《化工原理I》计算题 下载本文

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长1.5m的钢管所构成,空气在管内作湍流流动,其流量为740kg/h,比热为1.005kJ/kg·℃;饱和水蒸气在管间冷凝。已知操作条件下空气的对流传热系数为74 W/m2·℃,水蒸气的冷凝传热系数为10000 W/m2·℃,管壁及垢层热阻可忽略不计,试确定:

(1)所需饱和水蒸气的温度;

(2)若将空气量增加25%通过原换热器,在空气进口温度和饱和水蒸气温度不变情况下,空气能加热到多少度?(设在本题条件下空气温度有所改变时,其物性参数视为不变) 解:

(1) Q = m2Cp2(t2 – t1)= KoAoΔtm

Q?740?1.005?103??90?10??16.527(kW) 3600

?1doKo?1?????o?idi????1?25??1????58.85(W/(m℃))

1000074?20??2

2

Ao = πdoLn = 3.14×0.025×1.5×38 = 4.4745(m)

Q16.527?103?tm???62.76(℃)

KoAo58.85?4.4745?tm??t2?t1?lnT?t1?62.76

T?t2

lnT?t190?10T?t1??1.2747 ?3.58

T?t262.76T?t2 T – 10 = 3.58×(T - 90) T = 121℃ (10分)

0.8?du???(2) Nu??0.023???????

Pr0.4

?i??u??????i?u?0.8?1.250.8

2

αi′= 1.25×74 = 88.46(W/(m℃))

0.8

?1do?Ko?1?????o?i?di????1?25?1?????70.27(W/(m℃))

1000088.46?20??2

m2′Cp2(t2′– t1)= Ko′AoΔtm′

??t1??Ko?Ao?t2??t1?ln1.25m2Cp2?t2T?t1?T?t2

ln?AoKoT?t170.27?4.4745?3600???1.2176 3?T?t21.25m2Cp21.25?740?1.005?10 121 – 10 = 3.379×(121 - t2′)

T?t1?3.379?T?t2 33

t2′= 88.15℃ (10分)

42温度为20℃、比热为2.1kJ/kg℃的溶剂以5000kg/h的流率流过套管换热器内管,被加热至64℃,环隙内为170℃的饱和水蒸气冷凝。蒸气冷凝潜热为2054kJ/kg。换热器内管直径Ф108×10mm,总长12m,材料为钢。试计算 (1)蒸气用量; (2)总传热系数;(以内管外表面为基准)

(3)如果将内管改为Ф152×10mm的钢管,其它条件不变,求所需管长。

假设两种情况下蒸气冷凝传热系数均为10000W/m2K,液体在管内均为湍流,不考虑污垢热阻及热损失。钢的导热系数λ= 45W/m℃。

解:(1)Q = m2Cp2 (t2 – t1) = 5000/3600×2.1×10×(64 - 20) = 128.3 (kW)

3

m1?Q128.3??0.06246r12054?(kg/s) = 224.9 (kg/h) (4分)

(2) ?tm

t2?t164?20??126.7(℃)

ln??T?t1??T?t2??ln??170?20??170?64??Ao = πdol = 3.14×0.108×12 = 4.07 (m2) Q = KoAoΔtm 2

Q128.3?103Ko???248.8 (W/m℃) (6分)

Ao?tm4.07?126.7d?d11?o?0?Ko?idi?dm?o

(3)

dm = (di + do)/2 = (88 + 108)/2 = 98 (mm)

do?d1110.01?1081??0?????0.003646565?idiKo?dm?o248.845?8810000

?i?108?336.6(W/m℃)

88?0.0036465652

??diui??0.4??i?0.023i?Pr ??di???2?i?di?di?di????2???idi??didi??0.80.8?di????d????i?1.8?88?????132?1.8?0.482

?i?= 0.482×336.6 = 162.2 (W/m℃)

2

?= (di?+do?)/2 = (132 + 152)/2 = 142 (mm) dm???do??do1???1?Ko?????d??d????

m0??ii1520.01?1521????1?Ko????134.5(W/m℃)

45?14210000??162.2?1322

Q = Ko′Ao′Δtm

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Q128.3?103??Ao??7.5288( m)

??tm134.5?126.7Ko2

l???Ao7.5288??15.8(m) (14分) ?3.14?0.152?d043一套管换热器,外管为Ф83×3.5mm,内管为Ф57×3.5mm的钢管,有效长度为60m。用120℃的饱和水蒸气冷凝来加热内

管中的油。蒸气冷凝潜热为2205kJ/kg。已知油的流量为7200kg/h,密度为810kg/m3,比热为2.2 kJ/kg℃,粘度为5cP,进口温度为30℃,出口温度为80℃。试求: (1)蒸气用量;(不计热损) (2)总传热系数;(以内管外表面为基准)

(3)如油的流量及加热程度不变,加热蒸气压力不变,现将内管直径改为Ф47×3.5mm的钢管,求管长为多少? 假设两种情况下蒸气冷凝传热系数均为12000W/m2K,液体在管内均为湍流,管壁及污垢热阻不计。 解:(1)Q = m2Cp2 (t2 – t1) = 7200/3600×2.2×10×(80 - 30) = 220 (kW)

3

m1?Q220??0.09977r12205?(kg/s) = 359.2 (kg/h) (4分)

(2) ?tm

t2?t180?30??61.66(℃)

ln??T?t1??T?t2??ln??120?30??120?80??Ao = πdol = 3.14×0.057×60 = 10.74 (m2) Q = KoAoΔtm 2

Q220?103Ko???332.2 (W/m℃) (6分)

Ao?tm10.74?61.66d?d11?o?0?Ko?idi?dm?o

(3)

do1111?????0.0029269?idiKo?o332.212000

?i?57?389.5(W/m℃)

50?0.00292692

??diui??0.4? ?i?0.023i?Pr??di???2di?di?di????2???idi??didi??0.8?i?0.8?di????d????i?1.8?50?????40?1.8?1.4943

?i?= 1.4943×389.5 = 582 (W/m℃)

2

??do1????Ko?1??? ???d?0??ii1??47??1?Ko???475.7(W/m℃)

582?4012000??2

35

Q = Ko′Ao′Δtm

Q220?103??Ao??7.5( m)

??tm475.7?61.66Ko2

l???Ao7.5??50.8(m) (14分) ?3.14?0.047?d044有一套管换热器,内管为Ф54×2mm,外管为Ф116×4mm的钢管,内管中苯被加热,笨进口温度为50℃,出口温度为80℃,

流量为4000kg/h。环隙为133.3℃的饱和水蒸气冷凝,其汽化潜热为2168.1kJ/kg,冷凝对流传热系数为11630W/m2K。 苯在50℃~80℃之间的物性参数平均值为密度ρ= 880kg/m3,比热为1.86 kJ/kg K,粘度为0.39cP,导热系数为0.134W/m.K。管内壁污垢热阻为0.000265K.m2/W,管壁及管外污垢热阻不计。试求: (1)加热蒸气消耗量;

(2)所需的传热面积(以内管外表面计)。

(3)当苯的流量增加50%时,要求苯的进出口温度不变,加热蒸气的温度应为多少?

解:(1)Q = m2Cp2 (t2 – t1) = 4000/3600×1.86×103×(80 - 50) = 62 (kW)

m1?Q62??0.0286r12168.1?(kg/s) = 102.9 (kg/h) (4分)

(2) ?tm

t2?t180?50??67.2(℃)

ln??T?t1??T?t2??ln??133.3?50??133.3?80?? Q = KoAoΔtm

u?4m2?di2?diu??4?4000?0.643 (m/s)

3.14?0.052?880?36000.05?0.643?880?72543.6> 10 ?30.39?104

Re??Cp??

Pr??1.86?103?0.39?10?3??5.41

0.1340.8

??diui??0.4??i?0.023i?Pr ??di???

?i?0.023?0.134?72543.60.8?5.410.4?936.7(W/m℃) 0.052

dd11?o?Rsio?Ko?ididi?o

54541??Ko?1??0.000265????655.7(W/m℃)

5011630??936.7?502

Q62?103Ao???1.407( m) (10分)

Ko?tm655.7?67.22

?i??u??0.8???(3) ?i?u

?i?u?

0.8??m2???S??iim2Si?i????0.8??m2???m?2????0.8?1.50.8

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