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Atmospheric Thermodynamics –II The first Law of Thermodynamics and applications. Prof. Leila M. V. Carvalho Dept. Geography, UCSB. The first Law of Thermodynamics. Is the law that describes the relationships between heat, work and internal energy.

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## Atmospheric Thermodynamics –II The first Law of Thermodynamics and applications

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**Atmospheric Thermodynamics –IIThe first Law of**Thermodynamics and applications Prof. Leila M. V. Carvalho Dept. Geography, UCSB**The first Law of Thermodynamics**• Is the law that describes the relationships between heat, work and internal energy. • It establishes the physical and mathematical framework to understand heating processes in our atmosphere, the formation of clouds, the thermodynamical modifications in parcels in movement, etc… • Let’s explore these relationships and applications…**Motivation: Ex: formation of clouds**Pressure The first Law of Thermodynamics provides the physical concepts to understand cloud formation: Releases Latent heat: • Volume expands and the parcel’s temperature decreases. • As it cools the air becomes saturated • When that begins: Lifting Condensation Level (clouds are formed) LCL: Cloud base : Heat**Internal Energy u: measure of the total kinetic and**potential energy of a gas Kinetic energy: depend on molecular motions -> relationship with temperature Potential energy: changes in the relative position of the molecules due to internal forces that act between molecules (small changes)**Closed System definition:**• Is the one in which the total amount of matter, which may be in the form of gas, liquid, solid or a mixture of these phases, is kept constant**Suppose a closed system with one unity of mass**• Suppose that this volume receives certain quantity of thermal energyq (joules) by ‘conduction’ and/or radiation. • This system may do a certain amount of external work w(also measured in Joules) . Differences will cause changes in the internal energy Where 1 is before and 2 after the change**In the differential form**(34) • dq is the differential increment of heat added to the system, • dwis the differential element of work done by the system • du is the differential increase in internal energy of the system This is the First Law of Thermodynamics Changes in du depend only on the final and initial state: functions of state**Visualization**Volume is proportional to the distance x Frictionless Every state of the substance, corresponding to a given position of the piston, can be represented in this diagram below**AREA A**If the Piston moves outward with the same pressure p, then the work done by the substance in pushing the external force F through a distance dx is: F=pA If pressure is constant then: p pdV (35) V dV**If the substance passes from state A (V1) to B (V2) during**with its pressure changes than the work will be : (36) • IF V2>V1, W is positive, indicating that the SUBSTANCE DOES WORK ON ITS ENVIRONMENT • IF V2<V1, W is negative, indicating that the ENVIRONMENT DOES WORK ON THE SUBSTANCE**A few useful remarks**• If we are dealing with a unity mass of a substance, the volume V is replaced by the SPECIFIC VOLUME α. Therefore, the work dw that is done when the specific volume increases by dαis • dw= pdα(37) We can then rewrite the First Law of thermodynamics as: dq = du + pdα(38)**Some considerations**F=pA • Suppose that the force in the wall is supplied by the impact of the balls. If we do not supply kinetic energy to the gas (or balls) the work required to move the walls outward comes from a decrease in the kinetic energy of the balls that rebound from the walls with lower velocities than they struck them. • This decrease in kinetic energy is in accordance with the first law of thermodynamics under ADIABIATIC CONDITIONS (NO HEAT SUPPLIED). • In this case, the work done by the system by pushing the walls outward is equal to the decrease in the internal energy of the system. dq = du + pdα If dq = 0 pdα = - du**The Joule’s Experiment and the Joule’s Law**• Suppose that the gas is expanding within a Vacuum chamber. In this case, there is no F against the environment (since it is vacuum) • In this case, dw=0 (F=0) • The gas also does not take or gives out heat (dq=0) • As a consequence du =0 (First Law of the Thermodynamics) • Under these conditions the temperature of the gas does not change, which implies that the kinetic energy of the molecules remains constant (and potential energy too) • Big conclusion: The internal energy of a gas is independent of its volume if the temperature is kept constant**Specific Heat at constant volume**• Suppose that this volume with a unity of mass receives certain small quantity of heat dq (joules). The temperature T increases to T + dTwithout any changes in phase occurring within the material. If the volume is kept constant, We define the specific heat at constant volume as: (39)**If the volume of the material is constant, the first law of**the thermodynamics states that: dq = du + pdα If dα =0 dq = du For an ideal gas the Jaule’s law applies and therefore u depends only on temperature. Therefore, regardless of whether the volume of a gas changes, we may write (40)**FROM:**First Law of the Thermodynamics: and (40) dq = du + pdα(38) Therefore: First Law of the Thermodynamics: (41) Because u is a function of state, no matter how the material changes from state 1 to state 2, the change in its internal energy from (40) is:**Suppose that now pressure is kept constant**• The material is allowed to expand as heat is added to it and its temperature rises, as pressure remains constant. In this case, a certain amount of heat added to the material will have to be expended to DO WORK as the system expands against constant pressure of its environment • We can also define a specific heat at constant pressure cp**Implications…**• A larger quantity of heat must be added to the material to raise its temperature by a given amount than if the volume of the material was kept constant (it is because heat will be transformed into work and also into kinetic energy to increase the temperature of the gas). Mathematically speaking… First Law (Eq. 41) If we assume that p and α are changing, then (43) We saw before from the ideal gas equation (3) that: Therefore: (44)**At constant pressure…**Last term vanishes -> dp=0 Therefore: or: (45) The specific heats at constant volume and at constant pressure for dry air are 717 and 1004 JK-1kg-1,respectively. The difference between them is 287 J K-1 kg-1, which is the gas constant for dry air. (46)**Curiosities…**Are there processes in the atmosphere that occur at constant pressure (isobaric process)? All physical/dynamical processes that are considered at a given pressure level. For example air expanding at pressure p=800hPa) Are there processes in the atmosphere that occur at constant volume? Some horizontal movements may occur such that temperature varies, pressure varies but volume is constant.**Etymology**• The term enthalpy comes from the prefix ἐν-, en-, meaning "to put into", and the Classical Greek verb θάλπειν, thalpein, meaning "to heat". The original definition is thought to have stemmed from the adjective "enthalpos" (ἔνθαλπος). • enthalpy (denoted as H, or specific enthalpy denoted as h) is a thermodynamic property of a thermodynamic system. It can be used to calculate the heat transfer during a quasi-staticprocess taking place in a closed thermodynamic system under constant pressure (isobaric process). • Enthalpy is commonly referred to as ‘sensible heat’**Mathematical framework**• If heat is added to a material at constant pressure so that the specific volume of the material increases from α1to α2 the work done by a unity of mass of the material is p(α 2 - α1) Fromdq = du + pdα(38) Initial and final internal energy for a unity of mass**Therefore, at a constant pressure**h2 h1 h=Enthalpy of a unity mass u and p are functions of state, h is a function of state too.**Differentiating …**• The objective of computing differentials in these cases is to provide equations that can describe small variations (differentiations) and can be used later to integrate a given process over some large changes of a given variable (could be temperature, volume, pressure, etc) USING AND (48)**Some considerations**• Compare (46) and (48) (46) (48) (49) (50) • Where h is taken as zero when T=0.Eq (50) indicates that h corresponds to the heat required to raise the temperature of a material from 0 to T K at constant pressure**Examples:**• When a layer of air that is at rest and in hydrostatic balance is heated, for example, by radiative transfer, the weight of the overlying air pressing down on it remains constant (constant pressure). • The energy added to the air is realized in the form of an increase in enthalpy (or sensible heat) and**Enthalpy (sensible Heat)**1)internal energy increases at constant VOLUME Because the Earth’s atmosphere is made up mainly of the diatomic gases N2 and O2, the energy added by dq is partitioned between the increase in internal energy du and expansion work in the ratio 5:2 2) Work is done against the overlaying environment**General Expression: dry atmosphere (no clouds)**Pressure We can write a more general expression that is applicable to a moving parcel that changes pressure as it rises or sinks. Remember the definition of Geopotential: Dry Static Energy: constant provided that the parcel neither gain or loses heat (dq=0) Heat (This also means : no latent heat released or absorbed)

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