ME354
Thermodynamics 2
Name:
Quiz #6 - T02:
ID #:
Problem: Liquid propane (C3 H8 ) enters a
combustion chamber at 25 ◦ C at a rate
of 1.2 kg/min where it is mixed and
burned with 150 percent excess air that
enters the combustion chamber at 12 ◦ C.
If the combustion is complete and the exit
temperature of the combustion gases is
1200 K, determine:
(a) the mass flow rate of air and
(b) the rate of heat transfer from the combustion chamber
Assumptions
1.
2.
3.
4.
steady operating conditions
air and combustion gases are ideal gases
KE = P E = 0
combustion is complete
The balance equation for liquid propane and 150% excess air is
C3 H8 () + 12.5(O2 + 3.76N2 ) −→ 3CO2 + 4H2 O + 7.5O2 + 47N2
Part a)
The air fuel ratio can be calculated as
AF
=
=
mair
mf uel
(12.5 × 4.76 kmol)(29 kg/kmol)
(3 kmol)(12 kg/kmol) + (4 kmol)(2 kg/kmol)
= 39.22 kgair /kgf uel
and
ṁair = (AF )(ṁf uel ) = 39.22 kgair /kgf uel × 1.2 kgf uel /min = 47.1 kgair /min ⇐
Part b)
An energy balance over the combustion chamber can be written as follows:
Qout =
o
o
NR (hf + h − h )R −
o
o
NP (hf + h − h )P
o
Substance
hf
(kJ/kmol)
C3 H8 ()
O2
N2
H2 O(g)
CO2
-118,623
0
0
-241,820
-393,520
h285 K
(kJ/kmol)
h298 K
(kJ/kmol)
h1200 K
(kJ/kmol)
8,296.5
8,286.5
8682
8669
9904
9364
38,447
36,777
44,380
53,848
where we note from Tables A-26 and A-27
o
o
hf () = hf (g) − hf g
= −103, 850 kJ/kmol − (44.097 kg/kmol × 335 kJ/kg)
= −118, 623 kJ/kmol
Substituting we get
Qout = (1)(−118, 623) + (12.5)(0 + 8296.5 − 8682) + (47)(0 + 8286.5 − 8669)
−(3)(−393, 520 + 53, 848 − 9364) − (4)(−241, 820 + 44, 380 − 9904)
−(7.5)(0 + 38, 447 − 8682) − (47)(0 + 36, 777 − 8669)
= 190, 751.25 kJ/kmolC3 H8
The rate of heat transfer is
Q̇ = Ṅ Qout =
ṁ
N
Qout =
= 5, 202.3 kJ/min ⇐
1.2 kg/min
44 kg/kmol
(190, 751.25 kJ/kmolC3 H8 )