Thermodynamics lecture 8

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L t 8 Fi t L f C t l MLecture 8 – First Law for Control Mass

Work• Work is done by a system if the sole effect on the surroundings could be the lifting of a weight in the gravitational field• δW = Fsds, 1W2 = ∫ δW

W i t k iti if th di t f• W is taken as positive if the corresponding energy transfer is from the system to the surroundings, and negative if the transfer is from the surroundings to the systemg y• W = ∫δW is path dependent, and δW is inexact• Work at the moving boundary of a simple compressible

b t (PdV k k f i / i ) Wsubstance (PdV work, work of expansion/compression), W = ∫δW = ∫PdV if quasiequilibrium, W = ∫δW = ∫PextdV if not• Polytropic process PVn = constantPolytropic process PV constant

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Other types of work• Stretching of a wire under tension δW = -TdL where T is the tension, and dL the infinitesimal extension. In the elastic region, with the stress σ = T/A = Ee, where E is the modulus of elasticity and e the strain defined by de = dL/L0, δW = -AEeL0deAEeL0de• Expansion of a surface δW = -S dAS is the surface tension, and dA theinfinitesimal change in area• Work done in transporting charge dZ across potential diff Φ i δW ΦdZdifference Φ is δW = - ΦdZ• Shaft work• Flow work• Flow work


Internal Energy U• Joule experiment – For a closed system (control mass) in an adiabatic enclosure, W = -∆U where U is the internal energy, an extensive function of the state of a system (extension of the work-energy theorem• Internal energy sum total of energy of inter molecular• Internal energy – sum total of energy of inter-molecular interaction, molecular translational kinetic energy, and intra-molecular contributions such as electronic, vibrational, and rotational • Very generally, one may replace the internal energy above by E the total energy which will include the kinetic energy ofby E the total energy which will include the kinetic energy of the system as a whole, and the potential energy in the gravitational field E = U + KE + PE


Heat Q• State of control mass can be changed without the performance of work, specifically by placing it in thermal contact with another object at a different temperature• ∆U = Q (when W = 0), Q is called heat

T f f b d f th t• Transfer of energy across boundary surface that occurs solely due to a difference in temperature between two bodies in thermal contact• Energy transfer from the body at higher temperature to that at lower temperature

U it i J f W U E T k iti h• Unit is J, same as for W, U, or E. Taken as positive when transfer of energy is from the surroundings to the system• Q = ∫δQ is path dependent, and δQ is inexactQ ∫δQ is path dependent, and δQ is inexact


Work and Heat• Both heat and work are transient phenomena, not

properties

• Both are modes of energy transfer across system

boundary when system undergoes a process

• Both are path dependent and the infinitesimals δQ andBoth are path dependent, and the infinitesimals δQ and

δW are inexact differentials

•What can we say about the internal energy change of a

control mass in a process in which work and heat are both

non-zero?


First Law for Control MassAs seen above, for a closed system∆U = U2 – U1 = - 1W2 (adiabatic, Q = 0).∆U = U2 – U1 = 1Q2 (W = 0)What if both work and heat terms present? Since U is a state functionstate function∆U = U2 – U1 = 1Q2 - 1W2More generally since there may also changes in kinetic energy of the system, and its potential energy in the gravitational field, we use the total energy E = U + KE + PE, and write∆E = E2 – E1 = 1Q2 - 1W2, the conservation of energy principle extended to include thermodynamic variables and processespStrictly, we must indicate the path also for W and Q


First Law for Control Mass• What if a control mass undergoes a cyclic process? Then ∆U = 0, and so Q = W• For an infinitesimal change dU = δQ - δW• As mentioned above, E = U + KE + PE, where KE and PE

th b lk ki ti d it ti l t ti l i fare the bulk kinetic and gravitational potential energies of the system, dU + d(KE) + d(PE) = δQ – δW• KE = ½ mV2 and PE = mgZKE ½ mV and PE mgZ• dE = dU + d(KE) + d(PE) = δQ – δW• E2 – E1 = U2 – U1 +m(V2

2 – V12) + mg(Z2 – Z1) = 1Q2 - 1W2


Internal Energy U• U is an extensive property (as are KE and PE)• u = U/m is the specific internal energyp gy• Can be used to fix state of a phase• In the two-phase say liquid-vapour region, u = (1-x)uf + xugg= uf + xufg, and u can be used with a table to determine the quality ie., the relative amounts of the two phases


Adiabatic free expansion

Carry out a first law analysis of the process• Initial state specified by, say v and T. Then u(T,v) = u1 also knownknown• 1Q2 = 0 (adiabatic, 1W2 = 0 (rigid container), hence by I Law U2 – U1 = 0. Also V2 = 2V1. Hence the final state is specified by v2 = 2v1, and u2 = u1What about other properties such as the temperature or pressure? Yes, if we have the equation of state or other


pressure? Yes, if we have the equation of state or other equivalent information

Adiabatic free expansion – ideal gas

• The final state is determined by V2 = 2V1 and U2 = U1The final state is determined by V2 2V1, and U2 U1. • Joule expansion experiment: Joule carried out this experiment, and found that there was no measurable t t h i t b d th htemperature change, in part because under the chosen conditions, the gas closely approximated ideal behavior.• Conclusion: The internal energy of an ideal gas is a function only of the temperature, u = u(T). This can be proved from the EoS using the II Law, but for now we will take it as an additional postulate from observation


take it as an additional postulate from observation.• Hence T2 = T1, and so P2 = P1/2

Constant Pressure ExpansionFor the isobaric conversion of saturated liquid water at 100ºC to saturated vapour at the same temperature, find the heat required per unit mass.• The control mass is 1kg of water, in a piston-cylinder, with the pressure set at 101 3 kPa Draw a schematic sketchthe pressure set at 101.3 kPa. Draw a schematic sketch• All properties of the initial and final states are fixed, and can be read from the steam tables. Sketch the process p• The process is isobaric, sketch it in the P-v plane (or the T-v plane)

Th th d i d l i th i t b l f• The thermodynamic model is the eos in tabular form• Analyse the system using the I Law• Look up the internal energy values and the specific• Look up the internal energy values, and the specific volumes. Compute the work done, and thence the heat


Constant Pressure Expansion - Enthalpy• 1w2 = Psat(vg – vf), with Psat = 101.325 kPa = P1 = P2

• u2 – u1 = ug – uf2 1 g f

• 1q2 = ug – uf + Psat(vg – vf) = (u + Pv)2 – (u + Pv)1 (true for any constant pressure process for a control mass). • Define the enthalpy H = U + PV, an extensive state function with dimensions of energy• q = h – h where h is the enthalpy per unit mass and is• 1q2 = h2 – h1 where h is the enthalpy per unit mass, and is tabulated • It is useful to define the enthalpy H, since the combination U + PV appears naturally here. It will also arise in the discussion of flow work


Back to the example• Refer to the steam tables for the properties of

saturated water at 100 Csaturated water at 100 C.

Psat (kPa) vf (m3/kg) vg (m3/kg) vfg uf (kJ/kg) ug (kJ/kg) ufg101.325 0.001044 1.67290 1.67185 418.91 2506.50 2087.58

Values of u with reference to u at 0 01 C taken as 0Values of u with reference to ug at 0.01 C taken as 0Find 1Q2 = Δ(u + Pv) = hg – hf= hfgAt 100 C, hf = 419.02, hg = 2676.05 and hfg = 2257.03 kJ/kg


Thermodynamics lecture 8

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Transcript of Thermodynamics lecture 8