# The Efficiency of Heating Using a Flame

Topics: Temperature, Heat, Energy Pages: 5 (1088 words) Published: March 12, 2013
The Efficiency of Heating Using a Flame

Thermodynamics II

Contents
Introduction2
Observations2
Time/Temperature Plot3
Calculations and Results3
1)Enthalpy change of gas burned over the duration of the test3 2)Heat transfer to the water in the kettle (Qw)4
3)Heat transfer to the water which evaporates from the kettle (Qev)4 4)Heat transfer to the kettle from initial to final state (Qk)4 Find, as a fraction of 1, the values for 2, 3 and 4.4
5)Air–Fuel Ratio4
6) Enthalpy of combustion with the water formed by the combustion in the gas phase, (ΔhcoH2Ovap)5 Using Δhco H2Ovap, find as a fraction of 1), the values for 2), 3) and 4).6 Discussion and Conclusion6

Introduction
The purpose of this experiment was to learn about the heat transfer process and determine the energy transferred from a gas flame to the water contained in a stainless steel kettle. The value calculated for the energy transferred is compared with the energy change associated with burning the gas (from calorific values) and the molar enthalpy of combustion of the gas mixture. Observations

Atmospheric pressure: Pa = (767.15 – 2.49)mmHg = 764.66 mmHg
1 atm = 760 mmHg = 101.325 kPa
So, 764.66 mmHg = 101.946 kPa

Ambient temperature: θa = 20 oC
Can No.: = 5
Initial mass of can:mf1 = 114.06 g
Mass of kettle:mk = 338.43 g
Mass of kettle and water:m1 = 1321.26 g
Initial temperature of water: θ1 = 20.4 oC
Final temperature of water: θ2 = 80.0 oC
Final mass of kettle and water:m2 = 1320.26 g
Final mass of can:mf2 = 105.27 g
Heating time (s)
30 sec. intervalsTemp of water
(oC)
020.4
3025.1
6029.2
9034.1
12037.3
15042
18043.4
21048.6
24049.1
27053.2
30056.1
33058.5
36059.8
39061.7
42062.5
45063.5
48065.1
51067.2
54069
57070.1
Heating time (s)Temp of water (oC)
60072.6
63074.8
66076.5
69080

Cooling time (s)
(2 min intervals)Temp of water (oC)
81078.9
93078.8
105078.6
117078.4

Time/Temperature Plot

Figure 1. Temperature vs. Time plot
Calculations and Results
Enthalpy change of gas burned over the duration of the test ΔH=(m_f2-m_f1)(∆h_co)
Taking the enthalpy of combustion of the gas (∆h_co) as 49.6 MJ/kg; ΔH=((105.27-114.06)/1000)(-49.6×〖10〗^(-6) )
ΔH=436.0 kJ
Heat transfer to the water in the kettle (Qw)
Q_w=(m_2-m_k )(h_2-h_1 )
Values for h_2 and h_1 interpolated from table,
Q_w=((1320.26-338.43)/1000)(335.3-82.9)
Q_w=247.81 kJ
Heat transfer to the water which evaporates from the kettle (Q_ev) 〖Q_ev=〖(m〗_2-m_1)(h〗_g-h_1)
Values for h_g and h_1 interpolated from table,
Q_ev=((1320.26-1321.26)/1000)(2643.2-82.9)
Q_ev=-2.56 kJ
Heat transfer to the kettle from initial to final state (Q_k) Q_k= m_k×c_p×∆T
c_p value for the stainless steel kettle = 0.57 kJ/kg K
Q_k=(338.43/1000)(0.57×〖10〗^3)(80.0-20.4)
Q_k=11.50 kJ
Find, as a fraction of 1, the values for 2, 3 and 4.
Q_k⁄∆H= 247.81⁄436.0=0.568
Q_ev⁄∆H= 2.56⁄436=0.00587
Q_k⁄∆H= 11.50⁄436=0.02638
Air–Fuel Ratio
Fuel by mass: 30% Propane + 70% Butane
Combustion equation for propane;
C_3 H_8+5(O_2+3.76N_2 )→3〖CO〗_2+4H_2 O+18.8N_2+HEAT
Molecular masses for the compounds in the equation (in kgs); C_3 H_8=44.0962 kg
〖5×O〗_2= 159.99 kg
18.8×N_2=526.6632 kg
3〖×CO〗_2=132.027 kg
4×H_2 O= 72.059 kg
Combustion equation for butane;
C_4 H_10+13⁄2 (O_2+3.76N_2 )→4〖CO〗_2+5H_2 O+24.4N_2+HEAT Molecular masses for the compounds in the equation (in kgs); C_4 H_10=58.123 kg
〖13⁄2×O〗_2= 207.987 kg
24.4×N_2=683.5416 kg
4〖×CO〗_2=176.036 kg
5×H_2 O= 90.074 kg
Air – Fuel Ratio = m_A/m_F =([(159.99+526.6632)(0.3)+(207.987+683.5416)(0.7)])/((44.0962×0.3)+(58.123×0.7))= 15.396

6) Enthalpy of combustion with the water formed by the combustion in the gas phase, (Δh_co H_2 O_vap) Δh_co H_2 O_vap = Δh_co + mH_2 O_vap h_fg
Where:
Δh_co =...