Example 12.1

Heat balance:

• n-Propanol, Q = 60,000 X 285 = 17,100,000 Btu/hr

• Water, Q = 488,000 X l(120 - 85) = 17,100,000 Btu/h In Figure (2-60) : (tc-t) = 507-244= 263 λ = 285 Btu/Ib Assumed t2= 120 °f t2= (120 : 150) Maxi. w=488 M Ib/hr So, ta = (85+120)/2= 102.5 °F

∆t=LMTD=∆𝑡1+∆𝑡2

𝐿𝑛(∆𝑡1

∆𝑡2)

= 141 °𝐹

• No. of Tubes =Ά

𝐿𝑡 ∗𝑎”=

1213

8 ∗0.1963= 773