Thermos Flask (Project)

Adiabatic ContainerEMM3110

Wan Lokman Bin Wan Ahmad (177109)

Muhammad Redzuan Bin Kamarudin (180590)

Akmal Aizat Bin Mohamad Hanif (177831)

Muhammad Amri Bin Nasruddin (180400)

Muhammad Amirrullah bin Ahmad Suhari (178423)

Group Members

• Process is adiabatic if there is no exchange of heat between system and environment.

• Example:

– A thermos flask.

• Concept is to keep the heat

from leaving the system so that

the water inside the container

remains hot.

What is Adiabatic?

Work and Temperature

• First law: dU = dQ – dW

• Adiabatic process: dU = -dW

• If system does work (dW > 0), dU < 0 (system cools)

• If work is done on system (dW < 0), dU > 0 (system warms)

• Cylindrical shape is used for the adiabatic container.

• Reason.

– Easily handle.

– Hold large amount of water

– Stable design

Design

Inner Radius : 8.5 cmOuter Radius: 13 cmHeight: 20 cm

Governing equation

• For an adiabatic process of ideal gas equation we have

• Where γ is the ratio of specific heat (ordinary or molar) at constant pressure and at constant volume;

• Suppose in an adiabatic process pressure and volume of a sample of gas changs from (P1, V1) to (P2, V2) then we have P1(V1)γ=P2(V2)γ=KThus, P = K/Vγ

PVγ = K (Constant) (14)

γ = Cp/Cv

Results

• Recorded Observation (Temperature (oC)

Minutes Open Air Closed container

0 67.0oC 67.0oC

10 58.0oC 63.0oC

20 51.5oC 59.0oC

30 43.0oC 53.0oC

40 38.5oC 49.0oC

Calculation

Available parameters:– Surrounding temperature=27 °C

– Volume of air=6.5x10-3m3

– Volume of water=0.2m3

– Density ,• water =1000kg/m3

• air=1.2kg/m3

– In adiabatic container Ti=67 °C and Tf=49.0 °C

– In open surrounding Ti=67 °C and Tf=38.5 °C

– Time taken=40 minutes

Calculation

Heat transfer

Governing equation

Q= Heat Transfer Tf= Final Temperature

t= Time Take d= Thickness

k= Thermal Constant A= Area of conduction

Ti= Initial Temperature

Calculation

Heat loss in adiabatic container :

• Water:

– m,(mass of water)= pV

= (1000)(200×10^-6) =0.20 kg

– Q=-(0.20)(4200)(49-67) = -15120 J

Calculation

Heat loss in open surrounding:

– Q= -0.20×4200× (38.5-67) = -23940 J

Precentage heat loss reduction:

– 63.157%

Final Results

• Expected final temperature after 40min– Expected in closed container :60oC

– Obtained results : 52oC

• Total Heat Loss– Open Air : 39.7765 J

– Closed Container : 34.3334 J

• Total Work Done by water heat loss– Open Air : 𝟎. 𝟎𝟏𝟔𝟓𝑱𝒔−𝟏

– Closed Container : 𝟎. 𝟎𝟏𝟒𝟑𝑱𝒔−𝟏

• The heat escape from the container as the container is not properly sealed.

• The heat escape when the measurement is taken down.

• The material used is not proper material for thermal insulation.

Limitation

Conclusion

• The result of our experiment coherent to our expectation which is our design suppose to get less work done for our adiabatic container compared to open air controlled test.

• Although the experiment that we conducted follows ours expectation. The value of work done is too high for the container.