( h) ( ) Effect of Boiler Pressure (Using Molliar Diagram i.e., h-s diagram) We have, but W P << W T. = = h h h h

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1 Effect of Boiler Pressure (Using Molliar Diagram i.e., -s diagram) We ave, ( 3 ) ( 4 ) η t but W P << W T 3 ( ) S η t ( ) i.e., Rankine cycle η depends on, and S. From figure as P > P > P for te fixed maximum temperature of te steam t and condenser pressure P, Isentropic eat drops increases wit boiler pressure i.e., from te figure terefore it is evident tat as boiler pressure increases, te isentropic eat drop ( ) S increases, but te entalpy of te steam entering te turbine decreases, wit te result tat te Rankine η increases. But quality of te steam at te exit of te turbine suffers i.e., x 3 < x 3 < x 3, wic leads to serious wear of te turbine blades. VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS

2 Effect of Super Heating (Using Molliar Diagram i.e., -s diagram) Te moisture in te steam at te end of te expansion may be reduced by increasing te super eated temperature of steam t. Tis can be seen in figure were t > t > t, but x 3 < x 3 < x 3. It is, terefore, natural tat to avoid erosion of te turbine blades, an increase in te boiler pressure must be accompanied by super eating at a iger temperature and since tis raises te mean average temperature at wic eat is transferred to te steam, te Rankine η increases. Deviation of Actual Vapour Power cycles from Ideal cycle Te actual Vapour power cycle differs from te ideal Rankine cycle, as sown in figure, as a result of irreversibilities in various components mainly because of fluid friction and eat loss to te surroundings. VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS

3 Fluid friction causes pressure drops in te boiler, te condenser, and te piping between various components. As a result, steam leaves te boiler at a lower pressure. Also te pressure at te turbine inlet is lower tan tat at te boiler exit due to pressure drop in te connecting pipes. Te pressure drop in te condenser is usually very small. To compensate tese pressure drops, te water must be pumped to sufficiently iger pressure wic requires te larger pump and larger work input to te pump. Te oter major source of irreversibility is te eat loss from te steam to te surroundings as te steam flows troug various components. To maintain te same level of net work output, more eat needs to be transferred to te steam in te boiler to compensate for tese undesired eat losses. As a result, cycle efficiency decreases. As a result of irreversibilities, a pump requires a greater work input, and a turbine produces a smaller work output. Under te ideal conditions, te flow troug tese devices are isentropic. Te deviation of actual pumps and turbines from te isentropic ones can be accounted for by utilizing isentropic efficiencies, defined as WS S 4 η P Wa 4 Wa 3 And η t W Numerical Problems: S 3S. Dry saturated steam at 7.5 bar enters te turbine of a steam power plant and expands to te condenser pressure of 0.75 bar. Determine te Carnot and Rankine cycle efficiencies. Also find te work ratio of te Rankine cycle. Solution: P 7.5 bar P 0.75 bar η Carnot? η Rankine? a) Carnot cycle: At pressure 7.5 bar from steam tables, P t S f fg g S f S fg S g For P 7.5 bar, using linear interpolation VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 3

4 For t S, x C K Similarly, f kj/kg fg kj/kg g 794. kj/kg S f.3844 kj/kg 0 K S fg 4.00 kj/kg 0 K S g kj/kg K Also at pressure 0.75 bar from steam tables P t S f fg g S f S fg S g For 0.75 bar, using linear interpolation, t S C f fg g S f.6 S fg S g T T Te Carnot cycle η, η C T Steam rate or SSC δw W T W P Since te expansion work is isentropic, S S 3 But S S g and S 3 S f3 + x 3 S fg3 i.e., x 3 (6.453) x Entalpy at state 3, 3 f3 + x 3 fg (78.65) 7.63 kj/kg Turbine work or expansion work or positive work kj/kg Again since te compression process is isentropic i.e., S 4 S S f.3844 Hence.3844 S f4 + x 4 S fg4.6 + x 4 (6.453) x Entalpy at state 4 is 4 f4 + x 4 fg4 VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 4

5 (78.65) 8.79 kj/kg Compression work, W P kj/kg 3 SSC.9 x 0 kg / kj δw WT WP 456. work ratio rw ve work W 5.47 T b) Rankine cycle: ( 3 ) ( 4 ) ( ) WT WP η R QH Since te cange in volume of te saturated liquid water during compression from state 4 to state is very small, v 4 may be taken as constant. In a steady flow process, work W -v dp W P S 4 v fp (P P ) ( ) x 0 5 x (/000).737 kj/kg S kj/kg Hence, turbine work W T kJ/kg Heat supplied Q H S kj/kg η R SSC 904 x 0 kg / kj VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 5

6 Work ratio, r w If in problem (), te turbine and te pump ave eac 85% efficiency, find te % reduction in te net work and cycle efficiency for Rankine cycle. Solution: If η P 0.85, η T 0.85 WP.737 WP.0435kJ / kg W T η T W T 0.85 (5.47) kj/kg W net W T W P kj/kg % reduction in work output 5.% W P S 4 S kj/kg Q H S kj/kg η cycle % reduction in cycle efficiency 5.08% 0.6 Note: Alternative metod for problem using -s diagram (Mollier diagram) toug te result may not be as accurate as te analytical solution. Te metod is as follows Since steam is dry saturated at state, locate tis state at te pressure P 7.5 bar on te saturation line and read te entalpy at tis state. Tis will give te value of. As te expansion process -3 is isentropic, draw a vertical line troug te state to meet te pressure line, P 0.75 bar. Te intersection of te vertical line wit te pressure line will fix state 3. From te cart, find te value of 3. Te value of 4 can be found from te steam tables at pressure, P 0.75 bar, as 4 f4. After finding te values of, 3 and 4, apply te equation used in te analytical solution for determining te Rankine cycle η and SSC. VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 6

7 3. Steam enters te turbine of a steam power plant, operating on Rankine cycle, at 0 bar, C. Te condenser pressure is 0. bar. Steam leaving te turbine is 90% dry. Calculate te adiabatic efficiency of te turbine and also te cycle η, neglecting pump work. Solution: P 0 bar t C P 3 0. bar x η t? η cycle? Neglect W P From supereated steam tables, For P 0 bar and t C, 305. kj/kg, s 7.5 kj/kg From table A, For P 3 0. bar t S C f 9.8 fg 39.9 S f S fg Since x 3 0.9, 3 f4 + x 3 fg (39.9) kj/kg Also, since process -3s is isentropic, S S 3S i.e., 7.5 S fg4 + x 3S S fg x 3S (7.508) x 3S S (39.9) kj/kg VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 7

8 Turbine efficiency, η t 3 3S ηcycle W T 3 but QH 9.8 kj/kg i. e., 5% 4. A 40 mw steam plant working on Rankine cycle operates between boiler pressure of 4 MPa and condenser pressure of 0 KPa. Te steam leaves te boiler and enters te steam turbine at C. Te isentropic η of te steam turbine is 85%. Determine (i) te cycle η (ii) te quality of steam from te turbine and (iii) te steam flow rate in kg per our. Consider pump work. Solution: P 4 MPa 40 bar P 3 0 KPa 0. bar P 40000kW t C η t 0.85 η cycle? x 3? m? 35.7kJ kg and s kj/kg-k 0 / 40bar, 400 C f 0.bar 9.8kJ / kg 4 Process -3s is isentropic i.e., S S 3S x 3S (7.508) x 3S VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 8

9 3S f3 + x 3S fg (39.9) 45. kj/kg But η t 3 3S i. e., kj/kg W T kj/kg W P v dp (40 0.) 0 5 / kj/kg kj/kg ( i) η cycle Q W net ( ) 9.9% (ii) x 3? we ave x 3 (39.9) x (iii) P m W i.e., m (905.87) net m 44. kg/s 590 kg/r Ideal Reeat cycle: We know tat, te efficiency of te Rankine cycle could be increased by increasing steam pressure in te boiler and supereating te steam. But tis increases te moisture content of te steam in te lower pressure stages in te turbine, wic may lead to erosion of te turbine blade. Te reeat cycle as been developed to take advantage of te increased pressure of te boiler, avoiding te excessive moisture of te steam in te low pressure stages. In te reeat cycle, steam after partial expansion in te turbine is brougt back to te boiler, reeated by combustion gases and ten fed back to te turbine for furter expansion. VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 9

10 In te reeat cycle te expansion of steam from te initial state () to te condenser pressure is carried out in two or more steps, depending upon te number of reeats used. In te first step, steam expands in HP turbine from state to approximate te saturated vapour line (process -3s). Te steam is ten reeated (or resupereated) at constant pressure in te boiler (or in a reeater) process 3s-4 and te remaining expansion process 4s-5 is carried out in te LP turbine. Note: ) To protect te reeater tubes, steam is not allowed to expand deep into te two-pase region before it is taken for reeating, because in tat case te moisture particles in steam wile evaporating would leave beind solid deposits in te form of scale wic is difficult to remove. Also a low reeat pressure may bring down T m and ence cycle η. Again a ig reeat pressure increases te moisture content at turbine exaust. Tus reeat pressure is optimized. Optimum reeat pressure is about 0. to 0.5 of initial pressure. VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 0

11 We ave for kg of steam Q H ( S ) + ( 4 3S ); Q L 5S 6 W T ( 3S ) + ( 4 5S ); W P S 6 W W T P η R ; QH Steam rate 3600 ( W W ) T P kg / kw Since iger reeat pressure is used, W P work is appreciable. ) In practice, te use of reeat gives a marginal increase in cycle η, but it increases te net work output by making possible te use of iger pressures, keeping te quality of steam at turbine exaust witin a permissible limit. Te quality improves from x to x 5 S 5S by te use of reeat. VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS