# The Efficiency of Heating Using a Flame

Thermodynamics II

Contents

Introduction2

Observations2

Time/Temperature readings2

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

Time/Temperature readings

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 =...

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