Thermodynamics- Combustion -Conservation of Energy

Question

I was able to figure problem one, but I need help with problem 2. Thank you for the help!

Problem #1: Ethane (C2H6) gas at mass flow rate of 1.5 kg/min is mixed with air (O2 + 3.76N2) at mass flow rate of 1 kg/sec in a steady-state, steady-flow isobaric combustion process. All reactants enter the process at 298 K and 100 kPa, and the fuel burns to complete combustion.

a) Determine the Air-Fuel Ratio (AFR)

ANS: 40 kg air/kg

b) Write a balanced reaction equation based on one mole of fuel

ANS: C₂H₆+(7/2)(O₂+3.76N₂) ----> 2CO₂+3H₂O+(7/2)(3.76N₂)

c) Determine the excess air fraction (%)

ANS: 149.39%

Problem #2: Assuming that the combustion products exit the system at 1000 K, determine:

a) The volumetric flow rate of the combustion exhaust (m3 /s)

b) The rate of heat transfer to/from the system (kW)

Answer 1

Hey Raul,

Before answering problem #2, in regards to Question #1 (a), the question didn't specify whether it wanted theoretical or excess air/fuel ratio. This is just a minor observation.

Question 2 (a)

For this question we will use Charles Law to determine the exit volumetric flow rate. But before that can performed, inlet volumetric flow rate needs to be determined. Since the inlet mass flow rates are a combination of two different gases, a combined volumetric analysis is in order. Firstly, lets find the density of the air and look up the vapor density of ethane.

The density of air can be found using the ideal gas law: ρ = P/(RT) = 100,000/(286*298) = 1.17 kg/m³.

Consulting tables, ethane's density at the same conditions is 1.264 kg/m³.

The inlet volumetric flow rate is as follows: (m/ρ)_total = (m/ρ)_ethane + (m/ρ)_air.

Plugging in values: V₁ = (0.025/1.264)_ethane + (1/1.17)_air = 0.87m³/s.

Using Charles Law (eg. the law of volumes) to find V₂.

(V/T)₁ = (V/T)₂ , (0.87/298) = (V₂/1000), ∴ V₂ = 2.92 m³/s.

Question 2 (b)

To find the heat transferred to the system, enthalpy data is first acquired from data tables. The following is a list of enthalpies for products and reactants.

CO₂ = -393.5 kJ/mol

H₂O = -241.8 kJ/mol

O₂ = 0

C₂H₆ = -84 kJ/mol

N₂ = 0

And the equation used for this is: ∑h_(products) - ∑h_(reactants).

Using the balanced equation from question 1 (b) and plugging in enthalpies: [-2*393.5-3*241.8 - 3.5*3.76*0]_products-[-84-3.5*(0+3.76*0)]_reactants = -1428.4 kJ

In summary, the answers are 2.92 m³/s and -1428 kJ*.

*Question 2 (b) asked for thermal power, but considering the mass is flowing it already implies power. And fluid thermal power is defined as: m dot * ^Δh.