Modified Rankine cycle
1 Answer

Figures show the modified Rankine cycle on p-V and T-s diagrams (neglecting pump work) respectively. It will be noted that p-V diagram is very narrow at the toe i.e., point ‘2′’ and the work obtained near to e is very small. In fact this work is too inadequate to overcome friction (due to reciprocating parts) even. Therefore, the adiabatic is terminated at ‘2’; the pressure drop decreases suddenly whilst the volume remains constant. This operation is represented by the line 2-3. By this doing the stroke length is reduced; in other words the cylinder dimensions reduce but at the expense of small loss of work (area 2-3-2′) which, however, is negligibly small. The work done during the modified Rankine cycle can be calculated in the following way:

Let p1, v1, u1 and h1 correspond to initial condition of steam at ‘1’.

p2, v2, u2 and h2 correspond to condition of steam at ‘2’.

p3, h3 correspond to condition of steam at ‘3′’.

Work done during the cycle/kg of steam

=area l-1-2-3-m

=area'0-l-1-n' + area'1-2-q-n'- area'0-m-3-q'

=$p_1 u_1+(u_1-u_2 )-p_3 u_2$

Heat supplied=$h_1-h_(f_3 )$

The modified Rankine efficiency

=$\frac{(work done)}{(heat supplied)}$

=$\frac{(p_1 u_1+(u_1-u_2 )-p_3 u_2)}{(h_1-h_(f_3 )}$

Alternative method for finding modified Rankine efficiency:

Work done during the cycle/kg of steam

=area ' l-1-2-3-m'


=$(h_1-h_2 )+(p_2-p_3 ) u_2$

Heat supplied

=$h_1-h_(f_3 )$

Modified Rankine efficiency =$\frac{(work done)}{(heat supplied)}$

=$\frac{ (h_1-h_2 )+(p_2-p_3 ) u_2)}{(h_1-h_(f_3 )}$

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