Lab Report First Law of Thermodynamics
Lab Report First Law of Thermodynamics
Lab Report First Law of Thermodynamics
KEJURUTERAAN MEKANIK
KAMPUS KEJURUTERAAN
UNIVERSITI SAINS MALAYSIA
14300 NIBONG TEBAL, PULAU PINANG
---------------------------------------------------------
------
EML 211/2 –
2018/19
ENGINEERING
LABORATORY 1
----------------------------------------------------------
------
Title: First Law of Thermodynamics
12/11/2018
Signature: ...........................................................
Grader’s Comments :
…………………………………………………………………………………
………
…………………………………………………………………………………
………
…………………………………………………………………………………
………
I-OPEN SYSTEM (KETTLE)
Apparatus:
1. Kettle (2 liter kettle without automatic switch)
2. One thermocouple with temperature reader
3. Weighing scale (capacity 2 kg with resolution 0.001kg)
4. Timer
5. Clamp meter (watt meter)
Introduction
From the First Law of Thermodynamics that is based on the conservation of energy
and also net heat equals net work in a cycle, there are several corollaries that can be extracted
to explain more of the First Law of Thermodynamics.
Corollary I
There exist a property of a closed system where change in its value is the same as the
difference between heat supplied and work done for a non-flow process, du = δq - δw.
Corollary II
Internal Energy for an isolated system does not change.
From δq = du + δw
For a small system δq = 0 and δw = 0 , so du = 0
Corollary III
The change of internal energy in a thermodynamic cycle is zero
(δQ – δW)AC = 0 = dU
Corollary IV
Perpetual motion machine is impossible.
Since δQ = δW
If δQ = 0, so δW = 0
Theory:
A kettle is an perfect example of an open system. The application of The First Law of
Thermodynamics on kettle( open system) will results steady flow energy equation (SFEE)
which can be written in formula below:
1 2 2
Q – W = m( h 2−h1 ) + m ( V 2−V 1 ) + mg( Z 2−Z 1) ………. (1)
2
Where
m is the mass flow rate.
Q is the rate of heat transferred (J/s).
W is the rate of work done or power (J/s).
m( h 2−h1 ) is the rate of enthalpy change.
1
m ( V 22−V 21 ) is the rate of change of kinetic energy.
2
mg( Z 2−Z 1 ) is the rate of change of potential energy.
The subscript 1 and 2 denote quantities for inlet and exit, respectively. The mass flow rate is
denoted by m with the unit of kg/s.
With the absence of work , kinetic and potential energies for the water boiling process, Eq.
(1)
Can be reduced to
Q= m( h 2−h1 ) ………….(2)
Based on Eq. (3), the temperature rise of liquid water is due to the heat transfer supplied to
water. Expressing the mass flow rate as m/t, Eq. (3) can thus written as
Upon reaching boiling temperature , rate of heat transferred to evaporate the water from
boiling condition is based on
Q = me h fg/t e ……………(5)
Where
m e is mass of water evaporated
h fg is latent heat of vaporization of water
t e is time taken for the water to evaporate
The efficiency of the kettle is denoted by ῃ where it is a ratio between total rate of heat
transferred over electrical consumption.
Methodology
1. An electrical kettle is put on a weighing scale.
2. The weighing scale is set to zero value so that the weight of kettle is not included.
3. One liter of water is poured into the electrical kettle.
4. The temperature of the water is determined by using a thermocouple.
5. The initial temperature of the water and the initial mass of water are recorded.
6. The electricity is switched on.
7. The temperature and the time are recorded until the water boil with a 30 seconds time
interval.
8. The time and the reducing weight of water in the kettle are recorded as the water
evaporates until the weight of water left is 1.0litre.
9. The voltage and current are determined to determine the electrical consumption for
the heater.
10. A table is constructed to obtain the necessary data to determine the following output:
a) Rate of temperature rise to boil the water (table for Temperature and Time is
constructed and a graph of Temperature versus Time is plotted).
b) Rate of water evaporated from the kettle (table for weight of water left, water
evaporated and time is constructed; a graph of weight of water evaporated versus
time is plotted).
c) Rate of heat transferred to water to the boil.
d) Rate of heat transferred for 0.5 liter of water to evaporate.
e) Total power consumption.
f) The voltage and current are determined from the power supply to determine the
power input.
g) Efficiency of the kettle is calculated.
Result
Time (min) Temperature of water (°C) Mass of water (g)
0.5 34.00 1000
1.0 41.8 1000
1.5 48.3 1000
2.0 56.3 1000
2.5 68.2 1000
3.0 74.2 1000
3.5 80.7 998
4.0 88.2 993
4.5 94.4 987
5.0 98.6 981
5.5 98.8 970
6.0 98.8 962
6.5 98.8 962
7.0 98.9 950
7.5 98.9 938
8.0 99.0 929
8.5 98.9 925
9.0 98.9 918
9.5 98.9 907
10.0 98.9 896
10.5 98.9 884
11.0 98.9 869
11.5 98.9 848
12.0 98.9 830
12.5 98.9 815
13.0 98.9 809
13.5 98.9 798
14.0 98.9 790
14.5 98.9 775
15.0 98.9 752
15.5 98.9 730
16.0 98.9 719
16.5 98.9 705
17.0 98.9 698
17.5 98.9 680
18.0 98.9 675
18.5 98.9 650
19.0 99.0 633
19.5 99.1 621
20.0 99.1 590
20.5 99.0 572
21.0 99.1 558
21.5 99.0 536
22.0 99.1 519
22.5 99.1 496
Table 1: Table of time measured, temperature of water and mass of water remaining
Mass of water eavaporated against Time
1200
1000
800
Mass of water/g
600
400
200
0
5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5
0. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22.
Time/m
Graph 1
100
Water Temperature/ c⃘
80
60
40
20
0
5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5
0. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22.
Time/m
Graph 2
Calculation
From Graph 2, a linear portion is considered. The rate of temperature rise to boil the water,
∆ T /t is equal to the gradient of the graph in the linear portion, that is:
From Graph 1, the linear portion is considered. The rate of water evaporated from the kettle,
m e / t e is equal to the gradient of the graph in the linear portion, that is:
me ¿
=mass of evaporated water ¿ kettle evaporate ¿
te time taken for the water ¿
496−30
¿
1350−270
¿ 4.315 ×10−4 kg / s
T b=99.1 ℃
Power input:
P¿ =VI
¿ 240 × 4.81
¿ 1154.4 W
Efficiency of kettle:
P
η=¿ out ×100 %
P¿
1002.75
¿ ×100 %
1154.4
¿ 86.86 %
Discussion
Based on the experiment, we can measure the water boils at 94.4oC, which is slightly lower
than the actual boiling point of water which is 100oC. Time taken to boil 1000 g of water is
270 s, while time taken to evaporate 500 g of water is 1080 s. The experiment value of
boiling point is 94.4oC, which deviates from the theoretical value 100 0C. This is because the
atmosphere pressure is assumed to be 101.3kPa, where it might not be actual value of the
experiment environment. Latent heat of vaporization of 500 g water is almost the same as
sensible heat to boil 1000 g of water. This means that if the same amount of water was
evaporated and boiled, latent heat of vaporization of the water will be higher than sensible
heat needed to boil the same amount of water.There are several reasons why the measured
value has a difference with the theoretical value of this experiments.
First, the calibration error of the thermocouple in this experiment that cause bias error
during the experiment. The calibration process of thermocouples was not available at the
starting time.The usage of plastic outside the kettle can affects the difference in temperature
because plastic is a good heat insulator compared to metal. It can minimize the heat loss to
the surrounding in this experiment.Fluctuation of the heat transfer in the water molecule
inside the kettle and the fluctuation in the digital acquisition system of thermocouples
instrument.
Precautions that should be taken in this experiment are the fans and door in the
laboratory should be switched off and turned off to minimize the possible air flow inside
laboratory during this experiment. It is because inconsistent air flow causes fluctuating
reading and increase the heat loss from the kettle to the surrounding.Thermocouples should
be calibrated and set to zero before starting the experiment.Pure water should be used
throughout this experiment because the impurity of water can change its boiling point.
The ideal condition means the condition in which the results can be theoretical
predicted. Some assumption for the ideal condition for the boiling point of water is 100 oC,
atmospheric pressure is 101.3 kPa, the initial temperature of water is equal to ambient
temperature, the latent heat of vaporization is transferred after boiling temperature is achieved
and the mass of water remain unchanged (no evaporation) before reaching boiling point.
Furthermore, according to second law of thermodynamics, it is impossible for a heat
engine cycle to receive heat from a reservoir and produce a similar amount of work. Heat loss
must exist during the experiment and the efficiency of the kettle must be lower than 100%.
The water temperature remains constant when it reach 94.4 oC. This is because it reached its
boiling point. Heat supply is no longer to increase the temperature but to break the
intermolecular bond of the water molecules. This is named Latent Heat of Vaporization.
The efficiency of the electric kettle can be improved further by using entrapped air in
glass plates which act as heat insulators in the wall of the kettle to minimize heat lost to the
surrounding. The efficiency of electric heater is higher than the kettle that uses in stove.
Conclusion
The rate of sensible heat transferred for boiling 1000 g of water is, Q b = 1.02522 kW . The
efficiency of the kettle for boiling 1000g of water , η = 86.86 %. The rate of latent heat of
vaporization of water is Q e =¿ 0.98027 kW.
References
1. http://www.nrcan.gc.ca/energy/efficiency/industry/technical-info/tools/boilers/5431
2. http://automationwiki.com/index.php/Factors_Affecting_Boiler_Efficiency
3. https://greennav.wordpress.com/2008/02/12/understanding-%E2%80%93-how-a-
boiler-works-%E2%80%93-boiler-efficiency/
4. Thermodynamics notes 2016 edition, Prof Dr. Hj Zainal Alimuddin bin Zainal
Alauddin and Dr. Teoh Yew Heng
5. Thermodynamics Property Table, Steam Table, Refrigerant R134a Table Ideal Gas
Constants
6. Lab Manual EML 211/2 by Dr. Abdullah Aziz Saad.
II– HAIRDRYER EXPERIMENT
Introduction
Heat, work and mass all cross the boundary. From the first law of thermodynamics, the
energy into the system has to equal the energy out for steady state. From conservation of
mass, the mass in has to equal the mass out for steady state. This experiment requires all of
the energy terms associated with the hairdryer. The energy going in includes the electric
work, the total enthalpy of the incoming air, and the kinetic energy of the incoming air.
Energy out includes the total enthalpy of the outgoing air, kinetic energy of the outgoing air,
and any heat transfer from the case to the ambient. Potential energy differences between the
inlet and outlet are also considered.
Figure 1
The energy into the hairdryer includes the electric work, the total enthalpy of the incoming
air, and the kinetic energy of the incoming air. Energy out of the hairdryer includes the total
enthalpy of the outgoing air, kinetic energy of the outgoing air, and any heat transfer from the
case to the ambient. Potential energy differences between the inlet and outlet are also
considered. Some of the terms involved are very significant to the overall energy balance and
some are almost negligible.
Apparatus
The basic first law of thermodynamics for the hairdryer can be written as:
2
V2 V
( )
Ẇ elec −Q̇ loss + ṁ ( h¿ −hout ) + ṁ ¿ − out + ṁ g ( z ¿ −z out )=0
2 2
2
V2 V
( ) (
Ẇ elec −Q̇ loss + ṁ ¿ h¿ + ¿ + g z ¿ −m˙out h out + out + gz out =0
2 2 )
V 2
V 2out
Ẇ elec + ṁ (h + + g z )=Q ˙
¿ ¿
2
¿
¿ (
loss+ m˙out hout +
2 )
+ gz out (Equation 1)
A hairdryer is designed to produce heat, so a high power factor is desired. This means that the
reactance inside the circuit of the hairdryer must be kept as low as possible. Most hairdryers
have a power factor very close to one. A power factor of one is assumed for this experiment.
The electrical power input is used to supply the power to the electrical heater for the hair
dryer and the power to drive the fan.
Heat Loss:
˙ −T ) (Equation 3)
Q=hA(T s a
h is the convective heat transfer coefficient. This number is given as 5w/m²K. The area is
estimated based on our judgment about how much of the nozzle is actually warm. The surface
temperature T s is measured using an infrared thermometer. The temperature varies across the
surface, so we must make a judgment about what to use as an average temperature. T a is the
ambient temperature.
Cp is the specific heat of the incoming air, given as 1.005 kJ/kgK. The temperature T is the
absolute temperature of the incoming air (room temperature) in K.
Velocity Out:
Velocity of the exit air is determined directly using the hot wire anemometer. Hot wire
anemometers use a very fine wire electrically heated to some temperature above the ambient.
Air flowing past the wire cools the wire.
Air Density:
The air density is determined for the exit temperature measured using the thermodynamic
property table.
V is the average exit velocity and A is the area of the region of interest.
Velocity In:
Since the inlet area of the hairdryer is much larger than the exit, the velocity will be very low,
so velocity in is assumed as zero. However, continuity equation can be used to calculate the
input velocity. This equation states in a steady flow, when there is no accumulation of fluid,
the mass of fluid flow per unit time is equal to the fluid flow leaving a system at the same
time.
Potential Energy:
The vertical distance between the center of the inlet and the center of the outlet is measured.
The elevation change is used to calculate the potential energy change.
Methodology
2. The necessary parameters on the dimensions of the hair dryer is measured using a
5. A hot wire anemometer is placed at the exit of the hair dryer to measure the velocity
of air, while the thermocouple is also placed at the same place to measure the
A clamp meter is clamped over the electric cable of the hair dryer to determine the
current and the potential difference supplied. The temperature readings in the
experiment is repeated 8 times to obtain 8 sets of data in order to calculate the average
value for the data to increase the accuracy of data. All 8 sets of data and the averaged
Data analysis
DATA
PHYSICAL CHARACTERISTICS
Fixture Inside Diameter = 6.45 cm
Air Inlet Area of Dryer = 32.67 cm
Elevation Difference Inlet/Outlet = 0 cm
Air Exit Temperature = 95.2oC
Exit velocity = 10.67 m/s
ELECTRICAL CHARACTERISTICS
Input Voltage = 240 V
Input Current = 2.75 A
HEAT LOSS DATA
Average Temp. of Nozzle = 32.77oC
Length of Heated Surface = 12.45 cm
Outside Diameter of Nozzle = 4.551 cm
MISC. DATA
Room Temperature = 28.5oC
Barometric Pressure = 101.06 kPa
CALCULATE
Density of exit air = 0.9560 kg /m3
Mass flow rate = 0.0333 kg/s
Change of enthalpy = 2.2322 kW
Change of potential energy = 0 J
Change of Kinetic energy = 1.8956 J
Heat loss = 0.3800 W
DETERMINE
Electrical power output = 660 W
Total thermal power in = 10.755 kW
Total thermal power out = 12.329 kW
% difference power between thermal in and out = 14.63 %
Calculation
Change of enthalpy,
Cp = 1.005 kJ/kgK
Δ H =ṁ(h out – h¿ )=ṁC p (T out – T ¿ )
¿ 0.0333 kg /s ×(1.005 kJ /kgK )× [ 368.35−301.65 ] K
¿ 2.2322 kW
Assumption: C ¿ ≈ 0 m/s
Conclusion
The percentage difference between input power and output power is 14.63%. The
efficiency of hairdryer is 1.146.The rate of heat loss is 0.38 W. In order to prevent further
heat loss and increase the efficiency, hair dryer can be coated with heat insulator to prevent
heat loss in form of radiation and conduction to the surrounding.
Reference
1. https://sites.google.com/site/laurasapphysicslabportfolio/fluidthermolabs/20-
hairdryer-efficiency-lab
2. https://sites.google.com/site/kippsscienceclasses/physics/labs/efficiency-of-a-
hairdryer-lab
3. Thermodynamics notes 2016 edition, Prof Dr. Hj Zainal Alimuddin bin Zainal
Alauddin and Dr. Teoh Yew Heng
4. Thermodynamics Property Table, Steam Table, Refrigerant R134a Table Ideal Gas
Constants
5. Lab Manual EML 211/2 by Dr. Abdullah Aziz Saad