CHEMICAL ENGINEERING

HEAT TRANSFER 171

PROBLEM 9.39

A stirred reactor contains a batch of 700 kg reactants of specific heat 3.8 kJ/kg K initially
at 290 K, which is heated by dry saturated steam at 170 kN/m^2 fed to a helical coil. During
the heating period the steam supply rate is constant at 0.1 kg/s and condensate leaves at
the temperature of the steam. If heat losses are neglected, calculate the true temperature of
the reactants when a thermometer immersed in the material reads 360 K. The bulb of the
thermometer is approximately cylindrical and is 100 mm long by 10 mm in diameter with
a water equivalent of 15 g, and the overall heat transfer coefficient to the thermometer
is 300 W/m^2 K. What temperature would a thermometer with a similar bulb of half the
length and half the heat capacity indicate under these conditions?

Solution

The latent heat of dry saturated steam at 170 kN/m^2 and 388 KD2216 kJ/kg.

Therefore heat added to the reactorD 

2216 ð 0. 1 D 221 .6kJ/sD 221 .6kW

which is equal to the increase in enthalpy, dH/dt.
The enthalpy of the contents, neglecting the heat capacity of the reactor and lossesD
mCpdT/dtD
700 ð 3. 8 dT/dtor 2660 dT/dtkW

∴ 2660 dT/dtD 221. 6

and the rate of temperature rise, dT/dtD 0 .083 deg K/s.
At timet, the temperature of the reactants is:

TD 

290 C 0. 083 tK(i)

The increase in enthalpy of the thermometer is equal to the rate of heat transfer from
the fluid, or:

mCp (^) tdTt/dtDUAt
TTt (ii)
where the subscripttrefers to the thermometer.
∴
15 / 1000 ð 4. 18
dTt/dtD 0. 300
$ð 0. 010 ð 0. 100 
TTt
and: dTt/dtD 0. 0150
TTt deg K/s
At timets, the temperature of the thermometer is therefore:
TtD 290 C[0. 0150
TTt ]tK (iii)
WhenTtD360 K, then substituting from equation (i):
360 D 290 Cf 0 .0150[290C 0. 083 t360]gt