Chapter 5 Steam Engines – Thermal Engineering

Chapter 5

Steam Engines

5.1 ❐ INTRODUCTION

A steam engine uses steam as the working medium. It is the earliest prime mover developed to convert thermal energy into mechanical energy. It is called a prime mover because it is used to drive other devices like locomotives, compressors, lathe, shear presses, etc. On account of their low efficiency steam engines have become obsolete. However, they are still being used in locomotives and some industries.

A steam engine is called an external combustion engine as the fuel is burnt in the boiler and the steam raised is used to reciprocate the engine piston for generating power.

5.2 ❐ CLASSIFICATION OF STEAM ENGINES

A steam engine is of the reciprocating type, in which the piston moves to-and-fro in the cylinder due to the force applied by the expansion of steam. The reciprocating steam engines may be classified as follows:

  1. According to class of service:
    1. Stationary
    2. Marine
    3. Locomotive
    4. Pumping or hoisting
  2. According to speed:
    1. Low speed: below 125 rpm
    2. Medium speed: 125 to 300 rpm
    3. High speed: above 300 rpm

      Note that the speed of the engine means speed of the crankshaft.

  3. According to arrangement of cylinders:
    1. Horizontal
    2. Inclined
    3. Vertical
  4. According to type of valve design:
    1. Simple plate valve
    2. Balance plate valve
    3. Piston valve
    4. Riding cut-off valve
    5. Poppet valve
    6. Corliss valve
  5. According to method of governing:
    1. Manual or automatic governing
    2. throttle or cut-off governing
    3. Centrifugal or inertia type governor
  6. According to type of exhaust:
    1. Condensing type
    2. Non-condensing type
  7. According to steam condition:
    1. High or low pressure
    2. Saturated or superheated
  8. According to the action of steam upon the piston:
    1. Single acting
    2. Double acting
  9. According to the nature of expansion:
    1. Expansive engines
    2. Non-expansive engines
  10. According to the range of expansion of steam:
    1. Simple steam engine
    2. Compound steam engine
5.3 ❐ CONSTRUCTIONAL FEATURES OF A STEAM ENGINE

A schematic line diagram of a vertical steam engine is shown in Fig. 5.1,

The parts of a steam engine may be broadly classified as:

  1. Stationary parts
    1. Engine bedplate
    2. Engine frame
    3. Cylinder
    4. Steam chest
    5. Steam jacket
    6. Stuffing box
    7. Cross-head guides
    8. Main bearings
  2. Moving parts
    1. Piston
    2. Piston rod
    3. Cross-head

      Figure 5.1 Schematic diagram of a steam engine

    4. Connecting rod
    5. Crank and crank shaft
    6. D-slide valve
    7. Valve rod
    8. Eccentric and eccentric rod
    9. Flywheel
    10. Governor

5.3.1 Steam Engine Parts

The various parts are briefly described below:

Frame: It is a heavy casting made of cast iron and supports cross-head guides and main bearing. It generally rests on the engine foundation.

Cylinder: It is also a casting to the frame in which the piston reciprocates.

Its both ends are closed and made steam tight. The stuffing box is fixed in one end through which piston rod reciprocates.

Steam chest: It is compartment in a steam engine through which steam is delivered from the boiler to a cylinder. It is connected to the cylinder through the valve passage known as ports. It contains the D-slide valve. The steam is supplied alternately to the cylinder through the ports and exhausted to the condenser. The same port serves as inlet port and exhaust port in double acting engines.

D-slide valve: It connects the cylinder to the steam chest and to the exhaust side through the ports at the correct crank positions. The valve is operated by an eccentric.

Piston: It is made of cast iron. The steam pressure exerts force on it on both sides to reciprocate the piston rod. Piston rings are provided on the piston for steam tightness.

Piston Rod: One end of piston rod is connected to the piston and the other end to the cross-head. It transmits the force from the piston to the cross-head.

Stuffing Box and Gland: It is placed at the point where the piston rod passes through the cylinder cover. It prevents the leakage of steam from the cylinder to atmosphere, at the same time allowing the piston rod to reciprocate freely.

Cross-Head: It connects the piston rod and connecting rod. It guides the piston rod along a straight line. The thrust on the cross-head is taken of by the cross-head guides.

Connecting Rod: One end of it is connected to the cross-head by a gudgeon pin and the other end is connected to the crank. It transmits force from the cross-head to the crank.It has oscillatory motion.

Crankshaft: The crank converts the reciprocating motion to rotary motion of the crankshaft.The crank shaft is supported on main bearings and carries the flywheel and eccentric

Eccentric: It converts rotary motion of crankshaft into reciprocating motion of D slide valve.

Valve rod and Eccentric rod: They connect the eccentric to D-slide valve.

Flywheel: It is connected to the crankshaft and helps to maintain a constant angular speed of the engine by storing excess energy during idle stroke of the device being operated.

Governor: It helps in maintaining uniform speed of the engine by regulating supply of stream to the cylinder.

5.4 ❐ TERMINOLOGY USED IN STEAM ENGINE

A line diagram of steam engine mechanism is shown in Fig. 5.2.

Cylinder Bore, D: It is the inside diameter of the cylinder

Piston stroke, L: It is the distance travelled by the piston from the cover end of the cylinder to the crank end of cylinder. L = 2r.

Single Acting: Steam is supplied on one side of piston only

Double acting: Steam is supplied on both sides of piston.

Dead centres: The position of piston within the cylinder, when the crank and connecting rod are in the same straight line.

Crank radius, r: It is the distance between the centre of crankshaft and centre of crank pin

Figure 5.2 Line diagram of a steam engine

Swept volume, VS: It is the volume swept by the piston in one stroke

Mechanical Clearance, LC: The mechanical clearance is the distance between the cylinder cover and inner dead centre position of the piston. It is specified as percentage of stroke length.

Volumetric Clearance, VC: It is the volume between the cylinder cover and the piston at the inner dead centre position of piston.

Engine speed, N: It is measured in terms of revolutions of crankshaft per minute.

Piston speed: It is linear speed of the piston.

Compression Ratio, α: It is the fraction of stroke volume completed at the start of compression.

Expansion Ratio, r: It is the ratio of swept volume to col nine at cut-off of steam.

Cut off Ratio,: It is the ratio of volume at cut-off steam to the swept volume

Mean Effective Pressure, Pm: It defined as that hypothetical constant pressure which will produce same work for the same piston displacement with the variable conditions of temperature pressure and volume.

Where card area = area of indicator diagram, card length = length of stroke in indicator diagram.

Eccentric Throw: It is the distance between the centre of eccentric and centre of crankshaft.

Valve Travel: It is the maximum distance travelled by the valve along one direction. It is equal to twice the eccentric throw.

Back pressure, pb: The pressure of exhaust steam acting on the other side of the piston is known as back pressure.

5.5 ❐ WORKING OF A STEAM ENGINE

The working of a double acting steam engine is a shown in Fig. 5 3. As the piston reaches the inner dead centre (IDC) position A, the D-slide valve clears the inlet port C and admits the steam into the cylinder. This high pressure steam pushes the piston on the forward direction and performs work by rotating the crank and eccentric. The steam continues to enter into the cylinder maintaining a constant pressure inside. With the further rotation of crank and eccentric, D-slide valve is moved backwards and closes the inlet port C. The steam supply to the cylinder is stopped and this is known as cut-off of steam. After the cut-off the steam in the cylinder expands pushing the piston in the forward direction till the piston reaches the outer dead centre (ODC) position B. Just before the ODC position, the valve connects the cylinder through the port to exhaust E. Therefore the pressure on this side of cylinder falls to the atmospheric pressure or condenser pressure. After the ODC position the piston reverses its direction of motion and the return stroke starts. The steam is admitted through port D and exhausted on the IDC side. Just before the piston reaches the IDC position, the port D closes and low pressure steam in the cylinder is compressed and acts as a cushion to the piston at the end of the stroke. As the piston reaches the IDC position, the D-slide valve opens the port C and the cycle is repeated. When admission and expansion take place on head end side, exhaust and compression takes place on the crank end side.

Figure 5.3 Working of a steam engine

5.6 ❐ RANKINE CYCLE

Rankine cycle is the theoretical cycle on which the steam engine works. The Rankine cycle is shown in Fig. 5.4.

The p-v, T-s and h-s diagram for the Ranking cycle are shown in Fig. 5.5.

Figure 5.4 Rankine cycle

Figure 5.5 Rankine cycle in various ordinates: (a) p-v diagram, (b) T-s diagram, (c) h-s diagram

The various processes of Rankine cycle are:

Process 1-2: Reversible adiabatic expansion in the engine.

Process 2-3: Reversible constant pressure heat transfer in the condenser.

Process 3-4: Reversible adiabatic pumping process in the feed pump.

Process 4-1: Reversible constant pressure heat transfer in the boiler.

The steam on entry into the engine can be wet, dry saturated or superheated. Considering 1 kg of steam and applying steady flow energy equation to boiler, engine, condenser and pump, we have:

Boiler:

 

hf 4 + q1 = h1
q1 = h1hf 4

Engine:

 

h1 = we + h2

Engine work,     we = h1h2

Condenser:

 

h2 = q2 + hf 3
q2 = h2hf 3

Feed pump:

hf 3 + wp = hf 4

Pump work,     wp = hf 4hf 3

Net work,      wnet = wewp

Rankine cycle efficiency,

The feed pump handles liquid water which is incompressible. This implies that with the increase in pressure its density or specific volume undergoes a small change. For reversible adiabatic compression, we have

 

T ds = ahv dp
ds = 0
dh = v dp

or   Δh = v Δp           [∵ Δv = 0]

or    hf4hf3 = v3 (p1p2) × 105 J/kg

Now.    hf4hf3 = wp << we, and can be neglected

5.7 ❐ MODIFIED RANKINE CYCLE

In modified Rankine cycle to reduce friction the stroke length is reduced.

Figure 5.6(a) and 5.6(b) show the modified Rankine cycle on p-v and T-s diagrams respectively, by neglecting pump work.

It may be observed that p-v diagram is very narrow at the toe. i.e., point 3′. The work obtainable near to 3′ is very small which is too inadequate to overcome friction due to reciprocating parts of engine. Therefore the adiabatic process 2-3′ is terminated at point 3. The pressure drop decreases suddenly whilst the volume remains constant. This operation is represented by the line 3-4. By doing so the stroke length is reduced. The loss of work is represented by the hatched area 3-3′-4-3, which is negligibly small.

The work done during the modified Rankine cycle can be calculated as follows:

Let p1, v2, u2, h2 represents the initial conditions at point 2.

p2, v3, u3, h3 corresponds to condition of steam at point 3.

Figure 5.6 Modified Rankine cycle: (a) p-v diagram, (b) T-s diagram

p3, h4 correspond to condition of steam at point 4.

Work done during the cycle per kg of steam = area 1 – 2 – 3 – 4 – 5 – 1

= area 1 – 2 – b – 0 – 1 + area 2 – 3 – a – b – 2 area – 0 – b – a – 4 – 5 – 0

= p1 v2 + (u2u3) – p3 v3

Heat supplied, qs = h2hf4

5.8 ❐ HYPOTHETICAL OR Theoretical INDICATOR DIAGRAM

The indicator diagram is a representation of the variations of pressure and volume of steam inside the cylinder on p-V diagram for one complete cycle of operation. The theoretical p-v diagram for a single acting steam engine is shown in Fig. 5.7 without and with clearance.

The sequence of operations is as follows:

Process 1-2: The steam is supplied to cylinder at constant pressure p1. The steam is cut-off at point 2.

Process 2-3: The steam expands in the cylinder till the piston reaches the ODC position. The expansion process is hyperbolic isothermal, i.e. pv = const.

Process 3-4: The pressure falls from point 3 to point 4 at constant volume instantaneously due to the opening of the exhaust. The point 3 is called the point of release as the pressure is allowed to fall suddenly to the back pressure pb.

Process 4-5: It represents the exhaust of used steam at constant pressure

Process 5-1: It represents steam admission to cylinder at constant volume. The pressure suddenly rises form pb to p1.

The working cycle is completed within two strokes of piston or one revolution of crank.

Figure 5.7 Theoretical indicator diagram for a steam engine: (a) Without clearance, (b) With clearance

5.9 ❐ ACTUAL INDICATOR DIAGRAM

Indicator diagram is simply a graph between pressure of steam in a cylinder against the steam volume.

The following assumptions are made in drawing the theoretical indicator diagram:

  1. Clearance is neglected.
  2. Steam supplied to cylinder is at constant pressure.
  3. Ports are opened and closed instantaneously.
  4. There is no drop in pressure due to condensation.
  5. Expansion of steam in cylinder follows the hyperbolic law.

The theoretical and actual indicator diagrams are shown in Fig. 5.8. The actual indicator diagram is found to be different from the theoretical diagram due to following reason:

  1. A definite clearance is necessary to prevent the piston striking the cylinder head. Provide cushioning effect to piston and allow small quantity of water to be collected in the cylinder due to condensation of steam. Therefore the actual supply of steam starts from point ‘ainstead of point A. The admission pressure is also lower due to condensation of Steam on the cooler cylinder walls and throttling effect during flow of steam through the inlet port. Therefore, the supply steam pressure is represented by ab instead AB.
  2. The instantaneous cut-off is not possible as it takes place during few degrees of crank rotation. Therefore, the diagram gets rounded up near point b.
  3. The expansion of steam in the cylinder is not perfectly hyperbolic and is represented by some other curve bc instead of BC. This is due to condensation of steam during expansion.
  4. The exhaust opens before the piston reaches ODC position (say point C) as instantaneous drop in pressure is not possible. Therefore, fall in pressure starts from C and actual pressure drop is represented by ‘cd’.
  5. The exhaust pressure is slightly above the condenser or back pressure as the steam has to be forced out from cylinder. The actual exhaust pressure is represented by ‘de’.
  6. The supply of exhaust steam to exhaust port is stopped at point ‘c’ by closing the port before the piston reaches IDC positions. The entrapped steam is compressed during the remaining stroke and provides a cushioning effect. This is represented by ‘ef’. The inlet port opens just before the piston reaches the IDC position and the pressure inside the cylinder increases.

Figure 5.8 Theoretical and actual indicator diagrams

The area between the theoretical and actual indicator diagrams represents the lost work. The ratio of the area of the actual indicator diagram to the area of theoretical indicator diagram is called the diagram factor.

5.10 ❐ MEAN EFFECTIVE PRESSURE

The mean effective pressure is a quantity related to the operation of an internal combustion engine. It is a valuable measure of an engine’s capacity to do work that is independent of engine displacement.

5.10.1 Without Clearance

The theoretical p-v diagram for a single acting steam engine without clearance is shown in Fig. 5.9(a). It can be obtained by drawing a rectangle of area equal to the area of theoretical indicator diagram and base equal to the stroke volume, as shown in Fig 5.9(b).

Area under p-v diagram gives the work done per cycle by an engine, Work done per cycle w = (area l−2−2−0−l) + (area 2−3 −3−2−2) − area (5−4−4−0−5)

where p1 = steam admission pressure

p2 = release or cut-off pressure

Figure 5.9 p-v diagram without clearance

 

pb = back pressure
r = = expansion ratio
vs = v3 = swept volume
v2 = cut-off volume
v4 = v3, therefore

Now

Work done per cycle as represented by Fig. 5.9(b) is given by

 

w = pm. vs

or

From the above two equations, we get

5.10.2 With Clearance

The p-v diagram for a single acting steam engine with clearance is shown in Fig. 5.10(a). Work done per cycle is given by,

w = area 1-2-3-4-5 -1

=(area 1-2-2′-5′-1) + (area 2-3-3′-2′-2) − (area 5-4-3′-5′-5)

 

Clearance ratio, v3 = vc + vs = c vs + vs = (1 + c) vs

Expansion ratio,

Cut-off ratio,

Substituting the values of vc, v2 and v3, we get

Figure 5.10 p-v diagram with clearance

Mean effective pressure,

5.10.3 With Clearance and Compression

Figure 5.11 shows a hypothetical indicator diagram for a steam engine with clearance and compression. Clearance means the volume of gas left in the cylinder at the discharge end of the stroke. It includes the space between the piston and cylinder head, the volume of the valves, valve pockets etc. The various processes are:

Process 1-2: Steam admission at p1

Process 2-3: Hyperbolic expansion

Process 3-4: Steam released

Process 4-5: Exhaust of steam into condenser.

Process 5-6: Compression of remaining steam in cylinder

Process 6-1: entry of fresh from boiler into the cylinder rising the cylinder pressure suddenly to boiler pressure p1.

Work done per cycle w, = area l-2-3-4-5-6-1

= (area 1-2-2′-6′-1) + (area 2-3-3′-2′-2) − (area 5′-5-4-3′-5′) − (area 6′-6-5-5′-6′)

Figure 5.11 p-v diagram with clearance and compression

 

Let expansion ratio,

Swept volume, Vs = V3Vc

Clearance ratio,

 

V3 = Vc + Vs = cVs + Vs = (1+ c)Vs

Cut-off ratio

Compression ratio,

Vs = (α + c)Vs

5.10.4 With Clearance and Polytropic Expansion and Compression

A polytropic process can be expressed pvn = constant

The indicator diagram is shown in Fig. 5.12.

Work done per cycle.

w = area l-2-3-4-5-6-1

   = (area 1-2-2′-6′-1) + (area 2-3-3′-2′-2) − (area 5-4-3′-5′-5) − (area 6-5-5′-6′-6)

and

Let expansion ratio,

Swept volume, Vs = V3Vc

Clearance ratio,

V3 = Vc + Vs = cVs + Vs = (1+ c)Vs

Cut-off ratio

Compression ratio,

Figure 5.12 p-v diagram with clearance and polytopic expansion and compression

5.11 ❐ POWER DEVELOPED AND EFFICIENCIES

5.11.1 Indicated Power

The power developed by the steam engine based on the indicator diagram is the indicated power.

Work done by the steam engine per cycle is given by,

w = Force × distance travelled by the piston

Work done per second is the indicated power

where on cover end side.

on crank end side.

d = diameter of piston rod.

If area of piston rod is neglected, then

Let Ai = area of indicator diagram

Li = length of indicator diagram

The mean height of indicator diagram,

Mean effective pressure, Pm = h × s

where s = spring scale or spring number, (N/cm2)/cm.

5.11.2 Brake Power

A part of the power developed in the engine cylinder is lost in overcoming friction at different parts of the engine. Therefore the power available at the engine crankshaft is less than the indicated power. The power available at the crankshaft is known as the brake power of the engine because it is generally measured by some type of brake.

A rope is wound around the brake drum whose one end is connected to the spring balance suspended from a support and the other end carries the load W, as shown in Fig. 5.13:

Let W = weight on rope, N

S = spring pull, N

D = outer diameter of brake drum

dr = diameter of rope

The torque applied on brake drum,

Figure 5.13 Rope brake dynamometer

 

5.11.3 Efficiencies of Steam Engine

  1. Mechanical efficiency,

    Frictional power, FP = IPBP

  2. Thermal efficiency =

    Heat supplied to engine per kg of steam = h1 h2

    Where h1 = heat per kg of steam entering the engine

    hf 2 = heat per kg of condensate coming out of condenser

    ms = mass of steam supplied per hour

    Heat supplied per hour = ms (h1hf 2)

    Thermal efficiency of engine on IP basis, i.e., indicated thermal efficiency

    where msi = specific consumption of steam (kg/kWh) on IP basis

    where msb = specific consumption of steam (kg/kWh) on BP basis

  3. where mf = mass of fuel burnt per second in the boiler

        CV = calorific value of fuel in kJ/kg

5.12 ❐ GOVERNING OF STEAM ENGINES

The automatic variation of steam supply with the variation of load is called governing and the device used for this purpose is known as a governor. There are two methods commonly used for steam engine governing.

  1. Throttle governing: In this type of governing, the pressure of a steam admitted to the engine is reduced by throttling before it passes into the engine (Fig. 5.14) and the device used for this purpose is called the throttle valve. The lower pressure of steam during admission process reduces the work developed by the engine. It has been observed that the steam consumption rate is linearly proportional to the indicated power. This line is called Willams line.
  2. Cut-off governing: In this system of governing, the period of admission of steam entering the engine cylinder is reduced depending on the load on engine (Fig. 5.14b). The volume of steam admitted is proportional to the mass of steam.

The indicated power vs steam consumption per minute for throttle and cut-off governing are shown in Fig. 5.15. It is obvious from the figure that the steam consumption rate at full load is same for both types of governing, but part load steam consumption of throttle governing is higher than that of cut-off governing engine.

The governor used for throttle governing is simple and cheaper. But the thermal efficiency of cut-off governed engine is higher at part load conditions.

Figure 5.14 Methods of governing steam engine: (a) Throttle governing, (b) Cut-off governing

Figure 5.15 Effect of method of governing on indicated power

Example 5.1

The steam supply to a steam engine is at 15 bar dry and saturated. The condenser pressure is 0.4 bar. Calculate the Rankine efficiency of the cycle. Neglect pump work.

Solution

Given: p1 = 15 bar, p2 = 0.4 bar, x1 = 1.0

From steam tables at 15 bar, h1 = hg = 2789.9 kJ/kg, s1 = sg = 6.4406 kJ/kg.K

At 0.4 bar, hf 2 = 317.7 kJ/kg, hfg2 = 2319.2 kJ/kg
sf 2 = 1.0261 kJ/kg.K sfg2 = 6.6448 kJ/kg.K

For isentropic expansion of steam,

 

s1 = s2 = sf 2 + x2 sfg2
6.4406 = 1.0261 + x2 × 6.6448
x2 = 0.815
h2 = hf 2 + x2 hfg2 = 317.7 + 0.815 × 2319.2 = 2207.8 kJ/kg

Example 5.2

A simple Rankine cycle works between 28 bar and 0.06 bar. The initial condition of steam is dry and saturated. Calculate the cycle efficiency, work ratio and specific steam. consumption.

Solution

Given: p1 = 28 bar, x1 = 1.0 bar, p2 = 0.06 bar

The T-s diagram for the cycle is shown in Fig. 5.16.

From steam tables:

At 28 bar, h1 = 2802.0 kJ/kg, s1 = 6.2101 kJ/kg.K
At 0.06 bar, hf 2 = hf 3 = 151.5 kJ/kg, hfg2 = 2415.9 kJ/kg
sf 2 = 0.521 kJ/kg.K sfg2 = 7.809 kJ/kg.K
vf 3 = vf 2 = 0.001 m3/kg

Figure 5.16

For isentropic process 1-2, we have

 

s1 = s2 = sf 2 = x2 sfg 2
6.2104 = 0.521 + x2 × 7.809
x2 = 0.728
h2 = hf 2 + x2 hfg2 = 151.5 + 0.728 × 2415.9 = 1910.27 kJ/kg

Engine work, wc = h1h2 = 2802.0 − 1910.27 = 891.73 kJ/kg

Pump work, wp = hf 4hf 3 = vf 3 (p1 − p2)

= 0.001 (28 − 0.06) × 102 = 2.79 kJ/kg

Net engine work, wnet = wcwp = 891.73 − 2.79 = 888.94 kJ/kg

hf4 = hf3 + wp = 151.5 + 2.79 = 154.29 kJ/kg

Rankine efficiency,

Specific steam consumption,

Example 5.3

A single cylinder double acting engine has a bore of 250 mm and a stroke 300 mm. Steam is admitted at a pressure of 1.3 MPa. Cut-off occurs at 30% of stroke and the exhaust pressure is 0.12 MPa. The diagram factor is 0.82 and the mechanical efficiency is 80%. The engine runs at 150 rpm. Calculate of the power developed.

Solution

Given: D = 0.5 m, L = 0.3 m, p1 = 1.3 MPa, , Kd = 0.82

ηmech = 0.8, pb = 0.12MPa, N =150 rpm

Mean effective pressure,

Brake power developed,

Example 5.4

Dry saturated steam at 0.9 MPa is supplied to a single cylinder double acting steam engine developing 20 kW at 240 rpm. The exhaust is 0.15 MPa. Cut-off takes place at 40% of stroke. The diagram factor is 0.8 and stroke to bore ratio is 1.25. Assuming hyperbolic expansion determine (a) bore and stroke of engine, and (b) steam consumption per hour.

Solution

Given: p1 = 0.9 MPa, IP = 20 kW, N = 240 rpm, pb = 0.15 MPa

Mean effective pressure,

= [0.9× 0.4(1+ ln 2.5) − 0.15]× 0.8
= 0.4319 MPa

Power developed,

Steam consumption/h

Specific volume of dry saturated steam at 0.9 MPa = 0.2148 m3/kg from steam tables.

Example 5.5

A single cylinder double acting steam engine receives steam at 1.0 MPa and exhaust at 0.055 MPa. The dryness of inlet steam is 96%. The power developed by the engine when running at 220 rpm is 45 kW, with steam consumption of 450kg/h. The engine bore is 0.25 m and stroke is 0.375 m. The expansion ratio is 6. Calculate the diagram factor and the indicated thermal efficiency of the engine.

Solution

Given: p1 = 1.0 MPa, pb = 0.055 MPa, x = 0.96, IP = 45 kW, N = 220 rpm, ms = 450 kg/h, D = 0.25 m, L = 0.375 m, r = 6

Theoretical m.c.p. pmt

Actual m.e.p pm = pmt × Kd = 0.4103× Kd

Power developed, IP

 

Enthalpy of steam at 1 MPa and 0.96 dry, from steam tables,

h1 = hf 1 + x hfg = 762.6 + 0.96 × 2013.6 = 2717.6 kJ/kg

Enthalpy of water at exhaust pressure of 0.055 MPa, hf 3 = 350.6 kJ/kg

Indicated thermal efficiency

Example 5.6

A single cylinder double acting condensing type steam engine delivers 20 kW brake power at 240 rpm. The diameter and stroke of the engine are 0.2 m and 0.3 m respectively. The steam is supplied at 10 bar and cut-off takes place at 50% of stroke. The condenser vacuum is 56 cm of Hg while the barometer reads 76 cm of Hg. Mechanical efficiency is 80% clearance is 20 % of stroke and piston rod diameter is 5 cm. Determine the actual mean effective pressure and diagram factor.

Also determine the specific steam consumption on I.P. basis by neglecting clearance and piston rod area.

Solution

Given: BP = 20 kW, N = 240 rpm, D = 0.2 m, d = 0.5 m,

L = 0.3 m p1 = 10 bar, , c = 0.2, ηmech = 0.8 x = 0.95

Indicated, m.e.p

Indicated power,

Actual

Actual m.e.p., bar

Diagram factor,

Neglecting clearance and piston rod area, volume of steam supplied per revolution of engine at 10 bar.

supplied per hour

Steam supplied in kg/h

where vg = 0.198m3/kg at 10 bar from steam tables

Specific steam consumption on IP basis,

Example 5.7

Calculate the diameter and stroke of a single cylinder double acting steam engine developing 50 kW power at 120 rpm with mechanical efficiency of 80%. The steam is supplied at 8 bar pressure and back pressure of the engine is 1.2 bar. Cut-off takes place at 40% of the stroke and clearance is 15% of stroke volume.

Assume diagram factor 0.8, and stroke to bore ratio as 1.5.

Solution

Given: BP = 560 kW, N = 120 rpm, ηmech = 0.8, p1 = 8 bar, pb = 1.2 bar,

Indicated, m.e.p

= 9.8874 × 102 × D3 kW
62.5 = 9.8874 × 102 × D3
D = 0.398 m or 39.8 cm
L = 1.5 × 0.398 = 59.75 cm

Example 5.8

The steam enters a steam engine at 15 bar and exhausts at 1.5 bar.

The steam supply is cut-off at 40% of the stroke. The clearance volume is 5% of the swept volume. Calculate the mean effective pressure.

Solution

Given: p1 = 15 bar, pb = 1.5 bar, = 0.4, c = 0.05

Indicated, m.c.p

Example 5.9

In a steam engine, the clearance volume is 5% of swept volume and the back pressure is 1.15 bar. If the compression is at 0.3 of the stroke, find the pressure .at the end of compression stroke.

Find also the mean effective pressure, if the steam supply pressure is 13.7 bar and cut-off occurs at 40% of stroke.

Solution

The p-v diagram is shown in Fig. 5.17

Given: c = 0.05, p1 = 13.7 bar, pb = 1.15 bar, = 0.4, V5Vc = 0.3 Vs, a= 0.3 =

Figure 5.17

Example 5.10

The area of indicator diagram 25 cm2. The swept volume is 0.15 cm2. Calculated the theoretical mean effective pressure. The indicator diagram is drawn to the following scales:

 

1 cm = 1 bar along the pressure axis
1 cm = 0.02 m3 along the volume axis

Spring constant = 1 bar/cm.

Solution

Given: A1 = 25 cm2, Vs = 0.15 cm2, S = 1 bar/cm

Length of indicator diagram,

M.E.P.,

Example 5.11

Find the diagram factor for steam engine with the following data.

Inlet pressure = 10 bar

Back pressure = 1 bar

Expansion ratio = 3

Area of indicator diagram = 12.1 cm2

Length of indicator diagram = 7.5 cm

Pressure scale = 3 bar/cm

Solution

Given: p1 = 10 bar, pb = 1 bar, r = 3, A1 = 12.1 cm2, L1 = 7.5 cm

Theoretical m.c.p.,

Actual m.e.p.,

Diagram factor,

Example 5.12

Determine the brake power of a simple double acting steam engine having 400 mm diameter and 500 mm stroke length operating at 350 rpm. The initial and back pressure of steam are 9.5 bar and 1.5 bar respectively. The mechanical efficiency is 80% and the expansion ratio is 2.5.

Solution

Given: D = 0.4 m, L = 0.5 m, N = 350 rpm, p1 = 9.5 bar, pb = 1.5 bar

Indicated m.c.p.,

 

B.P = IP × ηmech = 423.843 × 0.8 = 339.074 kW

Example 5.13

A double acting steam engine has a cylinder bore 200 mm stroke 300 mm and cut-off takes place at 0.4 stroke. The steam admission and the exhaust pressure are 7 bar and 0.38 bar respectively. If the diagram factor is 0.8, calculate the indicated power at 200 rpm. Neglect clearance and assume hyperbolic expansion.

Solution

Given: D = 0.2 m, L = 0.3 m, = 0.4, p1 = 7 bar, pb = 0.38 bar, Kd = 0.8, N = 200 rpm

Theoretical m.e.p.,

Actual m.e.p.. pma = pmt × Kd = 4.9856 × 0.8 = 3.9885 bar

Power developed,

 

= 25.06 kW

Example 5.14

The following readings were taken during the test at full load on a single cylinder, double acting, condensing type, throttle governed steam engine

Diameter of cylinder = 400 mm

Stroke of engine = 600 mm

Cut-off = 50% of stroke

Pressure of steam supplied = 11 bar

Back pressure = 0.8 bar

Brake wheel diameter = 4.5 m

Net load on the bake = 4900 N

Speed of engine = 150 rpm

Diagram factor = 0.82

Find the indicated power, brake power and mechanical efficiency of the engine.

Solution

Given: D = 0.4 m, L = 0.6 m, = 0.5, p1 = 11 bar, pb = 0.8 bar, Db = 4.5 m, Wnet = 4900 N, N = 150 rpm, Kd = 0.82

Indicated theoretical m.e.p.,

 

= 11 × 0.5 (1 + ln 2) − 0.8 = 8.5123 bar

Actual m.e.p., pma = pml × Kd = 8.5123 × 0.82 = 6.98bar

Indicated Power,

Brake power,

Mechanical efficiency,

5.13 ❐ SATURATION CURVE AND MISSING QUANTITY

Saturation Curve: The saturation curve is the curve showing the volume the steam in the cylinder would occupy, during the expansion stroke if the steam is perfectly dry and saturated at all the points. It is plotted on p-V diagram and the wetness of stream can be seen on it at a glance.

Figure 5.18 shows a calibrated indicator. The total mass of steam in cylinder during expansion stroke = mC + m.

On the expansion curve, consider any point B and read off from the diagram the pressure (pB) and volume (VB) at this point. From steam tables, obtain the specific volume υ of dry steam at pressure pB. The volume the steam at B would occupy if dry saturated = mυ. Let this volume be represented by AC to the volume scale of the p-V diagram. Then, the point C represents the volume the steam at B would occupy if dry saturated.

Following this way, a number of points may be obtained and plotted. The curve passing through these points to known as saturation curve (because all the points on this line represent the condition of steam dry and saturated).

From this saturation curve, the dryness fraction for all points on expansion curve can be obtained. For instance, the dryness fraction at E will be given by

Dryness fraction at

From Fig. 5.18, it can be seen that the steam is wet at the beginning of the expansion stroke and becomes drier towards the end. This is owing to the fact that high pressure steam in the initial stage of expansion is better than the cylinder walls; this causes the steam to condense. During the expansion stroke the steam pressure falls and towards the end of the stroke the walls will be hotter than the steam; consequently the condensed steam re-evaporates and as a result the dryness fraction is improved. The improvement in dryness fraction will not be there if cylinder walls are not jacketed.

Missing quantity: The missing quantity is the horizontal distance between the actual expansion curve and the saturation curve (Fig. 5.18). At a pressure pD, the missing quantity is represented by EF (m3).

Figure 5.18 Calibrated indicator diagram

The missing quantity is mainly due to condensation of the steam, but a small amount will be due to leakage past the piston. Due to this missing quantity there is a loss of work represented by the area between the expansion curve and saturation curve.

The missing quantity can be reduced in the following ways

  1. By steam jacketing the cylinder walls efficiently.
  2. By reducing the temperature range of the steam during the stroke this can be accomplished by compounding the expansion of steam in two cylinders instead of allowing the whole pressure to drop to occur in one cylinder.
5.14 ❐ HEAT BALANCE SHEET

Procedure: For preparing a heat balance sheet for a steam engine cylinder, the engine should be tested over a period of time under conditions of constant load and steam supply. An indicator diagram should be taken and the steam pressure noted at regular intervals of time, an account should also he kept of the steam supply to the jackets. The mass of the steam supplied to the cylinder supplied can be obtained from the steam condensed by the condenser.

Analysis: The heat balance for the steam in engine is more easily drawn up than that of the internal combustion engine. The working fluid in the steam engine does not undergo any chemical change, and consequently changes in properties may be ascertained with reference to an arbitrary datum.

Figure 5.19 shows, diagrammatically, a steam engine and various quantities entering and leaving have been indicated there on. Treating the engine as a flow system as shown in Fig. 5.20, an energy equation may be written as follows:

where m = mass, W = work, h = specific enthalpy, Q = heat transferred, and suffices have the following meanings, s1 = steam supplied, s2 = condensate discharge, c = cooling water, R = radiation.

Figure 5.19 A steam engine

Figure 5.20

The Eq. (5.19) in tabular form is expressed as follows:

Heat in Heat out
In steam ms1 hs1 Heat as shift work W
Heat to condensate ms2 hs2
Heat to coolant mc (hc2hc1)
Heat to radiation QR

The value of enthalpy may be obtained from the steam tables. The heat to coding water is given by mc × cpw (touttin) and the heat to radiation obtained by the difference. In the absence of leakage ms1 = ms2.

The heat balance may therefore be written as follows:

Heat in Heat out
Heat in steam Heat as B.P.
Heat to condensate.
Heat to cooling water.
Heat to radiation, by difference.

It is to be noted that friction terms are absent, since any work done against friction is included in heat to the condensate, cooling water and radiation loss.

5.15 ❐ PERFORMANCE CURVES

Typical performance curves of a reciprocating steam engine under test conditions are shown in Fig. 5.21.

Most of the curves are self-explanatory and should be carefully scrutinised for analysis of the variables involved.

One salient fact is that the speed curve as shown is practically horizontal with a slight drop as the load is increased, this shows good governing. The Willian’s line as shown is straight when the governing is by throttling with a fixed cut-off, and indicates a linear relationship between the total steam consumption is kg/h and the power if steam pressure is varied to suit the load. This fact is also of great advantage in predicting the part load steam consumption of steam turbines, most of which use throttle governing. 1f the steam consumption is plotted for an automatic engine with cut-off governing, the line will not be strictly straight.

Figure 5.21

Example 5.15

In a single cylinder double acting steam engine steam is supplied at a pressure of 12 bar and exhaust takes place at 1.1 bar. The cut off takes place at 10% of the stroke which is equal to 1.25 times the cylinder bore and the engine develops and indicated power of 100 kW at 90 r.p.m., calculated the bore and stroke of the engine assuming hyperbola expansion and a diagram factor of 0.8. Also determine the theoretical steam consumption in m3/mm

Solution

Given: p1 = 12 bar, pb = 1.1 bar,v2 = 0.4v3, L = 1.25D, IP = 100 kW, N = 90 rpm, Kd =0.8

The p-v diagram is shown in Fig. 5.22.

Mean effective pressure without clearance

Expansion ratio,

 

pm = 0.8 [12 × 0.4 (1+ ln 2.5) + 1.1] = 6.4786 bar

Indicated power developed,

Theoretical steam consumption/min

Figure 5.22

Summary for Quick Revision

  1. A steam engine is an external combustion heat engine as the fuel is burnt in the boiler. The steam raised is used to reciprocate the piston in the cylinder for generating power.
  2. The mean effective pressure is the hypothetical constant pressure which will produce same work for the same piston displacement with the variable conditions of temperature, pressure and volume.

    where

    pb= back pressure

    p1= admission pressure of steam

    c = clearance ratio =

  3. Efficiency of simple Rankine cycle,

    where, Pump work, wp = hf 4hf 3 = v f3(p1p2)

  4. Efficiency of modified Rankine cycle
  5. The indicator diagram is a representation of the variation of pressure and volume of steam inside the cylinder on p-v diagram for one complex cycle of operation.
  6. Diagram factor is the ratio of the area of actual indicator diagram to the area of theoretical indicator diagram.
  7. Indicated power, kW for single acting

    kW for double acting

  8. Brake power,

    where D = drop drum dia, dr = rope dia, N = rpm, W = weight on rope, S = spring pull.

  9. Mechanical efficiency,

    Indicated thermal efficiency,

    Brake thermal efficiency,

    Overall efficiency,

  10. In throttle governing, the pressure of steam admitted to the engine is reduced by throttling before it passes into the engine.
  11. In cut-off governing, the period of admission of steam entering the engine cylinder is reduced depending on the load on engine.
  12. Saturation curve is the curve showing the volume the steam in the cylinder would occupy during the expansion stroke if the steam is perfectly dry and saturated at all the points.
  13. Missing quantity is the horizontal distance between the actual expansion curve and the saturation curve.
  14. Heat balance sheet shows the balance between the amount of heat input to the engine and heat utilised for various purposes.

Multiple-choice Questions

  1. Diagram factor is defined as the ratio of
    1. actual m.e.p. and swept volume
    2. theoretical m.e.p. and swept volume
    3. actual m.c.p. and theoretical m.e.p.
    4. theoretical m.e.p. and actual m.e.p.
  2. Choose the quantity which is varied during throttle governing of steam engines
    1. pressure
    2. volume
    3. temperature
    4. dryness fraction
  3. In a compound steam engine
    1. steam expands twice
    2. two engines are combined together
    3. steam expands in several stages
    4. two units are put together
  4. In the tandem compound steam engine, the axes of the two cylinders
    1. are at 90° to each other
    2. are inclined at 45°
    3. are at 0° to each other
    4. lie in different plane
  5. In order to reverse a steam engine, the eccentric to be shifted for an angle advance of 40°is
    1. 180°
    2. 200°C
    3. 160°
    4. 190
  6. A steam engine works on
    1. Carnot cycle
    2. Rankine cycle
    3. modified Rankine cycle
    4. reheat cycle
  7. The cross-head in a steam engine is used to connect
    1. piston rod and connecting rod
    2. valve rod and piston rod
    3. eccentric rod and piston rod
    4. connecting rod and valve rod
  8. A D-slide valve in a steam engine is used to
    1. admit steam into cylinder from steam chest.
    2. exhaust steam from cylinder to condenser
    3. to admit and exhaust steam from cylinder
    4. none of the above.
  9. The cut-off ratio in a steam engine is defined as the ratio of volume at cut off to
    1. swept volume
    2. cylinder volume
    3. clearance volume
    4. none of above.
  10. The clearance ratio in a steam engine is defined as the ratio of clearance volume to
    1. swept volume
    2. cut-off volume
    3. cylinder volume
    4. none of the above
  11. Willian’s line is the relation between steam mass flow late and
    1. indicated power
    2. brake power
    3. compression ratio
    4. thermal efficiency
  12. In receiver type compound steam engine, the cranks are placed at
    1. 90°
    2. 180°
    3. 270°
  13. In a Woolfe compound steam engine, the cranks are placed at
    1. 90°
    2. 180°
    3. 270°
  14. The function of piston in steam engine is to transfer motion in
    1. cross head
    2. connecting rod
    3. D-slide valve
    4. eccentric
  15. The function of eccentric in a steam engine is convert
    1. to-and-fro motion of D-slide valve to rotary motion of crankshaft
    2. oscillatory motion of connecting rod to rotary motion of crankshaft
    3. rotary motion of crankshaft to to-end-fro motion of D-slide valve
    4. rotary motion of crankshaft to oscillatory motion of eccentric rod
  16. The main function of a stuffing box in a steam engine is to
    1. prevent leakage of steam
    2. guide the piston rod
    3. guide the valve rod
    4. receive exhaust steam from engine
  17. The motion to D-slide valve is imparted by
    1. eccentric
    2. cam
    3. flywheel
    4. crank
  18. Rankine efficiency of a steam engine may be in the range of
    1. 15–20%
    2. 25–35%
    3. 70–80%
    4. 90–95%
  19. Rankine cycle comprises of
    1. two isentropic and two reversible constant volume processes
    2. two isentropic and two reversible constant pressure processes
    3. two isothermal and two reversible constant pressure processes
    4. none of the above

Review Questions

  1. Explain the difference between single acting and double acting steam engines
  2. Differentiate between condensing and non-condensing steam engines
  3. What are the functions of the following
    1. cross head
    2. Eccentric
    3. stuffing box
    4. D-slide valve
  4. Explain the following therms:
    1. Cut-off
    2. Release
    3. Dead centres
    4. Back pressure.
  5. Define diagram factor.
  6. Define indicated thermal efficiency and brake thermal efficiency.
  7. List the methods for governing of steam engines.
  8. Compare and contrast the methods for governing of steam engine
  9. How do you determine the brake power of a steam engine.
  10. What is frictional power of a steam engine.

Exercises

5.1 A single cylinder, double acting, condensing type steam engine 0.3 bore and 0.4 m stroke runs at 150 rpm. Steam at 10 bar is supplied upto 40% of the stroke. Back pressure of the engine is 0.35 bar. Neglecting clearance and assuming a diagram factor of 0.8 and a mechanical efficiency of 80% determine the brake power of the engine.

5.2 The following readings were taken from a test on a single cylinder double-acting steam engine having 0.25 m bore and 0.35 m stroke

Pressure of steam supplied = 9.5 bar

Engine speed = 240 rpm

Area of indicator card = 10.5 cm2

Length of indicator card = 8 cm

Spring number = 25 N/cm2/cm

Net load on brake wheel = 1000 N

Diameter of brake wheel = 3 m

Find IP, BP and mechanical efficiency

5.3 A single-cylinder, double-acting, non-condensing steam engine 25 cm bore 50 cm stroke develops 40 kW indicated power at 100 rpm. The clearance is 10% and cut-off takes place at 40% of stroke. The steam pressure at the point of cut-off is 5 bar. The compression starts at 80% of the return stroke. The pressure of steam on compression curve at 90% of the return stroke is 1.4 bar and steam is dry and saturated. Determine the actual and minimum theoretically possible specific steam consumption on indicated power basis.

5.4 Determine the brake power of a single cylinder double acting, condensing type steam engine having 0.3 m bore and 0.6 m stroke, running at 250 rpm. The steam 8 bar dry and saturated is supplied upto 50% stroke. The back pressure is 1 bar. Assume mechanical efficiency 70%, diagram factor = 0.75, and piston rod diameter = 5 cm. Also find the specific steam consumption on indicated power basis if indicated thermal efficiency = 15 %.

5.5 Single cylinder, double acting condensing type steam engine developed 2.10 indicated power at 120 rpm. Dry and saturated steam at 10 bars is supplied for 40% of the stroke. The exhaust pressure is 0.2 bar. Assuming a diagram factor of 0.85 and ratio of stroke to bore as 2 determine the bore and stroke of the engine.

5.6 A single-cylinder, double acting steam engine develops 150 kW indicated power at 240 rpm. The pressure of steam supplied is 10 bar and exhaust pressure is 0.15 bar. The cutoff is at 40 % of the stroke, clearance is 10 % of the stroke and compression starts at 80% of the return stroke. Expansion follows the low: PV1.25 = C, Compression follow law PV1.35 = C, ratio of stroke to bore diagram factor = 0.85. Calculate the bore and stroke of the engine.

5.7 The following data were obtained from the trail on a steam engine governed by throttle governor steam admission pressure = 10.5 bar dry and saturated

Back pressure = 0.2 bar

Steam consumption at no load = 44 kg/h

Indicated power developed at full load = 7 kW

Specific steam consumption at full load = 18 kg/kWh

Find the specific steam consumption and indicated thermal efficiency when the engine develops 5 kW indicated power.

5.8 A single cylinder, double acting steam engine having 25cm bore and 50 cm stroke is governed by cut-off governing. The maximum cut-off possible is 50 % of stroke. The steam is supplied at 8 bar and exhausts at 1 bar. Calculate the percentage decrease in power when the cut-off is reduce to 40% of stroke. Assume that the diagram factor is 0.8 and mechanical efficiency is 80%.

5.9 A single-cylinder, double-acting, steam engine develops 40 kW when running at 120 rpm. The steam is admitted at 5 bar and exhausted at 0.35 bar. Cut-off occurs at 1/3 of stroke. Assuming a diagram factor of 0.75 and mechanical efficiency of 0.85 determine the swept volume cylinder.

5.10 Steam is admitted to an engine for 30% of the stroke with pressure of 7 bar, the law of expansion followed is law pV1.5 = C the compression commences at 60% of return stoke and follows the law F = C. The clearance volume is 20% of the displacement volume and the back pressure is 6.18 mm of Hg vacuum, when barometer reads 760 mm of Hg. Estimate the mean elective pressure and the indicated power of a double acting engine with cylinder diameter 30 cm stroke 45 cm and speed 200 rpm.

ANSWERS TO MULTIPLE-CHOICE QUESTIONS
  1. c
  2. b
  3. d
  4. c
  5. d
  6. b
  7. d
  8. a
  9. a
  10. a
  11. a
  12. b
  13. c
  14. a
  15. c
  16. a
  17. a
  18. b
  19. b