Show description

So as long as some heat is discharged to surroundings colder than the geothermal source during its operation, there is no reason why this device should not work. A detailed analysis of the entropy changes associated with this device follows. 1 CV and Cp are the temperature dependent heat capacities of water INSTRUCTOR’S MANUAL 58 Three things must be considered in an analysis of the geothermal heat pump: Is it forbidden by the first law? Is it forbidden by the second law? Is it efficient? Etot = Ewater + Eground + Eenvironment Ewater = 0 Eground = −CV (Th ){Th − Tc } Eenvironment = −CV (Th ){Th − Tc } adding terms, we find that and Tc .

17. Use it to obtain an expression for the inversion temperature, calculate it for xenon, and compare to the result above. 21] ∂V p V − nb RV 3 so Substituting, nRT (V −nb) Cp,m − CV ,m = T (V −nb) − 2na RV 3 × (V − nb) = nλR with λ = 1 1− 2na RT V 3 × (V − nb)2 For molar quantities, Cp,m − CV ,m = λR with 2a(Vm − b)2 1 =1− λ RT Vm3 Now introduce the reduced variables and use Tc = 8a , Vc = 3b. 8 K. The perfect gas value for Vm may be used as any 1 error introduced by this approximation occurs only in the correction term for .

Will represent molar quantities in all cases. w=− C dV = −C Vn p dV = − dV Vn For n = 1, this becomes (we treat the case n = 1 later) (1) (2) w= final state,Vf C C = V −n+1 n−1 n −1 initial state,Vi = pi Vin × n−1 = pi Vi Vin−1 × n−1 w= pi Vi × n−1 1 − Vfn−1 Vfn−1 n−1 − − 1 Vin−1 [because pV n = C] Vin−1 1 Vi Vf But pV n = C or V = 1 1 Vfn−1 1 Vin−1 −1 = RTi × n−1 Vi n−1 −1 Vf C 1/n for n = 0 (we treat n = 0 as a special case below). So, p n−1 Vi n−1 pf n (3) = n=0 Vf pi Substitution of eqn 3 into eqn 2 and using ‘1’ and ‘2’ to represent the initial and final states, respectively, yields RT1 × (4) w = n−1 p2 p1 n−1 n −1 for n = 0 and n = 1 THE FIRST LAW: THE CONCEPTS 37 In the case for which n = 0, eqn 1 gives w= C × 0−1 1 Vf−1 − 1 = −C(Vf − Vi ) Vi−1 = −(pV 0 )any state × (Vf − Vi ) = −(p)any state (Vf − Vi ) (5) w = −p V for n = 0, isobaric case In the case for which n = 1 w=− C dV = − Vn p dV = − Vi Vf Vi = (RT )any state ln Vf w = (pV n )any state ln (6) w = RT ln (7) V1 V2 = RT ln C Vf dV = −C ln V Vi Vi Vf = (pV )any state ln p2 p1 for n = 1, isothermal case To derive the equation for heat, note that, for a perfect gas, Tf p f Vf − 1 = CV Ti −1 q + w = CV Ti Ti p i Vi Vin−1 = CV Ti − 1 [because pV n = C] n−1 Vf = CV Ti q = CV Ti pf pi pf pi n−1 n RTi = CV Ti − n−1 = n−1 n −1 −1 − [using eqn 3 (n = 0)] RTi n−1 × pf pi CV 1 − RTi R n−1 pf pi n−1 n −1 n−1 n CV pf 1 − RTi Cp − C V n−1 pi   CV 1  = − RTi Cp n − 1 C −1 = = (n − 1) − (γ − 1) RTi (n − 1) × (γ − 1) pf pi n−γ = RTi (n − 1) × (γ − 1) n−1 n n−1 n pf pi −1 pf pi CV pf pi −1 −1 n−1 n 1 1 − RTi γ −1 n−1 n−1 n pf pi = V U = q + w = CV (Tf − Ti ).