Thermodynamics lecture 9

BITS Pil iBITS PilaniPilani Campus

L t 9 Fi t L f C t l MLecture 9 – First Law for Control Mass

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

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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


Problem Analysis

• Identify system, sketch diagram, show forces, and heat and

work flows

• Identify (fix) the initial state and the final state• Identify (fix) the initial state, and the final state

• Characterize the process

• Sketch the process schematically

• Identify the thermodynamic model to be used y y

• What is the analysis to be done?

• What is the solution technique?


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 = U1. • Joule expansion experiment: Joule carried out this experiment and found that there was no measurableexperiment, and found that there was no measurable temperature change, in part because under the chosen conditions, the gas closely approximated ideal behaviour.• 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 as we will see


proved from the EoS using the II Law as we will see• 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 = ∆(u + Pv) = hg – hf= hfgAt 100 C, hf = 419.02, hg = 2676.05 and hfg = 2257.03 kJ/kgTh th l it i i t i t dThe enthalpy per unit mass is an intensive property, and may be used as one of the variables to fix the state of a phase, or to locate the position on a tie-line in the two


phase region

Specific Heat• Specific Heat C = 1/m (δQ/δT)

• Since Q path dependent, or δQ inexact, must specify the

path, or conditions

• Specific heat at constant volume Cv = 1/m (δQ/δT)V, which

for a simple compressible substance may be written as C =for a simple compressible substance may be written as Cv

1/m (∂U/∂T)V = (∂u/∂T)v

• Specific heat at constant pressure CP = 1/m (δQ/δT)P,

which for a simple compressible substance may be written

as CP = 1/m (∂H/∂T)P = (∂h/∂T)P


Ideal Gas – Specific Heats • Air at 1 bar and 300 K is heated (a) at constant volume, and (b) at constant pressure, to a final temperature of 420 K. In each case, calculate the work, heat, internal energy change, and enthalpy changeAside: Since u = u(T) for an ideal gas h = u + Pv = u + RT =Aside: Since u = u(T) for an ideal gas, h = u + Pv = u + RT = h(T). Also Cv = (∂u/∂T)v and (∂h/∂T)P are functions only of the temperature

CP – Cv = RP vCv and CP are functions of T


Ideal Gas Example (contd.)• (a) ∆h = ∫Cp0dTEspecially over small ranges of T, often Cp0 taken to be constant, in which case the integral is ≈ Cp0(T2 – T1) = 125 kJ/kgIf Cp0 not assumed constant, it is usually expressed as anIf Cp0 not assumed constant, it is usually expressed as an expansion in T, Cp0 = C0 + C1θ + C2θ2 + C3θ3 where θ = 10-3T(K) Using the values of the constants from the table, find ∆h = 133.2 kJ/kg∆h 133.2 kJ/kgThe enthalpies are also evaluated and tabulated from which∆h = ∫Cp0dT - ∫Cp0dT = 125.1 kJ/kg

T0

T2 T1

T 0∆u = ∆h - P ∆v = ∆h - R ∆Tw = 0q = ∆u

T0 T‐0

q = ∆u


Ideal Gas Example (contd.)• (b) ∆u and ∆h are both the same as in (a) since both u and h are functions of T alone, and the initial and final temperatures are the same in both cases.q = ∆h as this is a process at constant pressure

∆ R ∆Tw = q – ∆u = R ∆T


Liquids and Solids• Estimate the change in internal energy u and enthalpy h for liquid water when it is taken from the saturated liquid at 20 C to (a) 20 C and 500 kPa, (b) 40 C and 2000 kPaSolids, and liquids away from the critical point, are highly incompressible and so dv ≈ 0 is an excellent approximationincompressible, and so dv ≈ 0 is an excellent approximation. Also, the heat capacities at constant pressure and at constant volume are virtually the same, say C.Hence du ≈ C dT, and dh = du + d(Pv) ≈ C dT + vdPTaking C and v as constant, one has ∆ C ∆T d ∆h C ∆T ∆P∆u ≈ C ∆T, and ∆h ≈ C ∆T + v ∆P(a) ∆u ≈ 0, ∆h ≈ vf ∆P ≈ 0.5 kJ/kg(b) ∆u ≈ C ∆T = 84 kJ/kg ∆h ≈ C ∆T + v ∆P = 86 kJ/kg(b) ∆u ≈ C ∆T = 84 kJ/kg, ∆h ≈ C ∆T + vf ∆P = 86 kJ/kg


Example• A constant pressure piston-cylinder assembly

contains 0 20 kg water as saturated vapour at 400 kPacontains 0.20 kg water as saturated vapour at 400 kPa.

It is now cooled so that the water occupies half of the

i i l l Fi d Q i thoriginal volume. Find Q in the process


Solution• Initial state characterized by pressure and specific volume. Process is isobaric, so final state at same pressure, and v2= v1/2. The steam table will be used. By comparing v2 to vf(0.0010840, and vg 90.46138) at given P, determine that the final state is a mixture of saturated liquid and vapour with x2final state is a mixture of saturated liquid and vapour, with x2= (v2 – vf)/vfg, and then calculate u2 = uf + x2ufg

St t P (kP ) T (C) ( 3/k ) V ( 3) (kJ/k ) U (kJ)State P (kPa) T (C) v (m3/kg) V (m3) x u (kJ/kg) U (kJ)

1 400 143.6 0.46246 0.092492 1 2553.55

2 400 143.6 0.23123 0.046246 0.49882 1576.6

From the table above, ∆u = -977 kJ/kg, or ∆U = -195.4 kJW P(V V ) 18 5 kJ


1W2 = P(V2 – V1) = -18.5 kJ1Q2 = ∆U + 1W2 = -213.9 kJ

Example• A piston-cylinder arrangement has a linear spring and the outside atmosphere acting on the pistion. It contains water at 3 MPa and 400 C with a volume of 0.1 m3. If the piston is at the bottom, the spring exerts a force such that a pressure of 200 kPa inside is required to balance the forces Theof 200 kPa inside is required to balance the forces. The system now cools until the pressure reaches 1 MPa. Find the heat transfer for the process.


SolutionInitial state determined by P and v, it is superheated. Final state fixed by P (given) and v again. How do we find v?P = 200 kPa + ksV/A2, where ks is the spring constant, and A the area of cross-section. This is since the pressure to just balance when piston is at bottom is given as 200 kPa Sincebalance when piston is at bottom is given as 200 kPa. Since P1 = 3 MPa and V1 = 0.1 m3 given, ks/A2 = 2.8x104 kJ/m3, so V2 = 0.02857 m3, v2 = 0.02840 m3/kg, saturated mixture with x2 = 0.14107

State P (kPa) T (C) V (m3) v (m3/kg) Mass M (kg) x u (kJ/kg) U(kJ)

1 3000 400 0.1 0.09936 1.006 ‐ 2950

2 1000 143.6 0.02857 0.02840 1.006 0.14107 1019

W = ∫PdV = 200 (∆V) + 2 8x104(V 2 V 2)/2 = 143 kJ


1W2 = ∫PdV = 200 (∆V) + 2.8x104(V22 – V1

2)/2 = -143 kJ1Q2 = ∆U + 1W2 = -1925 -143 = -2068 kJ

Thermodynamics lecture 9

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