558 ENGINEERING THERMODYNAMICS

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\M-therm\Th12-1.pm5

= area l-1-2-3-m

= area ‘o-l- 1 -n’ + area ‘1-2-q-n’ – area ‘o-m-3-q’

= p 1 v 1 + (u 1 – u 2 ) – p 3 v 2

Heat supplied = h 1 – hf 3

∴ The modified Rankine efficiency

=

Work done

Heat supplied

=

pv u u pv

hhf

11 1 2 3 2

(^13)
+− −
−
bg
...(12.6)
Alternative method for finding modified Rankine efficiency :
Work done during the cycle/kg of steam
= area ‘l-1-2-3-m’
= area ‘l-1-2-s’ + area ‘s-2-3-m’
= (h 1 – h 2 ) + (p 2 – p 3 ) v 2
Heat supplied = h 1 – hf 3
Modified Rankine efficiency = Work done
Heat supplied

hh p pv
hhf
12 2 32
(^13)
−+ −
−
bgbg
...(12.7)
Note. Modified Rankine cycle is used for ‘reciprocating steam engines’ because stroke length and hence
cylinder size is reduced with the sacrifice of practically a quite negligible amount of work done.
+Example 12.10. (Modified Rankine Cycle). Steam at a pressure of 15 bar and 300°C
is delivered to the throttle of an engine. The steam expands to 2 bar when release occurs. The
steam exhaust takes place at 1.1 bar. A performance test gave the result of the specific steam
consumption of 12.8 kg/kWh and a mechanical efficiency of 80 per cent. Determine :
(i)Ideal work or the modified Rankine engine work per kg.
(ii)Efficiency of the modified Rankine engine or ideal thermal efficiency.
(iii)The indicated and brake work per kg.
(iv)The brake thermal efficiency.
(v)The relative efficiency on the basis of indicated work and brake work.
Solution. Fig. 12.13 shows the p-v and T-s diagrams for modified Rankine cycle.
From steam tables :

1. At 15 bar, 300°C : h 1 = 3037.6 kJ/kg, v 1 = 0.169 m^3 /kg,
   s 1 = 6.918 kJ/kg K.


2. At 2 bar : ts 2 = 120.2°C, hf 2 = 504.7 kJ/kg, hfg 2 = 2201.6 kJ/kg,
   sf 2 = 1.5301 kJ/kg K, sfg 2 = 5.5967 kJ/kg K,
   vf 2 = 0.00106 m^3 /kg, vg 2 = 0.885 m^3 /kg.


3. At 1.1 bar : ts 3 = 102.3°C, hf 3 = 428.8 kJ/kg, hfg 3 = 2250.8 kJ/kg,
   sf 3 = 1.333 kJ/kg K, sfg 3 = 5.9947 kJ/kg K,
   vf 3 = 0.001 m^3 /kg, vg 3 = 1.549 m^3 /kg.