Problem 4-14Purified medium-Btu gas at 77F is burned in 110
percent theoretical air. It leaves the combustion chamber at 1700F.
5 percent of the heat released is lost to the environment.
Calculate the useful heat added to the chamber in Btu per pound
mass and Btu per standard cubic foot (at 14.696psi and 60F) of the
gas.
GivenTR = 77F(298K)TP= 1700F (1200K)Fuel: Medium Btu-gas
(purified)
FindThe useful heat added to the chamber in: (1)Btu/lbm(2)
Btu/scf @ 14.696psi and 60F of the gas.
Derive the EquationThe feedstock is burned with a mixture of
oxygen (rather than air) and steam
In Air: C + O2 CO2 (Eq. 4-9)
CO2 from this reaction reacts further with additional C in the
rich mixture to give: C + CO2 2CO (Eq. 4-10)
In Steam: C + H2O CO + H2 (Eq. 4-11)
Stoichiometric ReactionCombination of products and combustion
with oxygen.
With 110% Theoretical Air
First Law EquationSee Table 4-7 and 4-8 for values of hf . 25oC
= 77oF , 1700oC = 1100K (no need for interpolation in Table
4-8)
Net Heat added to the system
Useful HeatSince 5% is released to the environment,
Standard Cubic FootEnglish Units
Standard Cubic Foot
Standard Cubic Foot
Standard Cubic Foot
2.2.2 MEDIUM-BTU GAS
In the production of medium-Btu gas, pure oxygen rather than air
is used as combusting agent, which results in an appreciable
increase in the heating value, by about 300400 Btu/scf. The product
gas predominantly contains carbon monoxide and hydrogen with some
methane and carbon dioxide. It is primarily used in the synthesis
of methanol, higher hydrocarbons via FischerTropsch synthesis, and
a variety of other chemicals. It can also be used directly as a
fuel to generate steam or to drive a gas turbine. The H2-to-CO
ratio in medium-Btu gas varies from 2:3 (CO-rich) to more than 3:1
(H2-rich). The increased heating value is attributed to higher
contents of methane and hydrogen as well as to lower concentration
of carbon dioxide, in addition to the absence of nitrogen in the
gasifying agent.*Tama until Q fuel184 Btu/lb
*A sfc is obtaoned at14.696psi and 60F of the ga. Since it is a
gas, the density at these conditions are obtained from the idel gas
equation PV=mRT *A sfc is obtaoned at14.696psi and 60F of the ga.
Since it is a gas, the density at these conditions are obtained
from the idel gas equation PV=mRT *A sfc is obtaoned at14.696psi
and 60F of the ga. Since it is a gas, the density at these
conditions are obtained from the idel gas equation PV=mRT *A sfc is
obtaoned at14.696psi and 60F of the ga. Since it is a gas, the
density at these conditions are obtained from the idel gas equation
PV=mRT *
