Liquid methanol is fed to a space heater at a rate of 12.0

Problem 4.71

Liquid methanol is fed to a space heater at a rate of 12.0 L/h and burned with excess air. The product gas is analyzed and the following dry-basis mole percentages are determined: CH30H = 0.45%, CO2 = 9.03%, and CO = 1.81 %. (a) Draw and label a flowchart and verify that the system has zero degrees of freedom. (b) Calculate the fractional conversion of methanol, the percentage excess air fed, and the mole fraction of water in the product gas. (c) Suppose the combustion products are released directly into a room. What potential problems do you see and what remedies can you suggest?

                                                       Step by step solution

Step 1 of 3

a)

The equations for the complete and incomplete combustion of propane are given below

The mole fractions of dry product gases are,

Convert the volumetric flow rate (12.0L/h) to a molar flow rate, using its density (0.792 g/Ml) and molar mass (32.04 g/mol).

Sketch and label the diagram for this process as shown below:

Here,  is molar flow rate of oxygen in the feed air, and n, is molar flow rate of water produced, x is the mole fraction of nitrogen in the dry product gas. The molar flow rate of nitrogen in the feed air is 3.76 (79/21) times the molar flow rate of oxygen in the feed air.

The mole fraction of  in the product gas is one minus the mole fractions of other dry gases.

The number of degrees of freedom is given by the equation.

Here, unknowns is number of unknowns and depen, is number of independent equations that

can be written for the process.

In this process, the unknowns are molar flow rates   and mole fraction (x), so the number of unknowns is 4. The atomic balance equations can be written for C, H, O each and the mole balance for nitrogen molecule can be written.

So, the number of degrees of freedom is 0.

= 4 - 4

= 0

Therefore, the number of degrees of freedom is 0.

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