Heat and Mass Transfer Fundamentals and.pdf

Thermodynamics and Heat Transfer 1-1C Thermodynamics deals with the amount of heat transfer as a system undergoes a process from one equilibrium state to another. Heat transfer, on the other hand, deals with the rate of heat transfer as well as the temperature distribution within the system at a specified time. 1-2C (a) The driving force for heat transfer is the temperature difference. (b) The driving force for electric current flow is the electric potential difference (voltage). (a) The driving force for fluid flow is the pressure difference. 1-3C The rating problems deal with the determination of the heat transfer rate for an existing system at a specified temperature difference. The sizing problems deal with the determination of the size of a system in order to transfer heat at a specified rate for a specified temperature difference. 1-4C The experimental approach (testing and taking measurements) has the advantage of dealing with the actual physical system, and getting a physical value within the limits of experimental error. However, this approach is expensive, time consuming, and often impractical. The analytical approach (analysis or calculations) has the advantage that it is fast and inexpensive, but the results obtained are subject to the accuracy of the assumptions and idealizations made in the analysis. 1-5C Modeling makes it possible to predict the course of an event before it actually occurs, or to study various aspects of an event mathematically without actually running expensive and time-consuming experiments. When preparing a mathematical model, all the variables that affect the phenomena are identified, reasonable assumptions and approximations are made, and the interdependence of these variables are studied. The relevant physical laws and principles are invoked, and the problem is formulated mathematically. Finally, the problem is solved using an appropriate approach, and the results are interpreted. 1-6C The right choice between a crude and complex model is usually the simplest model which yields adequate results. Preparing very accurate but complex models is not necessarily a better choice since such models are not much use to an analyst if they are very difficult and time consuming to solve. At the minimum, the model should reflect the essential features of the physical problem it represents. 1-7C Warmer. Because energy is added to the room air in the form of electrical work. 1-8C Warmer. If we take the room that contains the refrigerator as our system, we will see that electrical work is supplied to this room to run the refrigerator, which is eventually dissipated to the room as waste heat. 1-9C The claim is false. The heater of a house supplies the energy that the house is losing, which is proportional to the temperature difference between the indoors and the outdoors. A turned off heater consumes no energy. The heat lost from a house to the outdoors during the warming up period is less than the heat lost from a house that is already at the temperature that the thermostat is set because of the larger cumulative temperature difference in the latter case. For best practice, the heater should be turned off when no one is at home during day (at subfreezing temperatures, the heater should be kept on at a low temperature to avoid freezing of water in pipes). Also, the thermostat should be lowered during bedtime to minimize the temperature difference between the indoors and the outdoors at night and thus the amount of heat that the heater needs to supply to the house. 1-3 1-10C No. The thermostat tells an airconditioner (or heater) at what interior temperature to stop. The air conditioner will cool the house at the same rate no matter what the thermostat setting is. So, it is best to set the thermostat at a comfortable temperature and then leave it alone. Setting the thermostat too low a home owner risks wasting energy and money (and comfort) by forgetting it at the set low temperature. 1-11C No. Since there is no temperature drop of water, the heater will never kick into make up for the heat loss. Therefore, it will not waste any energy during times of no use, and there is no need to use a timer. But if the family were on a time-of-use tariff, it would be possible to save money (but not energy) by turning on the heater when the rate was lowest at night and off during peak periods when rate is the highest. 1-12C For the constant pressure case. This is because the heat transfer to an ideal gas is mcpT at constant pressure and mcvT at constant volume, and cp is always greater than cv. 1-13C The rate of heat transfer per unit surface area is called heat flux q  . It is related to the rate of heat transfer by   A Q qdA . 1-14C Energy can be transferred by heat and work to a close system. An energy transfer is heat transfer when its driving force is temperature difference. 1-15E A logic chip in a computer dissipates 3 W of power. The amount heat dissipated in 8 h and the heat flux on the surface of the chip are to be determined. Assumptions Heat transfer from the surface is uniform. Analysis (a) The amount of heat the chip dissipates during an 8-hour period is Q  Qt  (3 W)(8 h)  24 Wh  0.024kWh  (b) The heat flux on the surface of the chip is 2    37.5 W/in 2 0.08in 3 W A Q q   Logic chip Q   3 W 1-4 1-16 The filament of a 150 W incandescent lamp is 5 cm long and has a diameter of 0.5 mm. The heat flux on the surface of the filament, the heat flux on the surface of the glass bulb, and the annual electricity cost of the bulb are to be determined. Assumptions Heat transfer from the surface of the filament and the bulb of the lamp is uniform. Analysis (a) The heat transfer surface area and the heat flux on the surface of the filament are 2 As  DL   (0.05 cm)(5 cm)  0.785cm 6 2     1.9110 W/m 2 2 191 W/cm 0.785cm 150 W s s A Q q   (b) The heat flux on the surface of glass bulb is 2 2 2 As  D   (8 cm)  201.1cm 2 2 150 W 0.75 W /cm 201.1 cm s s Q q A = = = = 2 7500 W / m (c) The amount and cost of electrical energy consumed during a one-year period is  $35.04/yr      Annual Cost = (438kWh/yr)($0.08 / kWh) Electricity Consumptio n Q t (0.15 kW)(365 8 h/yr) 438kWh/yr  1-17 An aluminum ball is to be heated from 80C to 200C. The amount of heat that needs to be transferred to the aluminum ball is to be determined. Assumptions The properties of the aluminum ball are constant. Properties The average density and specific heat of aluminum are given to be  = 2700 kg/m3 and cp = 0.90 kJ/kgC. Analysis The amount of energy added to the ball is simply the change in its internal energy, and is determined from ( ) Etransfer  U  mc p T2 T1 where (2700kg/m )(0.15m) 4.77 kg 6 6 3 3 3        m V D Substituting, Etransfer  (4.77 kg)(0.90 kJ/kg  C)(20080)C = 515kJ