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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, pa = 10 bar Exhaust/Back pressure, pb = 1.3 bar Diagram factor = 0.78. Clearance and compression = Neglected |
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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, pam:
The actual mean effective pressure is given by
Pam (actual) = k ptm (theoretical)
Finding unknown, ptm:
The theoretical mean effective pressure is given by
ptm =
= 4.66 bar
Therefore, Pam (actual) = k ptm (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, pa = 10 bar;
Exhaust/Back pressure, pb = 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,
or
Finding unknown, pam:
The actual mean effective pressure is given by
Pam (actual) = k ptm (theoretical)
Finding unknown, ptm:
The theoretical mean effective pressure is given by
ptm =
= 6.615 bar
Therefore, Pam (actual) = k ptm (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 pV1.15 = C. The compression commences at 60% of the return stroke and follows the law pV1.3 = 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, pa = 8 bar;
Exhaust/Back pressure, pb = 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, Va = Vf = 20% of the stroke = 0.2Vs Expansion follows the law pV1.15 = C Compression follows the law pV1.3 = C Diameter of cylinder, D = 36 cm = 0.36 m; Stroke length, L = 50 cm = 0.5 m; Speed, N = 250 rpm;
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(a) Determine the mean effective pressure of the engine, ptm :
Formula: The theoretical mean effective pressure, ptm =
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
= Wac +Wcr – Wdk – Wkf
= pa (Vc – Va) + − pb (Vd–Vk) −
Finding unknown, Vc, Vr, Vd, Vk, Vf , Vk, pr , pf :
Vc = Va + ‘ac’ =0.2Vs + 0.30 Vs = 0.5Vs
Vr = Vd = Va + Vs = 0.2Vs + Vs = 1.2Vs
Vk = Va + ‘ak’ = 0.2Vs + 0.4 Vs = 0.6Vs
Finding unknown, pr ;
From expansion law, for process ‘cr’ we have
pcVc1.15 = prVr1.15
or pr = = 2.923 bar
Finding unknown, pf ;
From compression law, for process ‘kf’ we have
PkVk1.3 = pfVf1.3
or pf = = 0.611 bar
Therefore, W = pa (Vc–Va) + − pb (Vd–Vk) −
=
−
=
Answer: Therefore, ptm = 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, pa = 13 bar; Maximum cut off = 0. 80 of the stroke i.e., Therefore, ratio of expansion |
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Diagram factor, K = 0.81
Mechanical efficiency, ηm = 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)
or
or
where T = tractive effort, N ; pam = actual m.e.p., N/m2
Finding unknown, pam:
The theoretical mean effective pressure is given by
ptm =
The actual mean effective pressure is given by
Pam (actual) = k ptm (theoretical)
= 0. 81 x 11.42 = 9.25 bar
Answer: Tractive effort with this maximum cut-off, T
=
= 0.7139 x 105 N
(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 ηm x (No. of cylinder), kW
Therefore, x ηm x (No. of cylinder)
or 2 ×
or Total train load, W =
Finding unknown, pam:
The theoretical mean effective pressure is given by
ptm =
The actual mean effective pressure is given by
Pam (actual) = k ptm (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, pa = 12 bar; Exhaust/Back pressure, pb = 1.2 bar; Cut off = 40% of the stroke i.e. cut off ‘ac’ = 0.430 stroke ‘DF’ or Cut off ‘ac’ = (Vc-Va) = 0.40 Vs Clearance volume, Va = Vf = 10% of the stroke = 0.1Vs Volume ‘kd’= 0.85 of the return stroke i.e Volume ‘kd’= 0.85Vs Therefore, volume ‘ak’ = (Vk-Vf) = Vs - 0.85Vs = 0.15Vs |
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Determine the mean effective pressure.
Formula: With compression and clearance, the mean effective pressure
ptm
Finding unknown, c' , r', x' ;
c' = = 0.1
r' = = 0.4
x' = = 0.15
Answer: Therefore, the mean effective pressure
ptm = 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 ηm =
or
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 |
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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 hg = 2776.2 kJ/kg,
Back pressure = 1.4 bar
From Steam table (Dry saturated steam)
At pressure, p = 1.4 bar hf = 454.7 kJ/kg,
(a) Determine i.p.:
Formula: Total (i.p.) indicated power developed by the engine
= , kW
Finding unknown, Pam,1, Pam,2 , A1 and A2 :
Actual mean effective pressure for the cover end is given by
pam,1 (cover end) =
= 3.87 bar
Actual mean effective pressure for the crank end is
pam,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, ηm :
Formula: ηm =
Answer: Mechanical efficiency, ηm = × 100 = 91.68%
(d) Determine brake thermal efficiency:
Formula: The brake thermal efficiency is
ηb,th =
Answer: ηb,th = = 8.13%