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LESSON - 38 MASS OF STEAM IN THE CYLINDER
38.1. MASS OF STEAM IN THE CYLINDER DURING EXPANSION STROKE
The mass of steam in the cylinder at any point between cut-off ‘C’ and release ‘R’ is
m = ma + mc
where
ma = mass of the steam admitted into the cylinder from point of admission to point of cut-off per power cycle of engine (during process A-C as shown in Fig. 38.1).
mc = mass of steam left behind in the clearance space from the previous stroke i.e. cushion steam in clearance space (mass of steam between point K and point A).
Mass of steam admitted into the cylinder per power cycle (ma) can be calculated from
……………..(38.1)
where, M is mass of steam admitted per min (kg/min) also called actual steam consumption in steam engine.
n is number of power cycle per minute
For double acting engine, n = 2N and for single acting, n = N
Where, N = number of crankshaft revolutions per minute (rpm)
The actual steam consumption (M) can be determined experimentally by one of following methods:
i) Recording the quantity of feed water delivered to the boiler
ii) The quantity of steam condensed in the condenser
iii) Measuring the amount of steam supplied to the engine with the help of an orifice-meter.
Mass of cushion steam (mc ) can be calculated from any point on the compression curve ‘K-A’ of the indicated diagram. To do this it is first necessary to calibrate the pressure line and volume line of the indicator diagram so that the absolute pressure and volume of any point may be read off.
Calibration of indicator diagram:
Pressure line calibration:
Let ‘de’ represents the atmospheric pressure line on indicator diagram.
Let S be pressure in bar per cm of vertical ordinate.
Pressure scale = of vertical ordinate
The distance ‘od’ represents the atmospheric pressure to this pressure scale;
The line ‘ab’ is drawn at distance ‘od’ from atmospheric pressure line to represent the absolute zero pressure.
where, od = cm
The pressure may now be marked on the vertical ordinate to the scale of 1 cm = S bar absolute pressure, commencing with its zero at ‘o’.
Volume line calibration: The length ‘ab’ of the indicator diagram represents the stroke volume, hence ‘ab’ = Where, D is the diameter of the piston in meter L is the stroke in meter. Then volume scale = The distance ‘oa’ represents the clearance volume to this volume scale; The line ‘od’ is drawn at distance ‘oa’ from ‘aA’ to represent the absolute zero volume. |
Fig. 38.1. Mass of steam in cylinder |
where oa = , cm
The volume may now be marked on the horizontal line ‘ob’ to the volume scale, commencing with its zero at ‘o’.
The absolute pressure and volume at any point may now be determined from the calibrated p-V diagram.
Mass of cushion steam (mc) calculation from calibrated indicator diagram
The cushion steam is trapped in the cylinder during the compression stroke therefore the mass of cushion steam is calculated from the process ‘KA’ on the indicator diagram.
Select a point ‘f’ on the compression curve, towards the end of the stroke where the steam may be assumed to be just dry because compression tends to dry the steam. Read off the pressure and volume of the steam at point ‘f’ from the calibrated p-v diagram.
Let pf = absolute pressure of steam at ‘f’ , bar
Vf = volume of steam at ‘f’ , m3
vf = specific volume of dry steam corresponding to pressure pf, m3/kg (from steam table)
Mass of steam in cylinder at any point between cutoff and release is
(kg)
38.2. SPECIFIC STEAM CONSUMPTION (SSC):
The amount of steam consumed per unit power output is called the specific steam consumption. It is usually expressed in kg per kW hr. It is also called steam rate.
The specific steam consumption may be expressed on the basis of either i.p. or b.p. If specific steam consumption is based on i.p., it is known as indicated specific steam consumption. Similarly, if b.p. is taken as basis, it is known as brake specific steam consumption.
The mass of steam consumed is determined from eqn (38.1),
M = ma . n (kg/min)
= ma . n . 60 (kg/hr)
= ma . 2N . 60 (for double acting)
= ma . N . 60 (for single acting)
Problem 38.1: Estimate the specific Steam consumption in kg per i.p. per hour of a double acting engine from the following data:-
Cylinder diameter 60 cm, stroke 90 cm, rpm 88, admission pressure 8 bar, back pressure 1.8 bar, cut-off occurs at 20% of the stroke for both ends. The specific volume of steam at the point of cut off is 0.24 m3/kg. The diagram factor may be assumed equal to 0.8.
Solution: Given: Steam engine = double acting engine Diameter of cylinder, d= 60 cm = 0.6 m; Stroke length, L = 90 cm = 0.9 m; Speed, N = 88 rpm; Admission pressure, pa = 8 bar; Exhaust/Back pressure, pb = 1.8 bar; Cut off = 20% of the stroke or Volume at cut-off = Vc = 0.2V i.e., |
|
Therefore, ratio of expansion = 5
Specific volume at the point of cut-off = vc = 0.24 m3/kg
Diagram factor, k = 0.8
Determine the specific Steam consumption in kg per i.p. per hour of a double acting engine
Formula: Specific steam consumption = , kg/kWh
Finding unknown, i.p. and M ;
Finding unknown, i.p.;
Indicated power of double acting steam engine, i.p. = , kW
= 2 ×
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 =
= 2.375 bar
pam (actual) = k ptm (theoretical) = 0.8 x 2.375 = 1.9 bar
i.p. = 2 ×
= 2 ×
= 141.82 kW
Finding unknown, M;
We have, Volume at cut-off = Vc = 0.2V =
= 0.2 x (π/4) (0.6)2 x 0.9 = 0.05089 m3
Mass of steam admitted per stroke, ma = =
= 0.212 kg/cycle
Steam consumption per hour, M = ma . 2N . 60
= 0.212 x 2 x 88 x 60 = 2238.72 kg/h
Answer: Specific steam consumption = = 15.78 kg/kWh
Problem 38.2: Dry and saturated steam at 15 bar is admitted into a double acting steam engine. The steam exhausts at 0.4 bar. The original cut-off was 40% of the stroke. In order to effect a saving in economy it was decided that cut-off should occur earlier at 20% of the stroke. The power output was kept constant by increasing the speed. If diagram factor in both cases remains to be the same calculate (a) the ratio of new original speed and (b) saving in steam consumption.
Solution:
Given: Steam engine = double acting engine
Admission pressure, pa = 15 bar;
Exhaust/Back pressure, pb = 0.4 bar;
Original cut off = 40% of the stroke
i.e. Volume at 40 cut off, VC,40% = 0.4Vs
New cut off = 20% of the stroke
i.e. Volume at 20 cut off, VC,20% = 0.2Vs
Diagram factor for 40% (original) cut off = Diagram factor for 20% (new) cut off = k
Assume, Compression and Clearance = Neglected
(a) Determine the ratio of new and original speed, :
Formula: Since power developed remains the same for 40% (original) cut off and 20% (new) cut off,
i.e. i.p. 40% = i.p. 20%
Hence, 2 × = 2 ×
or
Finding unknown, pam,40%:
The actual mean effective pressure is given by
pam,40% (actual) = (k.ptm)40% = k x 11.09
Finding unknown, ptm,40% :
The theoretical mean effective pressure is given by
Finding unknown, r40%:
When cut-off occurs at 40% of the stroke,
i.e., =
= 0.40
Therefore, ratio of expansion r40% = = 2.5
Therefore, ptm, 40% = = 11.09 bar
pam,40% (actual) = (k.ptm)40% = k x 11.09
Finding unknown, pam,20%:
The actual mean effective pressure is given by
pam,20% (actual) = k.(ptm)20%
Finding unknown, ptm,20% :
The theoretical mean effective pressure is given by
ptm, 20% =
Finding unknown, r20% :
When cut-off occurs at 20% of the stroke,
i.e., = r20%′ = 0.20
Therefore, ratio of expansion r20% = = 5
Therefore, ptm, 20% = = 7.428 bar
pam,20% (actual) = (k.ptm)20% = k x 7.428
Answer: = 1.492
(b) Saving in steam consumption, S%:
Formula: Percentage saving in steam consumption is
Where, M40% is mass of steam supplied per hour when cut- takes place at 40% of the stroke is
and M20% is mass of steam supplied per hour when cut- takes place at 20% of the stroke is
Answer: Percentage saving in steam consumption is
× 100
= × 100 = 25.4%