4×300MW火力发电厂初步设计

锅炉给水量Dfw

Dfw?Db?Dbl?1.0285D0?0.01028D50?1.03878D50 （2-8）

即 ?fw?1.038785

排污冷却器计算；补充水温tma=20℃，取排污冷却端差为8℃，则

h?ma?h???28?4.1864?h''ma?hblhl?117.2192

有排污冷却器热平衡式：

?ma(h?ma?hma)??bl?(hbl??hbl??)?1 h?????blhbl??1?117.2192??mabl?ma???bl?1 ?0.0046399?742.64?0.98?117.2192?0.0149250.014929?0.0046399?0.99

?262.87kj/kg于是

h?ma?hbl???8?4.1864?262.87?33.49?229.38kj/kg 3.计算汽轮机各段抽汽量?Dj和凝汽流量Dc

（1） 由高压加热器H1热平衡计算求?1；如图2-2

图2-2

???fw(hw1?hw2)q?1.038785?135.212014.5?0.99?0.0704 1?h（2） 由高压加热器H2热平衡计算求?2；如图2-3

图2-3

2-9）

（

4×300MW火力发电厂初步设计

?fw?2??sg1(hsg1?hs2)?h??1(hs1?hs2)?h?2?q2?h1.038785?130.12?0.99?(3397?1034.7)?0.0043?0.0704??1063.5?1034.7??0.991982.3?0.99?0.0627?

H2的疏水?s2

?0.0627?0.0043?0.137 ?s2??1??2??sg1?0.0704?0.0627?.08669再热蒸汽量?rh ?rh?1??1??2?1?0.0704

（3）由高压加热器H3的热平衡计算求?3；如图2-4

已知给水在给水泵中的焓升为

图2-4

?3???fw?3??sg2(hsg2?hs3)?h??s2(hs2?hs3)?hq3?h1.038785?126.59?0.000575?(3532?798.64)?0.99??1034.7?798.64??0.99

2469.36?0.99?0.04H3的疏水?s3

?s3??s2??3??sg2 ?0.137?0.04?0.000575

?0.1776

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（4）除氧器H4热平衡计算求?4；如图2-5

由图所示，除氧器的物质平衡，求凝结水进水份额??c.除氧器出口水份额?fw；

图2-5

? ?fw??4??sg3??s3??f??c??(?fw??sg3??f??s3)??4?c?50.012?40.00564?50.177)6??4 ?(1.03878?0.843??4除氧器的热平衡式：

?4??fwhw4/?h??fhf??sg3hsg3??s3hs3?0.843hw5h4?hw50.99?0.005645?2776.4?0.0124?3017?0.1776?798.64?0.843?560.443090?560.441.038785?667.15?0.0129

??0.843?0.0129?0.83 故 ?c（5）低压加热器H5热平衡计算求?5；如图2-6

4×300MW火力发电厂初步设计

图2-6

?5???5?c0.83?105.33??0.038 q5?h2323.41?0.99（6）由低压加热器H6热平衡计算求?6；如图2-7

图2-7

?6????6??5(hs5?hs6)?h?cq6?h0.83?104.356?0.038?(560.59?468)?0.99

2410?0.99?0.0348H6疏水?s6??s5??6?0.036084?0.03489?0.070975 （7）由低压加热器H7热平衡计算求?7；如图2-8

图2-8

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