Question

Toluene Chiller A stream of liquid toluene at 100°C is fed to a chiller at a flow rate of 4.00 kg/min. The stream exits the chiller at 15°C. The heat capacity for liquid toluene can be estimated from the following equation (Cp in kJ/(mol*°C) and T in °C): = 0.1488 + (32.4 10−5)

a) Draw a flow sheet for this problem, labeling all known variables;

b) State assumptions required to solve the energy balance;

c) Write the mass balance for this problem;

d) Write the appropriate form of the energy balance and simplify;

e) Solve for the rate of heat transfer (kJ/min) for this process.

Answer 1

a)

b)

assumptions:

there is no heat loss/gained via "surroundings", that is, all heat from chiller is equal to the negative heat of toluene

this is steady state

c)

Mass balance:

for toluene:

Inlet = outelet

we know that

inlet = 4 kg/min, substitute:

4 kg/min = outlet

d)

Energy balance:

-Qchiller = Qtoluene

Qtoluene = dH

dH = n*Cp*dT

dH = n*(0.1488 + 32.5*10^-5)/T)dT

note, you stated 0.1488 and 32.4 10-5 as CP, this is most likely 1/T

mol of tolune = mass/MW = (4000)/92.14 = 43.4121 mol of toluene

so

H = (43.4121)(0.1488)(T2-T1) = (43.4121)(0.1488)(15-100) = -549.07 kJ/min

so

the total rate -->  = 549.07 kJ/min of cooling