11.5. Carbon tetrachloride flowing at 19,000 kg/h is to be cooled from 85 to 40°

Question

11.5. Carbon tetrachloride flowing at 19,000 kg/h is to be cooled from 85 to 40°C using 13,500 kg/h of cooling water at 20°C. The film coefficient for carbon tetrachloride, outside the tubes, is 1,700 W/m2. °C. The wall resistance is negligible, but h, on the water side, including fouling factors, is 11,000 W/m2. °C. (a) What area is needed for a counterflow exchanger? (b) By what factor would the area be increased if par allel flow were used to get more rapid initial cooling of the carbon tetrachloride?

Explanation / Answer

Heat gained by water = heat released by CCl4

Mass x Cpw x (T - 20) = mass x Cpc x (85 - 40)

13500 kg/h x 4184 J/kgC x (T - 20) = 19000 kg/h x 920 J/kgC x (45)C

56484000 (T - 20) = 786600000

T - 20 = 13.926

T = 33.926 C

heat released by CCl4 Q = 786600000 J/h x 1h/3600s

= 218500 W = 218.5 kW

Inside heat transfer coefficient hi = 11000 W/m2-C

Outside heat transfer coefficient ho = 1700 W/m2-C

Overall heat transfer coefficient

1/U = 1/ho + 1/hi

1/U = 1/1700 + 1/11000

U = 1472.44 W/m2-C

Part a

For counter current flow

LMTD = [(85-33.926) - (40-20)] / ln (51.074/20)

= 33.144 C

Area A = Q/U x LMTD

= 218500 / (1472.44 * 33.144)

= 4.477 m2

Part b

For parallel flow

LMTD = [(85-20) - (40-33.926)] / ln (65/6.074)

= 24.859 C

Area A = Q/U x LMTD

= 218500 / (1472.44 * 24.859)

= 5.969 m2

Area increases by factor = 5.969/4.477 = 1.33

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