Question 9.25: A vessel contains 1 tonne (1 Mg) of a liquid of specific hea......
A vessel contains 1 tonne (1 Mg) of a liquid of specific heat capacity 4.0 kJ/kg K. The vessel is heated by steam at 393 K which is fed to a coil immersed in the agitated liquid and heat is lost to the surroundings at 293 K from the outside of the vessel. How long does it take to heat the liquid from 293 to 353 K and what is the maximum temperature to which the liquid can be heated? When the liquid temperature has reached 353 K, the steam supply is turned off for 2 hours (7.2 ks) and the vessel cools. How long will it take to reheat the material to 353 K? The surface area of the coil is 0.5 m² and the overall coefficient of heat transfer to the liquid may be taken as 600 W/m² K. The outside area of the vessel is 6 m² and the coefficient of heat transfer to the surroundings may be taken as 10 W/m² K.
If T K is the temperature of the liquid at time t s, then a heat balance on the vessel gives:
(1000\times4000){\frac{\mathrm{d}T}{\mathrm{d}t}}=(600\times0.5)(393-T)-(10\times6)(T-293)or: 4,000,000\frac{\mathrm{d}T}{\mathrm{d}t}=135,480-360T
and: 11,111\frac{\mathrm{d}T}{\mathrm{d}t}=376.3-T.
The equilibrium temperature occurs when dT/dt = 0,
that is when: \underline{\underline{T=376.3\ K}}.
In heating from 293 to 353 K, the time taken is:
t=11,111\int_{293}^{353}\frac{\mathrm{d}T}{(376.3-T)}= 11,111\ln\left({\frac{83.3}{23.3}}\right)
= \underline{\underline{14,155\ s}} (or 3.93 h).
The steam is turned off for 7200 s and during this time a heat balance gives:
(1000\times4000){\frac{\mathrm{d}T}{\mathrm{d}t}}=-(10\times6)(T-293)66,700{\frac{\mathrm{d}T}{\mathrm{d}t}}=293-T
The change in temperature is then given by:
\int_{353}^{T}{\frac{\mathrm{d}T}{(293-T)}}={\frac{1}{66,700}}\int_{0}^{7200}\,\mathrm{d}t\mathrm{ln}\,{\frac{-60}{293-T}}={\frac{7200}{66,700}}=0.108
and: T = 346.9 K.
The time taken to reheat the liquid to 353 K is then given by:
t=11,111\int_{346.9}^{353}\frac{\mathrm{d}T}{(376.3-T)}= 11,111\ln\left({\frac{29.4}{23.3}}\right)
= \underline{\underline{2584\ s\ (0.72h).}}