The University of Liverpool: ENGG116 – Energy Science. Thermodynamics. PRACTICAL EXERCISE #2: OVERALL HEAT TRANSFER COEFFICIENT Introduction In this experiment, you will measure the heat loss from hot water in a cup. Heat transfer can occur by conduction, convection and radiation. We will assume that radiation is negligible and evaluate the heat transfer coefficient of the cup associated with convection and conduction. The energy loss from the hot water will be calculated based on its change of internal energy and the definition of specific heat capacity: the amount of heat per unit mass required to raise the temperature by one degree Celsius. Objective To determine by experiment the overall heat transfer coefficient for a mug and glass of your choice. On successful completion of this practical exercise, you should have improved your practical skills and your understanding of heat transfer, as well as your technical writing and reporting skills. Apparatus You will need: a measuring beaker, water, an electric kettle¹, a digital kitchen probe thermometer, a timer², and a mug and a glass that each hold about 0.25 litres. Procedure 1. Use the thermometer to measure the room temperature, T room and make a note of it. 2. Use the measuring beaker to fill the mug or glass three-quarters full of cold water with a known volume of water. Then pour the water into an empty kettle and switch on the kettle. 3. When the kettle boils pour the water back into the mug or glass. Start the timer and note the temperature, T water. 4. Record the temperature at 20 second intervals for 15 minutes [expand the results table to fit your data]. Plot the results on a graph with time on the abscissa and AT = Twater - Troom on the ordinate. 5. When the water has returned to room temperature, pour it back into the measuring beaker and note the final volume. Analysis We can express the first law of thermodynamics for a system undergoing a process as (1) (Win − Wout) + (Qin − Qout) + (Emass,in — Emass,out) = AU + AKE + APE where the terms on the left side of the equation represent energy flows across the system boundaries during the process and the terms on the right side represent the change in the energy content of the system during the process. Specifically, Q represents heat transfer across the system boundary, W work transfers and E mass energy transfers occurring as a consequence of mass flows.U, KE and PE are internal, kinetic and potential energies of the system respectively. A is the symbol for 'change in', while the subscripts 'in' and 'out' refer to energy transferred into and out of the system respectively. For this experiment, it is appropriate to define the water in the container as the thermodynamic system undergoing a process, i.e. being cooled to room temperature. The system is stationary through the process and so the changes in kinetic energy and potential energy are zero, i.e. AKE = 1 I used a 1.7litre, 3kW kettle in our kitchen that is similar to this one: http://www.amazon.co.uk/Russell-Hobbs- 18278-Deluxe-Stainless/dp/B0055426XKn 2 I used the one in my mobile phone. 1 The University of Liverpool: ENGG116 – Energy Science. Thermodynamics. ΔΡΕ = 0. There is no work done on or by the system, so Win = Wout = 0 and there is no mass transfer so Emass,in = Emass,out = 0. Consequently, equation (1) can be simplified to (QinQout) = AU (2) The temperature of the water, T is a measure of its internal energy, AU and, by definition, the specific heat capacity of water³, c₁ is the amount the temperature changes with internal energy, i.e. du Cv = dT (3) The differentials were used here to indicate that actually this is gradient of line on a graph of U plotted as function of T. The c is lower case because this is the change per unit mass, and the subscript v indicates the value for constant volume. It can be rewritten as AUH₂O = mc₁ATH₂O (4) Now, let's consider the left side of equation (2). Heat transfer occurs in three ways: conduction, convection and radiation. In this experiment, there will be some conduction to the bench from the bottom of the container and some convection to the surrounding air from the container. Fourier's law of conductive heat transfer can be expressed as Qcond = -kcond A dT dx (5) where Q is the rate of heat transfer across an area A. kcond is a material property, known as the thermal conductivity and is the temperature gradient along the direction of heat transfer. Similarly, Newton's law for convective heat transfer is given by dT dx Qconv = hconvAAT (6) where hconv is the heat transfer coefficient and AT = (Twater - Troom) is the temperature difference driving the heat transfer. In each case the heat transfer rate changes with time because the temperature of the hot system is changing. Hence, to find the total heat transfer, Qout we must integrate each equation over the time period of interest and sum the results. Hence, over time period, t₁ to t₂ for heat transfer occurring across an area A (see box at end of this document for an explanation of this integration) Qout = kcond A t1 Δχ AT dt + ft² hconvAAT dt (7) The first term on the right represents the heat transfer by convection and the second represents the contribution from heat transfer by conduction. Ax is the thickness of the container and together with A, hconv and kcond, does not change over time and so can be taken outside of the integrals. Hence, the equation can be rewritten as Qout = (*cond A+ hconvA) ft² AT dt = H H St₂ AT dt Δχ t1 tt (8) where H is the overall heat transfer coefficient for the container. Now, substituting equations (4) and (8) into (2) we have H = t2 mcvATH₂O St² ATdt (9) 3 I suggest using the following website where you will also find an alternative explanation of this analysis. http://www.engineeringtoolbox.com/specific-heat-capacity-d 339.html 2 The University of Liverpool: ENGG116 – Energy Science. Thermodynamics. The denominator in this equation is the area under the graph plotted from the data recorded in the experiment. In the numerator, which is the internal energy change of the water, you can find the total temperature change of the water, you can find the total temperature change experienced by the water, ATH2O and the mass of the water, m (using density, Pн20 = 1000 kg/m³) from the measurements made. Discussion What were the overall heat transfer coefficient for your mug and glass? Are the results significantly different? The contribution of radiation heat transfer was neglected because it is a function of surface temperature rather than the difference between the surface and the environment, as described by the Stefan-Boltzmann law: Qrad = AEσT (10) where & is the emissivity of the surface from which the radiation is occurring, T is the temperature of the surface and σ is the Stefan-Boltzman constant (=5.67x10-8 W/m²K4). Emissivity describes the deviation from ideal, or black body behaviour (0 ≤ ε ≤ 1) and is property of the surface. Hence estimate the amount heat transfer that occurred by radiation during the experiment and comment on its significance. What errors were present in your results? What is the magnitude of the errors? What steps could you take to reduce the errors? Notation Symbol Meaning A Area across which heat transfer occurs SI Unit m² Cv Specific heat capacity (fixed volume) Jkg-1K-1 Emass,in Emass,out Energy transfer as a consequence of mass flow into the system Energy transfer as a consequence of mass flow out of the system J J hconv Convective heat transfer coefficient W/(m²K) H Overall heat transfer coefficient W/K kcond Thermal conductivity W/(m K) m Mass kg Qin Heat transfer in to the system J Qout Heat transfer out of the system J Qcond Rate of conductive heat transfer Qconv Rate of convective heat transfer Qrad Rate of radiative heat transfer t1, t2 Ts Time at instant 1 and 2 (beginning & end of cooling) Surface temperature Work done on/ supplied to the system Work done by the system WWW sx >> K Win Wout Δ Change in ΔΚΕ Change in kinetic energy of system ΔΡΕ Change in potential energy of system AT ΔΤΗ20 AU Ax ε σ Temperature difference = (Twater - Troom) Temperature change of water over time period Change in internal energy of system Distance over which heat transfer occurs Emissivity of radiating surface Stefan-Boltzmann constant (=5.67x10-8) 3 J J m UUKKUE The University of Liverpool: ENGG116 – Energy Science. Thermodynamics. Reporting instructions Download the template for reporting this practical exercise, complete it and then submit your report online by the end of week 11. The report is worth 6% of the module. 1% will be awarded for an appropriate and original photograph of your experimental set-up; 2% will be awarded for a complete, appropriate and original table of results; 2% for a complete, appropriate and original graphical presentation of your results and 1% for a short comprehensible discussion in response to questions in the discussion section above. Note: while you might collaborate with another student in conducting this experiment, you must obtain your own data and conduct your own analysis. You must not copy material from another source, nor commit plagiarism, nor fabricate data, nor collude with another student in preparing the report. Students found to have committed academic malpractice are liable to receive a mark of zero for the assessment. When the temperature difference, AT is not a constant over a time period then the rate of heat transfer, Q will also vary with time. Hence to evaluate the total heat transfer, Q for a time period, it is necessary to add up the contributions from small increments of dt shown in the left figure, or to integrate AT with respect to time. The integral is the area under the graph of AT plotted as a function of time, t over the time period of interest, from t₁ to tz. 4 temperature difference dt temperature difference time, time, t