# Metodos de termodinamica

Páginas: 29 (7192 palabras) Publicado: 19 de marzo de 2011
12-1

Chapter 12 THERMODYNAMIC PROPERTY RELATIONS
Partial Derivatives and Associated Relations 12-1C

z
dz

∂x ≡ dx
(∂z)y (∂z)x

∂y ≡ dy dz = (∂z ) x + (∂z ) y

y x dx x +dx dy

y + dy

y

x
12-2C For functions that depend on one variable, they are identical. For functions that depend on two or more variable, the partial differential represents the change in the function withone of the variables as the other variables are held constant. The ordinary differential for such functions represents the total change as a result of differential changes in all variables. 12-3C (a) (∂x)y = dx ; (b) (∂z) y ≤ dz; and (c) dz = (∂z)x + (∂z) y 12-4C Only when (∂z/∂x) y = 0. That is, when z does not depend on y and thus z = z(x). 12-5C It indicates that z does not depend on y. Thatis, z = z(x). 12-6C Yes. 12-7C Yes.

12-2

12-8 Air at a specified temperature and specific volume is considered. The changes in pressure corresponding to a certain increase of different properties are to be determined. Assumptions Air is an ideal gas Properties The gas constant of air is R = 0.287 kPa·m3/kg·K (Table A-1). Analysis An ideal gas equation can be expressed as P = RT/v. Noting thatR is a constant and P = P(T,v),
R dT RT dv  ∂P   ∂P  dP =  −  dT +   dv = v v2  ∂T v  ∂v  T

(a) The change in T can be expressed as dT ≅ ∆T = 400 × 0.01 = 4.0 K. At v = constant,

(dP )v

=

R dT

v

=

(0.287 kPa ⋅ m 3 /kg ⋅ K)(4.0 K) 0.90 m 3 /kg

= 1.276 kPa

(b) The change in v can be expressed as dv ≅ ∆v = 0.90 × 0.01 = 0.009 m3/kg. At T = constant,

(dP )T=−

RT dv

v2

=−

(0.287 kPa ⋅ m 3 /kg ⋅ K)(400K)(0.009 m 3 /kg) (0.90 m 3 /kg) 2

= −1.276 kPa

(c) When both v and T increases by 1%, the change in P becomes
dP = (dP)v + (dP )T = 1.276 + (−1.276) = 0

Thus the changes in T and v balance each other.

12-9 Helium at a specified temperature and specific volume is considered. The changes in pressure corresponding to a certainincrease of different properties are to be determined. Assumptions Helium is an ideal gas Properties The gas constant of helium is R = 2.0769 kPa·m3/kg·K (Table A-1). Analysis An ideal gas equation can be expressed as P = RT/v. Noting that R is a constant and P = P(T, v ),
R dT RT dv  ∂P   ∂P  dP =  −  dT +   dv = v v2  ∂T v  ∂v  T

(a) The change in T can be expressed as dT ≅ ∆T =400 × 0.01 = 4.0 K. At v = constant,

(dP )v

=

R dT

v

=

(2.0769 kPa ⋅ m 3 /kg ⋅ K)(4.0 K) 0.90 m 3 /kg

= 9.231 kPa

(b) The change in v can be expressed as d v ≅ ∆ v = 0.90 × 0.01 = 0.009 m3/kg. At T = constant,

(dP )T

=−

RT dv

v2

=

(2.0769 kPa ⋅ m 3 /kg ⋅ K)(400 K)(0.009 m 3 ) (0.90 m 3 /kg) 2

= −9.231 kPa

(c) When both v and T increases by 1%, thechange in P becomes
dP = (dP) v + (dP) T = 9.231 + (−9.231) = 0

Thus the changes in T and v balance each other.

12-3

12-10 It is to be proven for an ideal gas that the P = constant lines on a T- v diagram are straight lines and that the high pressure lines are steeper than the low-pressure lines. Analysis (a) For an ideal gas Pv = RT or T = Pv/R. Taking the partial derivative of T withrespect to v holding P constant yields
P  ∂T    = ∂v  P R 
T P = const

which remains constant at P = constant. Thus the derivative (∂T/∂v)P, which represents the slope of the P = const. lines on a T-v diagram, remains constant. That is, the P = const. lines are straight lines on a T-v diagram. (b) The slope of the P = const. lines on a T-v diagram is equal to P/R, which is proportional toP. Therefore, the high pressure lines are steeper than low pressure lines on the T-v diagram.

v

12-11 A relation is to be derived for the slope of the v = constant lines on a T-P diagram for a gas that obeys the van der Waals equation of state. Analysis The van der Waals equation of state can be expressed as
T= a  1  P + 2 (v − b ) R v 

Taking the derivative of T with respect to...

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