Question

The synthesis reaction is highly exothermic; therefore, cooling water must be pumped through the jacket of the reactor to maintain a stable temperature. Find the enthalpies of the main reaction and side reaction. (Use Hess’ law if necessary.) Based on the heats of reaction, calculate the necessary rate of cooling, and the rate of cooling water flow required. (Cooling water is available at 5oC, with max ΔT = 15oC)

Main: C3H6 + 1.5O2 --> CO2 + H2O

side: C3H6 + 4.5O2 --> 3CO2 + 3H2O

Based on the data from the previous question, select an appropriate pipe size from the NPS schedule for your cooling water flow. Fluid velocity should be at least 1.5m/s to prevent the fouling of the pipes.

Answer 1

The enthalpy of reaction is given by:

Enthalpy of reaction = Enthalpy of Products - Enthalpy of Reactants

For the main reaction; the standard enthalpy of reaction is given by:

For the side reaction; the standard enthalpy of reaction is given by:

The enthalpy of reaction can be determined if we know the standard enthalpy of formation of the reactants and products.

ΔHoCO2 = - 393.5 kJ / mol

ΔHoH2O = - 292.74 kJ / mol

ΔHoC3H6 = 20.41 kJ / mol

ΔHoO2 = 0 (By definition)

Therefore; substituting these values in the above equations:

ΔHomain = - 393.5 - 292.74 - 20.41 = - 706.65 kJ / mol

ΔHoside = - 3 X 393.5 - 3 X 292.74 - 20.41 = - 2079.13 kJ / mol

The total heat released is given by:

H = ΔHomain + ΔHoside

H = - 706.65 - 2079.13 = - 2785.78 kJ/mol

The necessary rate of cooling = - H = - (- 2785.78) = 2785.78 kJ

The mass flow rate of water (M) can be calculated by

Cooling rate = M X Specific heat of water X Temperature difference

Specific heat of water = 0.075232 kJ/mol

Temperature difference = 15 oC

2785.78 = M X 0.07532 X 15

which gives

M = 2466 mol/s

Molar mass of water = 0.018 kg/mol

Therefore;

Mass flowrate of water = 0.018 X 2466 = 44.388 kg/s

In terms of density, area of flow and velocity; the mass flow rate of water can be expressed as:

M = Density X Area X Velocity

Here;

Density = 1000 kg/m3

Velocity = 1.5 m/s

Area = π X d2 / 4 (d = diameter of pipe)

Therefore;

44.388 = 1000 X 1.5 X π X d2 / 4

which on solving gives;

d = 0.19 m

0.0254 m = 1 inch

therefore;

d = 0.19 / 0.0254 = 7.48 inch

From the NPS schedule; a pipe of nominal size 8 inch should be used.