Thermodynamics II By: Engr. Lester Alfred M. Olasiman, M.E. The Regenerative Cycle The thermal efficiency of a simple power plant is less than fifty percent (50%). This means that more than half of the heat added to the water in the boiler is just wasted and rejected in the condenser. To utilize some of these heats that would have been wasted and rejected in the condenser, part of the throttle steam is extracted for feedwater heating after it has partially expanded in the turbine. The extraction points occur near the saturation state. This process of reheating/heating feedwater is called regeneration. Effects of Regenerative feedwater heating Increase in thermal efficiency. E=WNET/Q . Two ways of increasing the thermal efficiency are (a) by increasing the net cycle work and (b) by reducing the heat supplied, QA’ . The temperature of feedwater entering the boiler in the regenerative cycle is higher than that of the original Rankine cycle. Since the feedwater enters the boiler at a relative high temperature, a smaller quantity of heat is needed to transform it to steam than without the regenerative feedwater heating. This in effect tend to increase the thermal efficiency. Effects of Regenerative feedwater heating • Increase in thermal efficiency It is true that the net work done per kilogram of the throttle steam in the regenerative cycle is less than that of the Rankine cycle as the consequence of the extraction of steam for feedwater heating. This tends to decrease the thermal efficiency. But the rate of decreased in the heat supplied, QA’ is faster than the reduction rate in the net cycle work, WNET. Therefore, the net result of this is an increase in thermal efficiency. Effects of Regenerative feedwater heating • Decrease in the moisture content during the later stages of expansion. It is a fact that the quality of exhaust steam for both cycles are the same, i.e., x2 (Rankine cycle) = x3 (Regenerative cycle). But the quantity of exhaust steam decreases in the regenerative cycle as the result of the bleeding process. Therefore, the moisture content decreases. Open Feedwater Heater An open (or direct-contact) feedwater heater is basically a mixing chamber, where the steam extracted from the turbine mixes with the feedwater exiting the pump. Ideally, the mixture leaves the heater as a saturated liquid at the heater pressure. The schematic of a steam power plant with one open feedwater heater (also called single-stage regenerative cycle) and the T-s diagram of the cycle are shown in figure (next slide) Plant Layout of Regenerative Cycle With One Stage of Extraction for Feedwater Heating (Open heater) T – s Diagram Point 1 is at SH “m” can be determined By mass & energy balance At OH π 1 = π 2 = π 3 Considering isentropic expansion at Point 2 mass turbine extraction considering Mass total is 1 (100%) And mass extraction “m” Mass entering the condenser Is 1 – m Point 6 is where mass extracted ”m” And mass remain at condenser (1 – m) Will combined as 1 (mass total) At point 2 Considering mass extraction as “m” At point 1 Considering SH At table 3 At point 7 Considering the Equation to determine h7 At point 2 Considering properties of steam at pressure of OH Find the value of “ππ ” At table 2 At point 6 Considering OH P6 = POH h6 = hf at POH At point 3 Considering properties of steam Find the value of “π₯2 ” At table 2 At point 5 Considering the Equation to determine h5 At point 3 Considering the remaining mass after extraction “1 – m” At point 4 Considering constant pressure P3=P4 h4 = hf at Pcon Energy Balance at Boiler Ein = Eout • QA + h6 = h1 • QA = h1 – h6 But h6 = h5 + Wp2 Therefore, QA = (h1 – h5 – Wp2 ) mt Energy Balance at OH • Basis: 1kg of throttle steam βͺ Mass of Bled Steam, m Mass Balance: min = mout m5 + m = 1 Energy Balance: Ein = Eout mh2 + m5h5 = m6h6 mh2 + (1 - m)h5 = (1)h6 mh2 + (1)h5 - mh5 = (1)h6 mh2 - mh5 = + (1)h6 - (1)h5 m(h2 – h5) = h6 - h5 h6 − h5 π= h2 – h5 Energy balance Ein = Eout (1)h1 = mh2 + (1-m) h3 + W W = (1)h1 – mh2 – (1-m) h3 Energy Balance at Condenser Heat Rejected, QR Energy Balance: Ein = Eout (1 - m)h3 = QR + (1 - m)h4 QR = (1 - m)(h3 – h4) Energy balance at Pump A Ein = Eout • Wpa + h4 = h5 Considering the mass at Pump A Wpa= (h5 - h4 )(1-m) • Where: Wpa = vf4 (POH - PCON ) Energy balance at Pump B Ein = Eout • Wpb + h6 = h7 Considering the mass at Pump A Wpb= (h7 – h6 )(1) • Where: Wpb = vf6 (PB - POH ) Cycle analysis • Net Cycle Work, WNET WNET = W – ΣWP WNET = (1)h1 – mh2 – (1-m)h3 - ΣWP • Thermal efficiency, eC WNET ππ = π₯ 100% ππ (1)h1 – mh2 – (1−m)h3–ΣWP ππ = π₯ 100% (β1 −β7 )(1) The ideal regenerative Engine • Engine Analysis Work, W W = (1)h1 – mh2 – (1−m)h3 Energy Chargeable, EC The engine is charged with the enthalpy of steam entering the engine and credited with the enthalpy of feedwater leaving the last heater assuming that all the bled steam are used for feedwater heating. EC = Enthalpy of steam entering the turbine - Enthalpy of feedwater leaving the last heater For the given cycle EC = h1 – h5 • Thermal efficiency, ee The Actual regenerative Cycle • Any presence of the following conditions will make an ideal cycle an actual one. β’Pressure drop in the boiler. P1< PB5’ β’Pressure drop in the steam line (11’) P1’< P1 β’Pressure drop in the condenser. P4< P3’ β’Pressure drop in the bled steam line. P2’’< P2’ β’Pressure drop in the feedwater line. PB5’< PB5 β’Heat losses in the steam lines (1-1’) and (2’-2’’). β’Heat losses in the turbine β’Inefficient Pump β’Heat losses in the heaters. schematic diagram Plant Layout of Actual Regenerative Cycle with One Stage of Extraction for Feedwater Heating Cycle Analysis • Heat Added, QA’ QA’ = h1 – hB5’ • Heat Rejected, QR’ QR’ = (1-m’)(h3’ - h4) • Mass of Bled Steam • Engine Work, W’ W’ = (h1’ – h2’)+(1-m’)(h2’ – h3’) Cycle Analysis • Pump Work, WP’ WP’ = ΣWP = WP1’ + WP2’ • Net Cycle Work, WNET’ WNET’ = W’ – WP’ • Thermal Efficiency, EC’ Problem 1 Steam is delivered to an engine at 5.4 MPa and 600°C. Before condensation at 31°C, steam is extracted for feedwater heating at 0.60 MPa. For an ideal cycle, find (a) the amount of steam extracted (b) W, QA and e. (c) For an ideal engine and the same states, compute (d) W, QA and e and Schematic and T – s Diagram Problem 1 At point 1 considering • P1 =5.40 MPa • T1 = 600°C ; • π‘π ππ‘ = 268.84β At table 3 ππ½ h1 = 3663.3 ππ ππ½ s1 = 7.2206 ππ−πΎ Problem 1 at point 2 considering the open heater (OH) • P2 = 0.6 MPa • S1 = S2 = 7.2206 πΎπ½ ππ−πΎ π 2 = π π + π₯ π ππ 7.2206 − 1.9312 π₯1 = 4.8288 • x = 1.09539 β2 = βπ + π₯ βππ • β2 = 670.56 + 1.09539 h2 = 2955.86 2086.3 ππ½ ππ At table 2 considering pressure at partial expansion at turbine going to OH βπ = 670.56 Sf = 1.9312 βππ = 2086.3 Sfg = 4.8288 vf = 0.001006 Problem 1 at point 3 considering the condenser • TCON = T3 =0.6 MPa • S1 = S2 = S3 = 7.2206 πΎπ½ ππ−πΎ π 3 = π π + π₯ π ππ 7.2206 − 0.4369 π₯2 = 8.0164 • π₯2 = 0.84623 β3 = βπ + π₯ βππ • β3 = 129.97 + 0.84623 h3 = 2430.5 ππ½ 2186.73 ππ At table 1 considering temperature at condenser (0.004496 MPa) βπ = 129.97 Sf = 0.4369 βππ = 2430.5 Sfg = 8.0164 vf = 0.0010046 Problem 1 At point 4 considering • P4 = Pcon = 0.004496 MPa (31°C) • h4 = hf @P3 = ππ½ 129.97 ππ At Pump A considering point 5 WPa = Vf3 (POH - PCON) = 0.0010046(600 − 4.496) WPa = 0.59824 ππ½ ππ • h5 = Wpa + h4 = 0.59824 + 129.97 h5 = 130.57 kπ½/ππ Problem 1 At point 6 at open heater (OH) • P6 = POH = 0.6 MPa • h6 = hf @P2 = ππ½ 670.56 ππ At Pump B considering point 7 WPb = Vf2 (PB - POH) = 0.001006(5400 – 600) WPb = ππ½ 4.8288 ππ • h7 = WPb + h6 = 4.8288 + 670.56 h7 = ππ½ 675.40 ππ Problem 1 ππ½ ππ½ ; WPa = 0.59824 ππ ππ ππ½ ππ½ h2 = 2955.86 ; WPb = 4.8288 ππ ππ ππ½ h3 = 2186.73 ππ ππ½ h4 = 129.97 ππ ππ½ h5 = 130.57 ππ ππ½ h6 = 670.56 ππ ππ½ h7 = 675.40 ππ • h1 = 3663.3 • • • • • • (a) the amount of steam extracted (b) W, QA and e. (c) For an ideal engine and the same states, compute (d) W, QA and e and Problem 1 mass balancing at OH • Based on derive equation h 6 − h5 π= h 2 – h5 670.56 − 130.57 π= 2955.86 − 130.57 m = 0.191127, (1 – m) = 0.808873 ππ½ ππ½ ; WPa = 0.59824 ππ ππ ππ½ ππ½ h2 = 2955.86 ; WPb = 4.8288 ππ ππ ππ½ h3 = 2186.73 ; ππ ππ½ h4 = 129.97 ππ ππ½ h5 = 130.57 ππ ππ½ h6 = 670.56 ππ ππ½ h7 = 675.40 ππ • h1 = 3663.3 • • • • • • Problem 1 QA QA = (h1 – h6) mT QA = (3663.3 - 675.40)(1) QA = 2987.44 ππ½ ππ ππ½ ππ½ ; WPa = 0.59824 ππ ππ ππ½ ππ½ h2 = 2955.86 ; WPb = 4.8288 ππ ππ ππ½ h3 = 2186.73 ; m = 0.191127 ππ ππ½ h4 = 129.97 ; (1 – m) = 0.808873 ππ ππ½ h5 = 130.57 ππ ππ½ h6 = 670.56 ππ ππ½ h7 = 675.40 ππ • h1 = 3663.3 • • • • • • Problem 1 WNET WNET = W – Σ WP Based on derive equation, WNET = (1)h1 – mh2 – (1-m) h3 – (WPa + WPb) W =1329.57 ππ½ ππ WNET = 1329.57 – (4.8288 + 0.59824) WNET = 1324.14 ππ½ ππ ππ½ ππ½ ; WPa = 0.59824 ππ ππ ππ½ ππ½ h2 = 2955.86 ; WPb = 4.8288 ππ ππ ππ½ h3 = 2186.73 ; m = 0.191127 ππ ππ½ h4 = 129.97 ; (1 – m) = 0.808873 ππ ππ½ h5 = 130.57 ; ππ ππ½ h6 = 670.56 ; h7 = 675.40 ππ • h1 = 3663.3 • • • • • Problem 1 eC ππππ‘ ππ = π₯ 100% ππ΄π ππ½ ππ½ ; WPa = 0.59824 ππ ππ ππ½ ππ½ h2 = 2955.86 ; WPb = 4.8288 ππ ππ ππ½ h3 = 2186.73 ; m = 0.191127 ππ ππ½ h4 = 129.97 ; (1 – m) = 0.808873 ππ ππ½ ππ½ h5 = 130.57 ; WNET = 1324.14 ππ ππ ππ½ ππ½ h6 = 670.56 ; QA = 2987.44 ππ ππ ππ½ h7 = 675.40 ππ • h1 = 3663.3 • 1324.14 ππ = π₯ 100% 2987.44 • ππ =44.32% • • • • Problem 1 W for engine (start here) ππππ‘ = W = 1329.57 ππ½ ππ • Considering that work of turbine is equivalent to work net. Work of pump is negligible • • • • • • ππ½ ππ½ h1 = 3663.3 ; WPa = 0 ππ ππ ππ½ ππ½ h2 = 2955.86 ; WPb = 0 ππ ππ ππ½ h3 = 2186.73 ; ππ ππ½ h4 = h5 = 129.97 ; ππ ππ½ h6 = h7 = 670.56 ; ππ ππ½ ππππ‘ = 1329.57 ππ • (1 – m) = 0.808873 • m = 0.191127 Problem 1 for engine ee π ππ = π₯ 100% πΈπ πΈπ = (π π ) (β1 − β7 ) πΈπ = 2292.74 • • • • πΈπ = (3663.3 - 670.56)(1) • ππ½ ππ 1329.57 ππ = π₯ 100% 2292.74 ππ = 44.42% ππ½ ππ½ ; WPa = 0 ππ ππ ππ½ ππ½ h2 = 2955.86 ; WPb = 0 ππ ππ ππ½ h3 = 2186.73 ; ππ ππ½ h4 = h5 = 129.97 ; ππ ππ½ h6 = h7 = 670.56 ; ππ ππ½ ππππ‘ = 1329.57 ππ • h1 = 3663.3 • • (1 – m) = 0.808873 • m = 0.191127 Problem 2 Steam at 6.2 MPa and 480°C is received by a regenerative engine. Extractions for feedwater heating occurs at 3 MPa and again at 1.4 MPa, with the remaining steam expanding to 0.0065 MPa. For the ideal engine with a throttle flow of 54,500 kg/h, find (a) the amount of steam extracted (b) W, QA and e. (c) For an ideal engine and the same states, compute (d) W, QA and e and Regenerative with 2 OH mass ππ‘ if not given, ππ‘ =1 ππ mass extraction @ point 3 ππ mass extraction @ point 2 ππ mass extraction @ point 4 ππ = ππ‘ − ππ − ππ ππ‘ ππ + ππ T-s h10 = WPc + h9 h9 = hf @POH1 h8 = WPb + h7 h7 = hf @POH2 h6 = WPa + h5 h5 = hf @P4 h1 & s1 = assuming SH @table 3 h2 = considering pressure at OH1, Considering s1 = s2 = s3 find x1 β2 = βπ + π₯ βππ h3 = considering pressure at OH2, Considering s1 = s2 = s3 find x2 β3 = βπ + π₯ βππ h4 = considering pressure at condenser, considering s1 = s2 = s3 find x3 β4 = βπ + π₯ βππ Energy Balance at Boiler Ein = Eout • QA + h10 = h1 QA = (h1 – h10) mt But h10is based on energy balance at OH A Energy balance Ein = Eout mth1 = mah2 + mbh3 + mch4 + W W= mth1 - mah2 - mbh3 - mch4 Energy Balance at Condenser Heat Rejected, QR Energy Balance: Ein = Eout mc h4 = QR + mc h5 QR = mc h4 – mc h5 QR = mc (h4 –h5) Energy balance at Pump A Ein = Eout Wpa + h5 = h6 Considering the mass at Pump A Wpa= (h6 – h5 )(mc) • Where: Wpa = vf5 (POHA - PCON ) Energy balance at OH (2) mbh3 + mch6 = h7(1–ma) mbh3 + mch6 = h7(1–ma) Where: mc = 1 – ma – mb mbh3 + (1 – ma – mb) h6= (1–ma)h7 mbh3 + (1)h6 – mah6– mbh6= (1)h7–mah7 mbh3– mbh6 – mah6+ ma h7= (1)h7–(1)h6 mb(h3–h6) – ma(h6+ h7)= (h7–h6) (1) (h7–h6) (1) +ma(h6+ h7) mb = (h3–h6) Energy balance at Pump B Ein = Eout Wpb + h8 = h9 Considering the mass at Pump B Wpb= (h8 – h7)(1 – ma) Where: Wpb = vf7 (POHA – POHB) considering pressure at OH B Energy balance at OH (1) mah2 + (1 – ma)h8 = h9(1) mah2 + h8(1) – h8ma= h9(1) mah2– h8ma= h9(1) – h8(1) ma(h2– h8) = (h9 – h8) h9– h8 ma = h2– h8 Energy balance at Pump C Ein = Eout Wpc + h9 = h10 Considering the mass at Pump C Wpc= (h10 – h9)(1) Where: Wpb = vf9 (PB – POHA) considering pressure at OH A Regenerative with 2 OH mass ππ‘ if not given, ππ‘ =1 ππ mass extraction @ point 3 ππ mass extraction @ point 2 ππ mass extraction @ point 4 ππ = ππ‘ − ππ − ππ ππ‘ ππ + ππ Cycle analysis • Net Cycle Work, WNET WNET = W – ΣWP WNET = (mth1 - mah2 - mbh3 - mch4) – ΣWP a-b • Thermal efficiency, eC WNET ππ = π₯ 100% ππ (mth1 − mah2 − mbh3 − mch4) –ΣWPa−b ππ = π₯ 100% (β1 −β10 )(mt) Problem 2 At point 1 considering • P1 =6.2 MPa • T1 = 480°C ; • π‘π ππ‘ = 276.72β At table 3 ππ½ h1 = 3371.8 ππ ππ½ s1 = 6.7999 ππ−πΎ Problem 2 at point 2 considering the open heater (OH 1) • P2 =3.0 MPa • S1 = S2 = 6.7999 πΎπ½ ππ−πΎ π 2 = π π + π₯ π ππ 6.7999 − 2.6457 π₯1 = 3.5412 • x = 1.17311 β2 = βπ + π₯ βππ • β2 = 1008.42 + 1.17311 (1795.7) h2 = 3114.96 ππ½ ππ At table 2 considering pressure at partial expansion at turbine going to OH A βπ = 1008.42 Sf = 2.6457 βππ = 1795.7 Sfg = 3.5412 vf = 0.0012165 Problem 2 at point 3 considering the open heater (OH 2) • POH B = P3 = 1.40 MPa • S1 = S2 = S3 = 6.7999 πΎπ½ ππ−πΎ π 3 = π π + π₯ π ππ 6.7999 − 2.3150 π₯2 = 4.1850 • π₯2 = 1.07166 β3 = βπ + π₯ βππ • β3 = 830.30 + 1.07166 h3 = 1959.7 ππ½ 2930.43 ππ At table 2 considering pressure at OH B βπ = 830.30 Sf = 2.3150 βππ = 1959.7 Sfg = 4.1850 vf = 0.0011489 Problem 2 at point 4 considering the condenser • PCON = P4 = 0.0065 MPa • S1 = S2 = S3 = S4 = 6.7999 πΎπ½ ππ−πΎ π 3 = π π + π₯ π ππ 6.7999 − 0.5408 π₯2 = 7.7613 • π₯3 = 0.80644 β4 = βπ + π₯ βππ • β4 = 157.67 + 0.80644 h4 = 2412.4 ππ½ 2103.15 ππ At table 1 considering temperature at condenser (0.0065 MPa) βπ = 157.67 Sf = 0.5408 βππ = 2412.4 Sfg = 7.7613 vf = 0.0010069 Problem 2 At point 5 considering • P5 = Pcon = 0.0065 MPa • h5 = hf @P4 = 157.67 ππ½ ππ At Pump A considering point 6 WPa = Vf4 (POH2 - PCON) = 0.0010069(1400 – 6.5) ππ½ WPa =1.4031 ππ • h6 = Wpa + h5 = 1.4031 + 157.67 h6 = 159.073 kπ½/ππ Problem 2 At point 7 at open heater (OH) • P7 = POHB = 1.40 MPa • h7 = hf @P3 = ππ½ 830.30 ππ At Pump B considering point 8 WPb = Vf3 (POH1 – POH2) = 0.0011489(3000 – 1400) WPb = ππ½ 1.83824 ππ • h8 = WPb + h7 = 1.83824 + 830.30 h8 = ππ½ 832.14 ππ Problem 2 At point 9 at open heater (OH A) • P9= POHA = 3.0 MPa • h9 = hf @P2 = ππ½ 1008.42 ππ At Pump C considering point 10 WPc = Vf2(PB – POH1) = 0.0012165(6200 – 3000) ππ½ WPc= 3.8928 ππ • h10 = WPc + h9 = 3.8928 + 1008.42 h10 = ππ½ 1012.31 ππ Problem 2 • • • • • • ππ½ h1 = 3371.8 ; h = 159.073 ππ 6 ππ½ ππ½ h2 = 3114.96 ; h7 = 830.30 ππ ππ ππ½ ππ½ h3 = 2930.43 ; h8 = 832.14 ππ ππ ππ½ ππ½ h4 = 2103.15 ; h9 = 1008.42 ππ ππ ππ½ ππ½ h5 = 157.67 ; h10 = 1012.31 ππ ππ ππ½ WPabc = 7.113414 ππ (a) the amount of steam extracted (b) W, QA and e. (c) For an ideal engine and the same states, compute (d) W, QA and e and Problem 2 mass balancing at OH 1 • Based on derive equation h9– h8 ma = h2– h8 1008.42 − 832.14 ma = 3114.96 − 832.14 ma = 0.07722 • • • • • • ππ½ h1 = 3371.8 ; h = 159.073 ππ 6 ππ½ ππ½ h2 = 3114.96 ; h7 = 830.30 ππ ππ ππ½ ππ½ h3 = 2930.43 ; h8 = 832.14 ππ ππ ππ½ ππ½ h4 = 2103.15 ; h9 = 1008.42 ππ ππ ππ½ ππ½ h5 = 157.67 ; h10 = 1012.31 ππ ππ ππ½ WPabc = 7.113414 ππ Problem 2 mass balancing at OH 2 • Based on derive equation (h7–h6) (1) + ma(h6+ h7) mb = (h3–h6) (830.30 − 159.073)(1) + 0.07722 (159.073 + 830.30) mb = 2930.43 − 159.073 ππ½ • h1 = 3371.8 ma = 0.07722 mb = 0.2269769 mc = 0.653011 • • • • • ; h = 159.073 ππ 6 ππ½ ππ½ h2 = 3114.96 ππ; h7 = 830.30 ππ ππ½ ππ½ h3 = 2930.43 ππ; h8 = 832.14 ππ ππ½ ππ½ h4 = 2103.15 ππ; h9 = 1008.42 ππ ππ½ ππ½ h5 = 157.67 ππ; h10 = 1012.31 ππ ππ½ WPabc = 7.113414ππ • Problem 2 QA QA = (h1 – h10) mt QA = (3371.8 - 1012.31)(1) QA = ππ½ 2359.49 ππ • • • • ππ½ h1 = 3371.8 ; h = 159.073 ππ 6 ππ½ ππ½ h2 = 3114.96 ; h7 = 830.30 ππ ππ ππ½ ππ½ h3 = 2930.43 ; h8 = 832.14 ππ ππ ππ½ ππ½ h4 = 2103.15 ; h9 = 1008.42 ππ ππ ππ½ ππ½ h5 = 157.67 ; h10 = 1012.31 ππ ππ ππ½ WPabc = 7.113414 ππ • ma = 0.07722 mb = 0.2269769 mc = 0.653011 Problem 2 WNET WNET = W – Σ WP Based on derive equation, WNET = (mth1 - mah2 - mbh3 - mch4)– (WPabc ) WNET = 1085.63 ππ½ ππ ππ½ • h1 = 3371.8 ; h = 159.073 ππ 6 ππ½ ππ½ • h2 = 3114.96 ; h7 = 830.30 ππ ππ ππ½ ππ½ • h3 = 2930.43 ; h8 = 832.14 ππ ππ ππ½ ππ½ • h4 = 2103.15 ; h9 = 1008.42 ππ ππ ππ½ ππ½ • h5 = 157.67 ; h10 = 1012.31 ππ ππ ππ½ • WPabc = 7.113414 ππ ma = 0.07722 mb = 0.2269769 mc = 0.653011 ππ½ QA = 2359.49 ππ Problem 2 eC ππππ‘ ππ = π₯ 100% ππ΄ 1085.63 ππ = π₯ 100% 2359.49 ππ =46.01% ππ½ • h1 = 3371.8 ; h = 159.073 ππ 6 ππ½ ππ½ • h2 = 3114.96 ; h7 = 830.30 ππ ππ ππ½ ππ½ • h3 = 2930.43 ; h8 = 832.14 ππ ππ ππ½ ππ½ • h4 = 2103.15 ; h9 = 1008.42 ππ ππ ππ½ ππ½ • h5 = 157.67 ; h10 = 1012.31 ππ ππ ππ½ • WPabc = 7.113414 ππ ma = 0.07722 mb = 0.2269769 mc = 0.653011 ππ½ QA = 2359.49 ππ WNET = 1085.63 ππ½ ππ Problem 2 W for engine ππππ‘ = W = 1092.7434 ππ½ ππ • Considering that work of turbine is equivalent to work net. Work of pump is negligible • h1 = • h2 = • h3 = • h4 = • h5 = • h7 = • h9 = ππ½ 3371.8 ; ππ ππ½ 3114.96 ; ππ ππ½ 2930.43 ; ππ ππ½ 2103.15 ; ππ ππ½ h6 = 157.67 ; ππ ππ½ h8 = 830.30 ππ ππ½ h10 = 1008.42 ππ • WPabc = 0 Problem 2 for engine ee • h1 = π ππ = π₯ 100% πΈπ • h2 = πΈπ = (π π ) (β1 − β10 ) • h3 = • πΈπ = (3371.8 - 1008.42 )(1) ππ½ 2363.38 ππ πΈπ = 1092.7434 ππ = π₯ 100% 2363.38 ππ = 46.24% • h4 = • h5 = • h7 = • h9 = ππ½ 3371.8 ; ππ ππ½ 3114.96 ; ππ ππ½ 2930.43 ; ππ ππ½ 2103.15 ; ππ ππ½ h6 = 157.67 ; ππ ππ½ h8 = 830.30 ππ ππ½ h10 = 1008.42 ππ • WPabc = 0; W = 1092.7434 ππ½ a ππ Closed Feedwater Heater Another type of feedwater heater frequently used in steam power plants is the closed feedwater heater, in which heat is transferred from the extracted steam to the feedwater without any mixing taking place. The two streams now can be at different pressures, since they do not mix. The schematic of a steam power plant with one closed feedwater heater and the T-s diagram of the cycle are shown in figure (next slide). In an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure. In actual power plants, the feedwater leaves the heater below the exit temperature. Plant Layout of Regenerative Cycle With One Stage of Extraction for Feedwater Heating (Closed Heater) At point 1 Considering SH Steam @ table 3 At point 2 Considering pressure At CH, find π₯1 to solve β2 using table 2 Enthalpy At point 7 Considering the Energy balance at Mixing Chamber to determine h7 At point 6 Considering CH P6 = PCH h6 = hf at PCH At point 9 Considering the Equation of pump 2 to determine h9 At point 3 Considering properties of steam Find the value of “π₯2 ” At table 2 At point 4 Considering constant pressure P3=P4 h4 = hf at Pcon At point 8 h8 = h6 At point 5 Considering the Equation of pump 1 to determine h5 Mass Balancing At point 1 Mass Total equivalent to 1 (100%) At point 2 Considering the mass extraction “m” At point 3 Considering the remaining mass after extraction “1 – m” At point 6 “1 – m” At point 4 “1 – m” At point 7 Mass Total equivalent to 1 (100%) before Entering boiler At point 9 “m” At point 5 “1 – m” At point 8 “m” NOTE: MC – Mixing Chamber CH – Closed Heater (non – mixing) T – s Diagram “m” can be determined By mass & energy balance At CH In CH no mixing of Steam will happen Enthalpy at 6 & 8 are Equal considering ideal Heat transfer at CH Point 1 is at SH m 1- m π 1 = π 2 = π 3 Considering isentropic expansion at turbine Point 2 mass extraction considering Mass total is 1 (100%) And mass extraction “m” Mass entering the condenser Is 1 – m Energy Balance at Boiler Ein = Eout • QA + h7 = h1 QA = (h1 – h7) mt But h7 is based on energy balance at CH Energy Balance at CH • Basis: 1kg of throttle steam βͺ Mass of Bled Steam, m Mass Balance: min = mout m + (1 – m) = (1 – m) + m Energy Balance: Ein = Eout mh2 + (1 – m)h5 = (1 – m)h6 +mh8 mh2 + (1)h5 – mh5 = (1)h6 – mh6 +mh8 (1)h5 –(1)h6 =mh8 + mh5 – mh2 –mh6 (h5 –h6) =m(h8 + h5 – h2 –h6) (h5 – h6) π= (h8 + h5 – h2 –h6) Energy balance Ein = Eout (1)h1 = mh2 + (1-m) h3 + W W = (1)h1 – mh2 – (1 – m) h3 Energy Balance at Condenser Heat Rejected, QR Energy Balance: Ein = Eout (1 - m)h3 = QR + (1 - m)h4 QR = (1 - m)(h3 – h4) Energy balance at Pump A Ein = Eout • Wpa + h4 = h5 Considering the mass at Pump A Wpa= (h5 - h4 )(1-m) • Where: Wpa = vf4 (PCH - PCON ) Energy balance at Pump B Ein = Eout • Wpb + h8 = h9 Considering the mass at Pump B Wpb= (h9 – h8 )(m) Where: Wpb = vf2 (PB - PCH ) considering pressure at CH Cycle analysis • Net Cycle Work, WNET WNET = W – ΣWP WNET = (1)h1 – mh2 – (1-m)h3 - ΣWP • Thermal efficiency, eC WNET ππ = π₯ 100% ππ (1)h1 – mh2 – (1−m)h3–ΣWP ππ = π₯ 100% (β1 −β7 )(1)
