Thermodynamik I/II PDF

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Summary

Thermodynamik I/II Zusammenfassung 2016/2017 ...

Description

z1 = 1 m V˙ = 1 L/s P = 50 W Q˙

g z2 z1

Pel z2 d1 Q˙

• • • • •

ρ

= 1 kg/L

•

g = 9,81 m/s2

Qab

n1 = 40 mol 2

2

V = 1 m3

T1 = 298,15 K T2 = T1

p1

p2 p1 = 1,137 bar

p2 = 0,634 bar Q12

• • • • • •

f,0 ∆hH = −20,50 kJ/mol 2S

p1 = 15

t2 = 900 ◦ ηST = 0,9

tFW,aus Pab = 50

t3 = 40 ◦C

tFW,ein = 8 ◦C = 30 ◦C

T s .

mW

.

mW = 30,123 kg/s

◦

◦ tneu FW,ein = 10 C

tFW,aus

.

mFW

t1

. neu

neu Pab

mW

t4

p1

. neu mW

η neu = 23,123 kg/s

neu Pab = 42,123 MW η neu = 0,423 .

∆EV

• • if cW = 4,18 kJ/(kg K)

• • • • •

tU = 25 ◦

pU = 1

K w1,W = 30 gew%

t1

V1K = 4 L = 80 ◦C

K = 70 gew% w1,E

Q˙ = 1 000 W V 2K = 1 L ab = 1 800 kJ tA = 25 ◦C Q12 A w 2,E = 80 gew% ′′ K = 39,36 mol % x 2,E

T, x

T2 T2K = 364 K τ12

p = 1 bar

• • • • • ln

B ps0i =A− kPa T /K − C

A = 16,664 B = 3 667,7 C = 46,966

• ρE = 789 kg/m3

• ρW = 998 kg/m3

if = 2,43 kJ/(kg K) cE

• if cW

• •

= 4,18 kJ/(kg K)

p1 = 100 bar T1 = 600 K g (T, p) = aT − bT 2 + RT

(p/p0 )

mit a = 2 J/(mol K) b = 0,06 J/(mol K2 ) .

T2

p2 = 1 bar...