

Given: Steam engine type = Double acting

Diameter of cylinder, D = 60 cm = 0.6 m;

Stroke length, L = 70 cm = 0.7 m;

Mean piston speed = 2LN = 20 m/min;

Admission pressure, p_a = 10 bar

Exhaust/Back pressure, p_b = 1.3 bar

Diagram factor = 0.78.

Clearance and compression = Neglected

Cut off = 25% of the stroke

i.e.,

Therefore, Ratio of expansion,

Determine the i.p. of the engine:

Formula: We have i.p. of double acting steam engine = , kW

or i.p. =

Finding unknown, p_am:

The actual mean effective pressure is given by

P_am (actual) = k p_tm (theoretical)

Finding unknown, p_tm:

The theoretical mean effective pressure is given by

p_tm =

= 4.66 bar

Therefore, P_am (actual) = k p_tm (theoretical) = 0.78 x 4.66 = 3.634 bar

Answer: Therefore, i.p. = = 34.24 kW

Problem 37.2: A double acting steam engine has a stroke equal to 1.3 times the diameter and a diagram factor of 0.80. Dry and saturated steam is supplied at 10 bar and exhausts at 1.05 bar. If the speed is 250 rpm and ratio of expansion 2.5, indicated power 185 kW, calculate the dimensions of the cylinder. Neglect clearance and compression.

Solution:

Given: Steam engine type = Double acting;

Stroke length, L = 1.3 x (diameter of cylinder) = 1.3D;

Speed, N = 250 rpm;

Admission pressure, p_a = 10 bar;

Exhaust/Back pressure, p_b = 1.05 bar;

Diagram factor = 0.8;

Clearance and compression = Neglected;

Ratio of expansion, r = 2.5;

Indicated power, i.p. = 185 kW.

Determine the dimensions of the cylinder, D and L:

Formula: We have, indicated power of double acting steam engine,

i.p. = , kW

Therefore,

Finding unknown, p_am:

The actual mean effective pressure is given by

P_am (actual) = k p_tm (theoretical)

Finding unknown, p_tm:

The theoretical mean effective pressure is given by

p_tm =

= 6.615 bar

Therefore, P_am (actual) = k p_tm (theoretical) = 0.8 x 6.615 = 5.2952 bar

Answer: Therefore, D =

= 0.354 m = 34.5 cm

and L = 1.3 D = 1.3 x 34.5 = 44.85 cm

Problem 37.3: Dry and saturated steam at 8 bar is supplied to a single cylinder double acting steam engine. The cut-off occours at 30% of the stroke. The expansion follows the law pV^1.15= C. The compression commences at 60% of the return stroke and follows the law pV^1.3= C. The clearance volume is 20% of the stroke. The back pressure is 65 cm of Hg vaccum when barometer reads 76 cm of Hg. Calculate the mean effective pressure and i.p. of the engine if the cylinder diameter is 36 cm, stroke 50 cm and speed 250 rpm.

Solution:

Given: Steam engine type = Double acting;

Admission pressure, p_a = 8 bar;

Exhaust/Back pressure, p_b = 65 cm of Hg vaccum =

= 0.1466 bar;

Cut off = 30% of the stroke i.e. cut off ‘ac’ = 0.30 stroke ‘df’

or cut off ‘ac’ = 0.30 Vs

Volume ‘kd’ = 60% of the return stroke i.e Volume ‘kd’ = 0.6 Vs

Therefore, volume ‘ak’ = Vs - 0.6 Vs = 0.4 Vs

Clearance volume, V_a = V_f= 20% of the stroke

= 0.2Vs

Expansion follows the law pV^1.15= C

Compression follows the law pV^1.3= C

Diameter of cylinder, D = 36 cm = 0.36 m;

Stroke length, L = 50 cm = 0.5 m;

Speed, N = 250 rpm;

(a) Determine the mean effective pressure of the engine, p_tm :

Formula: The theoretical mean effective pressure, p_tm =

Finding unknown, W:

Since, the Area of the hypothetical indicator diagram represents the theoretical work done per cycle, so

W = area acrdkf = area acCF + area crDC − area dkKD – area kfFK

= W_ac +W_cr – W_dk – W_kf

= p_a (V_c– V_a) + − p_b (V_d–V_k) −

Finding unknown, Vc, Vr, V_d, V_k, V_f, V_k, p_{r ,}p_f:

Vc = V_a+ ‘ac’ =0.2Vs + 0.30 Vs = 0.5Vs

Vr = V_d = V_a+ Vs = 0.2Vs + Vs = 1.2Vs

V_k = V_a+ ‘ak’ = 0.2Vs + 0.4 Vs = 0.6Vs

Finding unknown, p_r;

From expansion law, for process ‘cr’ we have

p_cV_c^1.15 = p_rV_r^1.15

or p_r = = 2.923 bar

Finding unknown, p_f;

From compression law, for process ‘kf’ we have

P_kV_k^1.3 = p_fV_f^1.3

or p_f = = 0.611 bar

Therefore, W= p_a (V_c–V_a) + − p_b (V_d–V_k) −

= −

Answer: Therefore, p_tm = 5.53 bar

Determine the i.p. of the engine:

Formula: i.p. (double acting) steam engine = , kW

Answer: Therefore, i.p. =

= 221.68 kW

Problem 37.4: A steam locomotive has two cylinders of 50 cm diameter and 65 cm stroke. The diameter of the driving wheels are 2 m. Steam is admitted into the cylinder at 13 bar. The maximum cut-off occurs at 0.80 of the stroke. The diagram factor is 0.81. Calculate the tractive effort with this maximum cut-off. Mechanical efficiency may be assumed as 0.95.

If the train is running at 90 km/h then the resistances is 120 N per tonne. Calculate the total train load that can be hauled at this speed if the cut- off occurs at 25% of the stroke. Assume that diagram factor. 0.73.

Solution:

Given: No. of cylinders = 2;

Diameter of cylinder, d= 50 cm = 0.5 m;

Stroke length, L = 65 cm = 0.65 m;

Admission pressure, p_a = 13 bar;

Maximum cut off = 0. 80 of the stroke

i.e., = r′ = 0.80

Therefore, ratio of expansion = 1.25

Diagram factor, K = 0.81

Mechanical efficiency, η_m = 0.95

The diameter of the driving wheels, D = 2 m

(a) Determine the tractive effort with this maximum cut-off.

Formula: Work done per revolution of the wheel = Work done available at the shaft due to two cylinders per revolution (i.e. N = 1)

where T = tractive effort, N ; p_am= actual m.e.p., N/m²

Finding unknown, p_am:

The theoretical mean effective pressure is given by

p_tm=

The actual mean effective pressure is given by

P_am (actual) = k p_tm (theoretical)

= 0. 81 x 11.42 = 9.25 bar

Answer: Tractive effort with this maximum cut-off, T

= 0.7139 x 10⁵N

(b) Determine the total train load (W) that can be hauled at 90 km/h speed if the cut- off occurs at 25% of the stroke.

Given: For, 0.25 cut-off, the expansion ratio,

Speed of train, S = 90 km/h

Resistances, R = 120 N/tonne

Formula: Work done required per sec to haul loaded train = , kW

where, F is the tractive effort required = R x W, N

W is the train load to be hauled, tonnes

This required work to haul loaded train will be supplied by the two cylinders and is available at the crank shaft

= x η_m x (No. of cylinder), kW

Therefore, x η_m x (No. of cylinder)

or 2 ×

or Total train load, W =

Finding unknown, p_am:

The theoretical mean effective pressure is given by

p_tm =

The actual mean effective pressure is given by

P_am (actual) = k p_tm (theoretical)

= 0. 73 x 6.45 = 4.7 bar

Finding unknown, N:

Speed in m/min,

N = = rpm

Answer: Therefore, Total train load, W =

= 302.3 tonnes

Problem 37.5: Dry and saturated steam at a pressure of 12 bar is supplied to a double acting steam engine cylinder. The cut-off occurs a 40% of the stroke and the exhaust pressure is 1.2 bar. The clearance is 10% of the stroke and the compression commences at 0.85 of the return stroke. Find the mean effective pressure.

If the b.p. of the engine is 21 kW running at 120 rpm, calculate the cylinder dimension. The mechanical efficiency may be assumed as 80%. Assume the expansion of steam is hyperbolic. The mean speed of the engine is 70 m/min.

Solution:

Given:

Steam engine cylinder = Double acting

Admission pressure, p_a = 12 bar;

Exhaust/Back pressure, p_b = 1.2 bar;

Cut off = 40% of the stroke i.e. cut off ‘ac’ = 0.430 stroke ‘DF’

or Cut off ‘ac’ = (V_c-V_a) = 0.40 Vs

Clearance volume, V_a = V_f= 10% of the stroke

= 0.1Vs

Volume ‘kd’= 0.85 of the return stroke i.e Volume ‘kd’= 0.85Vs

Therefore, volume ‘ak’ = (V_k-V_f) = Vs - 0.85Vs = 0.15Vs

Determine the mean effective pressure.

Formula: With compression and clearance, the mean effective pressure

p_tm

Finding unknown, c' , r', x' ;

c' = = 0.1

r' = = 0.4

x' = = 0.15

Answer: Therefore, the mean effective pressure

p_tm = 8.2 bar

Determine the cylinder dimension Length (L) and diameter of cylinder (d):

Given: b.p. of the engine = 21 kW

Speed, N = 120 rpm

Mechanical efficiency = 80%

Expansion of steam is hyperbolic.

Mean speed of the engine, S = 70 m/min.

Determine length of stroke, L:

Formula: Since the mean piston speed = 2LN = S m/min.

where, L is piston stroke

or L =

Answer: = = 0.2616

Determine diameter of the cylinder, d:

Formula: i.p. of double acting steam engine

, kW

and η_m=

or 2 ×

or d =

Answer: d =

= 0.1874 m = 18.74 cm

Problem 37.6: Following data were obtained during a trial on a single cylinder double acting steam engine.

Cylinder diameter , D = 25 cm	Radius of brake wheel = 0.75 m
Stroke =36 cm	Radius of rope=0.12 cm
Piston rod diameter, d =5 cm	Dead load on brake = 2650 N
rpm = 240	Reading on spring balance = 150 N
Length of indicator diagram =5.5 cm	Steam used = 15 kg/min
Area of indicator diagram for cover end =14.2 cm²	Initial steam pressure = 10 bar
Area of indicator diagram for crank end 15.5 cm²	back pressure = 1.4 bar
Scale of the spring=1.5 bar/cm

Determine i.p., b. p., mechanical efficiency and brake thermal efficiency.

Solution:

Given:

Cylinder diameter , D = 25 cm = 0.25m

Stroke, L =36 cm = 0.36m

Piston rod diameter, d =5 cm = 0.05m

Rpm, N = 240

Length of indicator diagram =5.5 cm = 0.055m

Area of indicator diagram for cover end =14.2 cm²

Area of indicator diagram for crank end 15.5 cm²

Scale of the spring =1.5 bar/cm

Radius of brake wheel, R = 0.75 m

Radius of rope, r =0.12 cm = 0.0012 m

Dead load on brake, W = 2650 N

Reading on spring balance, S = 150 N

Steam used, M= 15 kg/min

Initial steam pressure = 10 bar

From Steam table (Dry saturated steam)

At pressure, p = 10 bar h_g= 2776.2 kJ/kg,

Back pressure = 1.4 bar

From Steam table (Dry saturated steam)

At pressure, p = 1.4 bar h_f= 454.7 kJ/kg,

(a) Determine i.p.:

Formula: Total (i.p.) indicated power developed by the engine

= , kW

Finding unknown, P_am,1, P_am,2 , A₁and A₂:

Actual mean effective pressure for the cover end is given by

p_am,1 (cover end) = = 3.87 bar

Actual mean effective pressure for the crank end is

p_am,2 (crank end) = = = 4.227 bar

Effective area on which pressure acts on the cover end side is given by

A₁ (cover end) = = 0.0452 m²

Effective area on which pressure acts on the crank end side is

A₂ (crank end) = = 0.0432 m²

Answer: Total indicated power developed by the engine,

i.p. =

= 25.18 + 26.29 = 51.47 kW

(b) Determine b. p.:

Formula: b.p. developed by the engine;

Brake power (bp) = (kW or kJ/S)

Finding unknown, T;

T = (W - S) x (R + r)

= (2650 - 150) x (0.75 + 0.0012) = 1878 N-m

Answer: Therefore, Brake power (bp) = = 47.19 (kW or kJ/S)

(c) Determine mechanical efficiency, η_m :

Formula: η_m=

Answer: Mechanical efficiency, η_m = × 100 = 91.68%

(d) Determine brake thermal efficiency:

Formula: The brake thermal efficiency is

η_b,th =

Answer: η_b,th= = 8.13%

Last modified: Wednesday, 11 September 2013, 4:39 AM

Thermodynamics and Heat Engine4 (3+1)

LESSON - 37 NUMERICAL PROBLEMS ON STEAM ENGINE INCLUDING PROBLEMS ON CALCULATIONS OF CYLINDER DIMENSIONS

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