Question 3.15: In the batch preparation of an aqueous solution the water is...

In the batch preparation of an aqueous solution the water is first heated to 80ºC in a jacketed, agitated vessel; 1000 Imp. gal. (4545 kg) is heated from 15ºC. If the jacket area is 300 ft2 (27.9 m2) and the overall heat-transfer coefficient can be taken as 50 Btu \ ft^{-2} h^{-1} ºF^{-1} (285 W \ m^{-2} K^{-1} ), estimate the heating time. Steam is supplied at 25 psig (2.7 bar).

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The rate of heat transfer from the jacket to the water will be given by the following expression (see Volume 1, Chapter 9):

\ \frac{dQ}{dt} =UA(t_{s} -t)       (a)

where dQ is the increment of heat transferred in the time interval dt, and

U = the overall-heat transfer coefficient,

\ t_{s} = the steam-saturation temperature,

t = the water temperature.

The incremental increase in the water temperature dt is related to the heat transferred dQ by the energy-balance equation:

\ dQ=WC_{p} dt         (b)

where \ WC_{p} is the heat capacity of the system.

Equating equations (a) and ( b)

\ WC_{p}\frac{dt}{dt} =UA(t_{s} -t)

Integrating            \ \int\limits_{0}^{t_{B} }{dt} =\frac{WC_{p}}{UA}\int\limits_{t_{1} }^{t_{2}}{\frac{dt}{(t_{s} -t)} }

Batch heating time

\ t_{B} =-\frac{WC_{p}}{UA}\ln \frac{t_{s}-t_{2}}{t_{s}-t_{1}}

For this example

\ WC_{p}=4.18\times 4545\times 10^{3} JK^{-1}

\ UA=285×27 WK^{-1}

\ t_{1}=15ºC,  t_{2}=80ºC,  t_{s}=130ºC  

\ t_{B} =-\frac{4.18\times 4545\times 10^{3}}{285\times 27.9} \ln \frac{130-80}{130-15} =1990s=\underline{\underline{33.2min} }

In this example the heat capacity of the vessel and the heat losses have been neglected for simplicity. They would increase the heating time by 10 to 20 per cent.

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