40.1. GOVERNING OF SIMPLE STEAM ENGINE
The purpose of governing of simple steam engine is to maintain the speed of an engine fairly constant irrespective of load. Commonly used methods employed for the governing of the simple steam engines are as following.
(a) By throttling the steam.
(b) By varying cut-off point of the steam
40.1.1. By throttling the steam:
In this, governing is achieved by throttling the steam with the help of centrifugal governor which is mounted on steam engine and gets drive from steam engine shaft through belt and pulley arrangement (Fig. 40.1).
Suppose the engine is running at its rated speed and load and, corresponding to this the area ACRKF on the p-v diagram is shown in Fig. 40.2. Now suppose the load on the engine decreases, then the power developed (area ACRKF on the p-v diagram)in the engine will be more than the demand and, as a consequence of this the engine speed will increase. As thefly-balls revolve with a spindle, which is driven by the engine through bevel gears, the increase in engine speed will result in the increase of centrifugal force on the fly-balls and hence fly-balls moves outwards and upwards. The sleeve which is connected to the fly-balls through links and can only slide up and down on spindle also rises upwards with fly-balls. Finally, the upward movement of the sleeve partly closes the throttling valveat the other end of the bell crank lever and. This throttles the steam from pressure pato a suitable lower pressure pa',as shown in p-v diagram in Fig. 40.2,so that the heat drops and load for doing work is reduced from (h1 –h2) to (h1' –h2'),as shown in h-s diagram,and consequently, the work developed in the engine is reduced from ACRKF to A'C'R'KF which is just equal to the demanded load. Under this condition the speed of the engine drops to the normal value required and the speed is maintained constant.If the load is still further reduced, the steam is throttled to the required, lower pressure pa'', so that, the power developed is equal the demand, and the speed is maintained constant. With increase in load reverse action takes place.
Fig. 40.1. Centrifugal Governor.
Fig. 40.2. p-v and T-s diagram of throttle governing
During throttling process:
(i) Steam cut-off point ‘C’ remains constant,
(ii) there is no change in condenser vacuum ‘pb’,
(iii) expansion‘C-R’ follows the hyperbolic law,
(iv) engine runs at constant speed and
(v) the enthalpy of steam 1-1′ or 1-1′′ remains constant.
This type of governing is simple and low in cost but it has the disadvantage of low thermal efficiency at part load. It is suitable for small engines where engine efficiency is of no importance.
In Fig. 40.3, the steam consumption has been plotted against the indicated power of the engine; a straight line graph is obtained. This line is known as Willan’s line. Thus showing the directly proportionallybetween the steam consumption and indicated power.
It is expressed as
M = A + B . (i.p.) ……………. (40.1)
Where M is steam consumption in kg/h. A and B are constants of proportionality.
The equation (40.1) may be derived in the following manner We know that the mean effective pressure neglecting clearance and compression is given by pim = where, p is admission pressure at which steam is admitted into cylinder and pb is back pressure. |
Fig. 40.3. Willian’s Line |
and i.p. for double acting steam engine is given by i.p. = 
or i.p. = k1 p - k2 …………………………………….. (40.2)
where k1 = 
= constant and k2 = 
= constant
Due to throttling, dry and saturated steam becomes superheated. Therefore, gas laws can be applied. This fact may be verified using steam tables. Thus at cut-off point
pv = constant or v = 
Where p and v are pressure and specific volume of admitted steam respectively.
and V= volume of steam admitted per stroke
Mass of steam admitted per stroke, 
For throttling governing, at all admission pressures (pa, pa' and pa'') volume of steam admitted per stroke is also constant (i.e. Vc = Vc’ = Vc’’) and hence we may write.
ma = p x constant
Now, the mass of steam consumed per hour for a double acting steam engine will be
M = ma x 2N x 60 , kg/h = 2N x p x constant = k3 p
or 
where, k3 is another constant
From eqn (40.2), 
From above two eqn. , we get 
or M = 
or M = A + B (i.p.)
Where, A and B are constants.
40.1.2. Cut-off Governing:
In the cut-off governing, the governing is achieved by varying the amount of steam supplied to the engine i.e. by varying the steam cut-off point.
Suppose an engine is working at its rated load and speed and the power developed is given by the indicator diagram area ACRKF with steam cut-off the point at ‘C’ as shown in Fig. 40.4.Now suppose the load on the engine is reduced, then the power developed (area ACRKF on the p-v diagram) in the engine will be more than the demand and, as a consequence of this the engine speed will increase. To maintain constant speed, the cut-off point of steam is varied to a suitable point C', so that, the steam consumption is reduced causing the power developed shown by area AC'R'KF is just equal the load demanded. Under this condition the speed will decrease to the normal value, and thus, by this method the speed is maintained constant. If the load is further reduced, then the cut-off is varied to another suitable point C'', so that, the power developed is just equal to the load demanded. If the load increases, the reverse action takes place.
In cut-off governing, the admission pressure remains constant. The variation of the rate of steam consumption against the i.p. developed is found as follows: Neglecting clearance and compression effects, the mean effective pressure, pim = where p is admission pressure at which steam is admitted into cylinder and pb is back pressure.
In the cut-off governing, p and pb remain constant. Hence pim = |
Fig. 40.4. Cut-off Governing |
Again, i.p. for double acting steam engine = 
But 2 LAN is a constant for an engine.
………………………………..(40.3)
The mass of steam consumed per hour for a double acting steam engine will be
M = ma x 2N x 60 kg/h
where Mass of steam admitted per stroke, 
where V is volume of steam admitted per stroke and not constant for cut-of governing (i.e. Vc ≠ Vc’ ≠ Vc’’).
v = specific volume of admitted steam and a constant for cut-of governing (i.e. vc = vc’ = vc’’ ).
………………………………..(40.4)
Note that the volume of steam admitted per stroke not constant in Cut- off governing but varies with 
But the swept volume Vr is constant hence
Putting the value of V in equation (40.4), we have
( as N = constant)
or 
………………………………..(40.5)
From equations (40.3) and (40.5), we get
or 
or 
………………………………..(40.6)
where,
The equations (40.3) and (40.4) show the variation of steam rate consumption (M) with expansion ratio (r) and indicated power (i.p.).This variation is also plotted in Fig. 40.5. A smooth curve is obtained for steam consumption versus indicated power plot. A straight line can be obtained if the ratio of expansion is constant; this condition is not fulfilled in a cut off governed engine as it is fulfilled by throttled governed engine. |
Fig. 40.5. Variation of M with i.p. and r |
In cut-off governing, the part load performance is better, but to meet very light loads, the cut-off has to be extremely small and this promotes condensation. From experiment, it has been found that if the cut-off is reduced below 10%,(as shown by the dotted line C′′′R′′′ in Fig. 40.4).then towards the end of expansion R′′′D, the piston has to do work on the steam.
If the cut-off is 100% (as shown by the dotted line DC′′′′ in Fig. (40.4), then there is no time left for expansion.
40.2. PERFORMANCE COMPARISON OF NON-EXPANSION ENGINES AND EXPANSION ENGINES
Fig. 40.6 shows the indicated diagram of non expansion and expansion steam engines.
The area ACRDF is the hypothetical indicated work when the cut-off occur at 50% of the stroke (expansion engines) and area AC’RDF is the hypothetical indicated work thewhen cut-off is at 100% i.e., the steam is admitted throughout the stroke length (non expansion engines). In case of non expansion engines, the increase in work-done is given by the area CC’R as compare to expansion engines whereas the steam consumption of non expansion engines is twice the expansion engines i.e. in case of non expansion engines, the increase in work-done is not proportional to the increase in steam consumption. So, non expansion engines are not at all economical as work done per kg of steam consumption is more as compare to expansion engines. However, non-expansive engines are used in rolling mill where large and intermittent power is needed.
The effect of cut-off on power, steam consumption and work-done per kg of steam consumption is also shown in Fig. 40.7.
Fig. 40.6. Indicated diagram of steam engine. |
Fig. 40.7. Performance of steam Engine |
Problem 40.1: A throttle governed steam engine consumes 350 kg of steam per hour when delivering i.p. 20 kW. The steam consumption at no load condition is 60 kg/h. Calculate the specific steam consumption per i.p. when the engine develops i.p. 12 kW.
If the steam is supplied to the engine is at 11 bar dry and saturated and the back pressure is 0.3 bar, calculate the indicated and brake thermal efficiencies of the engine assuming 90% mechanical efficiency.
Solution:
Given: Steam consumption per hour, M1 = 350 kg/h when delivering Indicated power, i.p.1 = 20 kW;
Steam consumption per hour, M2 = 60 kg/h when delivering Indicated power, i.p.2 = 0 kW;
Indicated power, i.p. = 12 kW;
Admission pressure, pa = 11 bar;
Back pressure, pb = 0.3 bar;
ηmech = 90%
(a) Determine the specific steam consumption per i.p. when the engine develops i.p. 12 kW;
Formula: Specific steam consumption per i.p.
s.s.c. = 
= 
s.s.c. when the engine develops i.p. 12 kW = 
Finding unknown, M(12) ;
We have from Willian’s line, M(i.p.) = A + B (i.p.) ………. (40.7)
Where, M(i.p.) is the steam consumption per hour when delivering i.p. indicated power and
A and B are constants.
Finding constants ‘A’ and ‘B’
Steam consumption per hour, M1 = 350 kg/h when delivering Indicated power, i.p.1 = 20 kW, then from Willian’s line, we have
350 = A + 20 x B
Steam consumption per hour, M2 = 60 kg/h when delivering Indicated power, i.p.1 = 0 kW, then from Willian’s line, we have
60 = A + 0 x B
By solving above two equations, we have,
A = 60 and B = 
= 14.5
By substituting constants A and B in equation (40.7), we have
M(i.p.) = 60 + 14.5 (i.p.)
M(12) = 60 + 14.5 (i.p.)
= 60 + x 12 = 234 kg/h
Answer: Specific steam consumption per i.p. when the engine develops i.p. 12 kW
s.s.c. = 
= 19.5 kg/kw.h
(b) Determine the indicated thermal efficiencies of the engine ηi,th ;
Formula: Indicated thermal efficency (ηith) = 
Finding unknown, 
From steam tables (saturated steam) , we have
At 11 bar → hg,11bar = 2797.7 kJ/kg
At 0.3 bar → hf,0.3bar = 289.3 kJ/kg
Answer: Indicated thermal efficiency, ηi,th = 
= 
= 7.35 %
(c) Determine the brake thermal efficiencies of the engine assuming 90% mechanical efficiency ηb,th .
Formula: Brake thermal efficiency, ηb,th = ηi,th x ηmech
Answer: Brake thermal efficiency, ηb,th = ηi,th x ηmech
= 7.35 x 0.9 = 6.62%
40.3. HEAT BALANCE SHEET
Refer Fig. 40.8. In order to prepare a heat balance sheet for steam engine cylinder, the engine should be tested over a period of time under the conditions of constant load and steam supply. The following heat quantities should he included in the heat balance:
Fig. 40.8. Heat balancing of steam engine
Heat in steam supplied to engine, Qsteam
Heat supplied to engine = 
hsteam , kW ……………….. (40.1)
where 
= mass of steam supplied to the cylinder obtained from the steam condensed by the condenser, kg/s
= mass of steam supplied to steam jacket measured from the jacket drains, kg/s
hsteam = total heat of steam supplied, kJ/kg (from steam table with respect to pressure of steam and condition (dry, superheated or wet) of steam entering the cylinder)
Heat energy converted into indicated power of the engine, Qi.p.
The heat energy converted into indicated power should be obtained from the indicator diagram by the following equation given in lesson 36.
For
- Single acting engine,
Qi.p. = I.P. = 
, kW
- Double acting engine having mean effective pressure and effective piston area different on head and crank end sides of the piston,
Qi.p. = I.P. = 
, kW
- Double acting engine having mean effective pressure and effective piston area same on head and crank end sides of the piston
Qi.p. = I.P. = 
, kW ……………….. (40.2)
Heat energy rejected in exhaust steam, Qexhaust
This is obtained from condenser and hot well
Heat rejected to exhaust steam,
Qexhaust = Heat given to condensing water + Heat to hot well
= mc .Cp (t2 - t1) + mcyl hwell , kW .....................................………(40.3)
Where mc = mass of circulating cooling water in the condenser, kg/s
Cp = specific heat of cooling water, kJ/kg K
(t2 - t1) = temperature rise of cooling water in condenser; K
hwell = total heat of condensed steam in hot well, kJ/kg (from steam table with respect to condenser pressure and saturated water condition of steam)
Heat energy rejected in steam jacket drain, Qjacket
Heat drained from steam jacket,
Qjacket = mjacket x hjacket , kW ................. .................... (40.4)
where mjacket = mass of steam in jacket, kg/s
hjacket = total heat of water drained from the steam jackets at the temperature of the steam in the steam jacket
Heat unaccounted, Q.unaccounted
Heat lost by radiation and other unaccounted for
Q.unaccounted = Qsteam – [Qi.p. + Qexhaust + Qjacket] ; kW
Heat balance sheet
Heat balance sheet may be put in tabular form as : —
40.4. PERFORMANCE OF STEAM ENGINES Fig. 40.8 shows the performance of steam engine. The diagram shows that specific steam consumption decreases with increasing load to the point of maximum thermal efficiency. In the lower range of load the mechanical efficiency increases faster but in the middle portion of the load i.e. (40-90% load) the increase is slower and at about 90 to 100% load, the curve is more or less flat. The same trend found in the case of thermal efficiency.
|
Fig. 40.8. Performance of steam engine. |
Problem: 40.2: The following data were obtained during a test on a single cylinder double acting steam engine:-
Engine stroke = 30 cm Piston diameter = 22 cm Piston rod diameter = 5 cm | Coal consumption = 0.4 kg/minute Calorific value of coal (CV)= 35170 kJ/kg of coal |
Mean speed, N = 124 rpm | Steam consumption = 4 kg/min |
Indicator area of cover end = 7 cm2 Indicator area of crank end = 8 cm2 | Steam pressure = 3 bar (dry and saturated) |
Length of indicator card = 7 cm Strength of indicator spring = 2 bar/cm of compression. | Condition of steam at the exit of steam engine, x2 = 0.939 Condenser pressure = 1 bar |
Load on brake = 1250 N Spring balance reading = 70 N | Cooling water flow rate = 51 kg/minute Rise in temperature of cooling water (t2-t1) = 20°C |
Radius of brake wheel = 50 cm Radius of rope = 2 cm |
|
Calculate the i.p., b.p., mechanical, thermal, Rankine, relative and overall efficiencies. Also draw up a heat balance sheet for engine cylinder.
Solution:
Given: Engine type = Single cylinder and double acting
Engine stroke, L = 30 cm = 0.3m
Piston diameter, D = 22 cm= 0.22
Piston rod diameter, d = 5 cm= 0.05m
Mean speed, N = 124 rpm
Indicator area of cover end = 7 cm2
Indicator area of crank end = 8 cm2
Length of indicator card = 7 cm
Strength of indicator spring = 2 bar/cm of compression.
Load on brake, W = 1250 N
Spring balance reading, S = 70 N
Radius of brake wheel, R = 50 cm = 0.5m
Radius of rope, r = 2 cm = 0.02m
Coal consumption, mf = 0.4 kg/minute
Calorific value of coal (C.V.) = 35170 kJ/kg of coal
Steam consumption, mcyl = 4 kg/min
Cooling water flow rate, mc = 51 kg/minute
Rise in temperature of cooling water, (t2-t1) = 20°C
Steam pressure, p1 = 3 bar (dry and saturated)
From Steam table (Dry saturated steam)
At pressure, p1 = 3 bar → hg = 2724.7 kJ/kg,
Condition of steam at the exit of steam engine, x2 = 0.939
Condenser pressure, p2 = p3 = 1 bar
From Steam table (Dry saturated steam)
At pressure, p2 = p3 = 1 bar → hg = 2675.4 kJ/kg; hf= 417.15 kJ/kg ; hfg = 2257.9 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) = 
= 
= 2 bar
Actual mean effective pressure for the crank end is
pam,2 (crank end) = 
= 
= 2.285 bar
Effective area on which pressure acts on the cover end side is given by
A1 (cover end) =
= 0.03803 m2
Effective area on which pressure acts on the crank end side is
A2 (crank end) =
= 0.03606 m2
Answer: Total indicated power developed by the engine,
i.p. = 
= 
= 4.71 + 5.107 = 9.817 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 Reff
= (W - S) x (R + r)
= (1250 - 70) x (0.5 + 0.002) = 613.6 N-m
Answer: Therefore, Brake power (bp) = 
= 7.967 (kW or kJ/S)
(c) Determine mechanical efficiency, ηm:
Formula: ηm = 
Answer: Mechanical efficiency, ηm = 
× 100 = 81.15%
(d) Determine indicated thermal efficiency:
Formula: The indicated thermal efficiency is 
= 
Answer: Indicated thermal efficiency, ηi,th =
= 
= 6.38%
(e) Determine brake thermal efficiency:
Formula: The brake thermal efficiency is ηb,th = 
Answer: The brake thermal efficiency, ηb,th =
=
= 5.17%
(f) Determine Rankine thermal efficiency:
Formula: The Rankine thermal efficiency is 
= 
Finding unknown, 
;
After expansion in the steam engine, the condition of steam at the exit of steam engine is x2 = 0.939 :
For wet steam, we have = hf,1 bar + x2 hfg, 1bar
= 417.15 + 0.939 (2257.9)
= 2537.3 kJ/kg
Answer: Rankine efficiency, ηth = 
= 
= 9.21%
(g) Determine Relative efficiency:
Formula: Relative efficiency = 
Answer: Relative efficiency = = 
= 9.27%
(h) Determine overall efficiency
Formula: Overall efficiency (ηo) = 
Answer: Overall efficiency (ηo) = = 
= 3.39%
(i) Heat balance sheet for engine cylinder.
Heat supplied in steam, Qsteam = mcyl x hg,3bar = 
= 181.64 kW
Heat equivalent of i.p., Qip= 9.817 kW
Heat rejected to cooling water = mc cp (t2-t1) = 
= 71.06 kW
Heat to hot-well = mcyl x hf,1bar = 
= 27.83 kW
Heat rejected in exhaust steam, Qexhaust = heat to cooling water + heat to hot well
= 71.6 + 27.83 = 98.8 kW
Heat unaccounted, Q.unaccounted = Qsteam – [Qip + Qexhaust]
= 181.64 – [9.817+98.8] = 72.93
Heat balance sheet may be put in tabular form as : —
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