Termodinamica

7-148

7-198 Refrigerant-134a is vaporized by air in the evaporator of an air-conditioner. For specified flow rates, the exit temperature of air and the rate of entropy generation are to be determined for the cases of an insulated and uninsulated evaporator. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3There are no work interactions. 4 Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The constant pressure specific heat of air at room temperature is cp = 1.005 kJ/kg.K (Table A-2). The properties of R-134a at the inlet and the exit states are (Tables A-11 through A-13)
P1 = 120 kPa  h1 = h f + x1 h fg =22.49 + 0.3 × 214.48 = 86.83 kJ/kg  x1 = 0.3  s1 = s f + x1 s fg = 0.09275 + 0.3(0.85503) = 0.3493 kJ/kg ⋅ K T2 = 120 kPa  h2 = hg @ 120  sat. vapor  s2 = hg @ 120
P V&
3

kPa kPa

= 236.97 kJ/kg = 0.9478 kJ/kg ⋅ K
R-134a 6 m3/min AIR

Analysis (a) The mass flow rate of air is
& mair =
3 3

3

RT3

=

(100 kPa )(6 m /min ) (0.287 kPa ⋅ m3/kg ⋅ K )(300 K ) = 6.97 kg/min

1
2kg/min

We take the entire heat exchanger as the system, which is a control volume. The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance ( for each fluid stream):
& & & min − mout = ∆msystem
0 (steady)

2 4
sat. vapor

& & & & & & & & = 0 → min = mout → m1 = m2 = mair and m3 = m4 = mR

Energy balance (for the entire heat exchanger):Rate of net energy transfer by heat, work, and mass

& & Ein − Eout 1 24 4 3

=

Rate of change in internal, kinetic, potential, etc. energies

& ∆Esystem 0 (steady) 1442443

=0

& & Ein = Eout & & & & & & m1h1 + m3h3 = m2 h2 + m4 h4 (since Q = W = ∆ke ≅ ∆pe ≅ 0)

Combining the two, Solving for T4, Substituting,

& & & mR (h2 − h1 ) = mair (h3 − h4 ) = mair c p (T3 − T4 )

T4 =T3 −

& m R (h2 − h1 ) & m air c p
(2 kg/min)(236.97 − 86.83) kJ/kg = −15.9°C = 257.1 K (6.97 kg/min)(1.005 kJ/kg ⋅ K)

T4 = 27°C −

Noting that the condenser is well-insulated and thus heat transfer is negligible, the entropy balance for this steady-flow system can be expressed as
Rate of net entropy transfer by heat and mass

& & Sin − Sout 1 24 4 3

+

Rate of entropy generation& Sgen {

& = ∆Ssystem 0 (steady) 144 44 2 3
Rate of change of entropy

& & & & & m1s1 + m3 s3 − m2 s2 − m4 s4 + Sgen = 0 (since Q = 0) & & & & & mR s1 + mair s3 − mR s2 − mair s4 + Sgen = 0

or,

7-149
& & & Sgen = mR (s2 − s1 ) + mair (s4 − s3 )

where
s 4 − s 3 = c p ln T4 P T 257.1 K − R ln 4 = c p ln 4 = (1.005 kJ/kg ⋅ K) ln = −0.1551 kJ/kg ⋅ K T3 P3 T3 300 K
0Substituting, & Sgen = (2 kg/min )(0.9478 - 0.3493 kJ/kg ⋅ K ) + (6.97 kg/min)(−0.1551 kJ/kg ⋅ K)
= 0.116 kJ/min ⋅ K = 0.00193 kW/K

(b) When there is a heat gain from the surroundings at a rate of 30 kJ/min, the steady-flow energy equation reduces to
& & & Qin = mR (h2 − h1 ) + mair c p (T4 − T3 ) & & Qin − mR (h2 − h1 ) & air c p m

Solving for T4, Substituting,

T4 = T3 +

T4 = 27°C +

(30kJ/min) − (2 kg/min)(236.97 − 86.83) kJ/kg = −11.6°C = 261.4 K (6.97 kg/min)(1.005 kJ/kg ⋅ K)

The entropy generation in this case is determined by applying the entropy balance on an extended system that includes the evaporator and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surrounding air at all times. The entropy balance for theextended system can be expressed as
Rate of net entropy transfer by heat and mass

& & Sin − Sout 1 24 4 3

+

Rate of entropy generation

& Sgen {

& = ∆Ssystem 0 (steady) 144 44 2 3
Rate of change of entropy

Qin & & & & & + m1s1 + m3 s3 − m2 s2 − m4 s4 + Sgen = 0 Tb,out Qin & & & & & + mR s1 + mair s3 − mR s2 − mair s4 + Sgen = 0 Tsurr & Q & & & Sgen = mR (s2 − s1 ) + mair (s4 − s3...