文档介绍：Problem
(i) 1 ft = m
(ii) 1 lb m = kg
(iii) 1 lb f = N
(iv) 1 HP = 746 W
(v) 1 psi = kN m -2
(vi) 1 lb ft s -1 = N s m -2
(vii) 1 poise = N s m -2
(viii) 1 Btu = kJ
(ix) 1 CHU = kJ
(x) 1 Btu ft -2 h-1 oF-1 = W m -2 K-1
Examples:
o
(viii) 1 Btu = 1 lb m of water through 1 F
= g through oC
= cal
= ()()
= J = kJ
−2
J − m
(x) 1 Btu ft -2 h-1 oF-1 = 1 Btu x x 10 3 x ft1 12x x 10x 3
Btu ft
−1 −1
s o C
o
x xh1 3600 x 1 xF o
h F
= W m -2 oC-1
= W m -2 K-1
Problem
W2, t1
T2
W1, T1
t2
Variables , M :
1. Duty, heat transferred, Q
2. Exchanger area, A
3. Overall coefficient, U
4. Hot-side flow-rate, W1
5. Cold-side flow-rate, W 2
6. Hot-side inlet temperature, T1
7. Hot-side outlet temperature, T2
8. Cold-side inlet temperature, t1
9. Cold-side outlet temperature, t2
Total variables = 9
Design relationships, N:
1. General equation for heat transfer across a surface
Q = UA ∆Tm (Equation )
Where ∆Tm is the LMTD given by equation ()
= ( −)
2. Hot stream heat capacity Q W1C p T1 T2
= ( −)
3. Cold stream heat capacity Q W2C p t2 t1
4. U is a function of the stream flow-rates and temperatures (see Chapter 12)
Total design relationships = 4
So, degrees of freedom = M – N = 9 – 4 = 5
Problem
Number ponents, C = 3
Degrees of freedom for a process stream = C + 2 (see Page 17)
Variables:
Streams 4( C + 2)
Separator pressure 1
Separator temperature 1
Total 4C + 10
Relationships:
Material balances C
v-l-e relationships C
l-l-e relationships C
Equilibrium relationships 6
Total 3C + 6
Degrees of freedom = (4 C + 10) –(3 C + 6) = C + 4
For C = 3, degrees of freedom = 7
The feed str