exweek3

Upload: adiyaksapermana

Post on 03-Jun-2018

218 views

Category:

Documents


0 download

TRANSCRIPT

  • 8/12/2019 exweek3

    1/4

    T h e r m o d y n a m i c s o f M a t e r i a l s 3 . 0 0 F a l l 2 0 0 1

    E x a m p l e P r o b l e m s f o r W e e k 3

    E x a m p l e P r o b l e m 3 . 1

    O n e m o l e o f a n i d e a l g a s , i n a n i n i t i a l s t a t e P = 1 0 a t m , V = 5 L , i s t a k e n , r e v e r s i b l y , i n

    a c l o c k w i s e d i r e c t i o n , a r o u n d a c i r c u l a r p a t h g i v e n b y ( V ; 1 0 )

    2

    + ( P ; 1 0 )

    2

    = 2 5 . C a l c u l a t e

    t h e a m o u n t o f w o r k d o n e b y t h e g a s a s a r e s u l t o f t h i s p r o c e s s , t h e c h a n g e i n i n t e r n a l e n e r g y

    o f t h e g a s , a n d t h e m a x i m u m a n d m i n i m u m t e m p e r a t u r e s a t t a i n e d b y t h e g a s d u r i n g t h e

    c y c l e . H o w d o t h e s e a n s w e r s c h a n g e i f t h e n u m b e r o f m o l e s o f g a s d o u b l e f o r t h e s a m e P - V

    c y c l e ?

    S o l u t i o n 3 . 1

    T h e i n t e r n a l e n e r g y i s a s t a t e f u n c t i o n a n d o n l y d e p e n d s o n t h e i n i t i a l a n d n a l s t a t e s

    o f t h e s y s t e m . T h i s i s a c l o s e d i n t e g r a l s o t h e c h a n g e i n i n t e r n a l e n e r g y i s z e r o .

    H

    d U = 0 .

    T h e r e a r e t w o w a y s t o c a l c u l a t e t h e w o r k d o n e b y t h e s y s t e m d u r i n g t h e c y c l e . T h e

    r s t i s t o a n a l y t i c a l l y i n t e g r a t e

    R

    P d V , t h e s e c o n d i s t o a s s e s s t h e a r e a u n d e r t h e c u r v e

    g r a p h i c a l l y . ( P ; 1 0 )

    2

    + ( V ; 1 0 )

    2

    = 2 5 i s t h e e q u a t i o n o f a c i r c l e i n t h e P - V p l a n e w i t h

    c e n t e r a t ( 1 0 1 0 ) a n d r a d i u s 5 . T h e i n i t i a l s t a t e i s a t t h e l o w e s t a l l o w e d v o l u m e f o r t h e

    s y s t e m , ( V

    0

    P

    0

    ) .

    P

    V10

    10=P_o=P_1

    V_o V_1

    F i g u r e 1 : P - V d i a g r a m f o r t h e w o r k c y c l e .

    F i r s t w e m u s t n d P = P ( V ) i n o r d e r t o p e r f o r m t h e i n t e g r a t i o n . P = 1 0

    q

    2 5 ; ( V ; 1 0 )

    2

    w h e r e t h e p l u s s i g n c o r r e s p o n d s t o t h e u p p e r h a l f o f t h e c i r c l e ( P

    +

    ( V ) ) a n d t h e m i n u s s i g n

    c o r r e s p o n d s t o t h e l o w e r h a l f ( P

    ;

    ( V ) ) . N o w l e t V

    0

    = V ; 1 0 i n t h e f o l l o w i n g .

    W

    s

    =

    Z

    P d V

    W

    s

    =

    Z

    V

    1

    = 1 5

    V

    0

    = 5

    P

    +

    ( V ) d V +

    Z

    V

    0

    = 5

    V

    1

    = 1 5

    P

    ;

    ( V ) d V

    W

    s

    =

    Z

    V

    0

    1

    = 5

    V

    0

    0

    = ; 5

    1 0 +

    p

    2 5 ; V

    0 2

    d V

    0

    +

    Z

    V

    0

    0

    = ; 5

    V

    0

    1

    = 5

    1 0 ;

    p

    2 5 ; V

    0 2

    d V

    0

  • 8/12/2019 exweek3

    2/4

    W

    s

    =

    2 5

    2

    ( + 8 ) +

    2 5

    2

    ( ; 8 ) = 2 5 L a t m

    B y t h i s m e t h o d t h e w o r k d o n e b y t h e s y s t e m i s 2 5 1 0 1 : 3 J = 7 9 5 6 J . T h e s e c o n d m e t h o d

    i n v o l v e s e x a m i n i n g t h e a r e a u n d e r t h e c u r v e , a c i r c l e w i t h r a d i u s 5 . S o t h e a r e a i s 2 5 L a t m .

    T h i s a g a i n i s t h e w o r k d o n e b y t h e s y s t e m e q u a l t o 7 9 5 6 J .

    T h e m a x i m u m a n d m i n i m u m t e m p e r a t u r e s c a n b e f o u n d f r o m t h e i d e a l g a s l a w f o r o n e

    m o l e o f g a s , P V = R T . T

    m a x

    a t t a i n s w h e n ( P V ) i s a m a x i m u m a n d T

    m i n

    a t t a i n s w h e n

    ( P V ) i s a m i n i m u m . L e t P

    0

    = P ; 1 0 a n d V

    0

    = V ; 1 0 s o P V i s a n e x t r e m a l w h e n P

    0

    V

    0

    i s e x t r e m a l .

    P

    0

    V

    0

    = f ( P

    0

    ( V

    0

    ) V

    0

    ) =

    p

    2 5 ; V

    0 2

    V

    0

    d f

    d V

    0

    = 0 =

    p

    2 5 ; V

    0 2

    V

    0

    1

    2

    ; 2 V

    0

    p

    2 5 ; V

    0 2

    0 =

    "

    2 5 ; V

    0 2

    ; V

    0 2

    p

    2 5

    ;V

    0 2

    #

    T h e e x t r e m a a r e f o r V

    0

    =

    q

    2 5

    2

    . E x a m i n a t i o n o r c h e c k i n g t h e s i g n o f t h e s e c o n d d e r i v a t i v e ,

    d

    2

    F

    d V

    2

    > 0 f o r a m i n i m u m a n d

    d

    2

    F

    d V

    2

  • 8/12/2019 exweek3

    3/4

    C a l c u l a t e t h e v a l u e o f P w h i c h m a k e s t h e w o r k d o n e o n t h e g a s i n t h e r s t c y c l e e q u a l t o

    t h e w o r k d o n e b y t h e g a s i n t h e s e c o n d c y c l e .

    S o l u t i o n 3 . 2

    F o r o n e m o l e o f a n i d e a l g a s P V = R T a n d W =

    R

    P d V . W i s t h e w o r k d o n e b y t h e

    g a s . T h e r e a r e 3 k i n d s o f p r o c e s s e s h e r e a n d t h e w o r k m u s t b e c a l c u l a t e d a c c o r d i n g l y .

    1 . F o r a n i s o t h e r m a l p r o c e s s ( T i s c o n s t a n t ) : P = R T = V s o W = R T l n

    V

    1

    V

    0

    = R T l n

    P

    0

    P

    1

    .

    2 . F o r a n i s o b a r i c p r o c e s s ( P i s c o n s t a n t ) : W = P ( V

    1

    ; V

    0

    ) = R ( T

    1

    ; T

    0

    )

    3 . F o r a c o n s t a n t v o l u m e p r o c e s s ( o n e o f t h e i s o c h o r i c p r o c e s s e s ) W = 0 b u t t h e p r e s s u r e

    a n d t e m p e r a t u r e w i l l c h a n g e .

    L e t ' s b e g i n b y s u m m a r i z i n g t h e t r a n s f o r m a t i o n s o f t h e g a s b y r e c o r d i n g t h e T , P , V b e f o r e

    a n d a f t e r e a c h p r o c e s s i n l i t e r s , a t m o s p h e r e s a n d K e l v i n a n d a l s o t h e f o r m o f t h e w o r k d o n e

    b y t h e s y s t e m .

    T

    0

    P

    0

    V

    0

    T

    1

    P

    1

    V

    1

    W

    i

    2 9 8 1 . 2 9 8 0 : 5 . W

    1

    = R T l n

    P

    0

    P

    1

    2 9 8 0 : 5 . 3 7 3 0 : 5 . W

    2

    = R ( T

    1

    ; T

    0

    )

    3 7 3 0 : 5 . 3 7 3 1 . W

    3

    = R T l n

    P

    0

    P

    1

    3 7 3 1 . 2 9 8 1 . W

    4

    = R ( T

    1

    ; T

    0

    )

    2 9 8 1 . 3 7 3 1 V

    u

    W

    0

    1

    = R ( T

    1

    ; T

    0

    )

    3 7 3 1 V

    u

    T

    3

    P

    u

    V

    u

    W

    0

    2

    = 0

    T

    3

    P

    u

    V

    u

    T

    4

    P

    u

    2 4 : 5 W

    0

    3

    = R ( T

    1

    ;T

    0

    )

    . P

    u

    2 4 : 5 . 1 2 4 : 5 W

    0

    4

    = 0

    F o r t h e r s t r e v e r s i b l e p r o c e s s t h e w o r k d o n e b y t h e s y s t e m i s W =

    P

    4

    i = 1

    W

    i

    .

    W =

    4

    X

    i = 1

    W

    i

    = 2 9 8 R l n 2 + R ( 3 7 3 ; 2 9 8 ) + 3 7 3 R l n 0 : 5 + R ( 2 9 8 ; 3 7 3 )

    W = ( 2 9 8 ; 3 7 3 ) R l n 2 = ; 7 5 R l n 2

    F o r t h e s e c o n d c y c l i c p r o c e s s t h e w o r k d o n e b y t h e s y s t e m i s W

    0

    =

    P

    4

    i = 1

    W

    0

    i

    .

    W

    0

    =

    4

    X

    i = 1

    W

    0

    i

    = R ( 3 7 3 ; 2 9 8 ) + 0 + R ( T

    4

    ; T

    3

    ) + 0

    W

    0

    = 7 5 R + R ( T

    4

    ; T

    3

    )

    S o w e m u s t n d t h e u n k n o w n t e m p e r a t u r e s u s i n g T = P V = R a n d V = R T = P .

    T

    3

    =

    P

    u

    R

    V

    u

    =

    P

    u

    R

    3 7 3 R

    1

    = 3 7 3 P

    u

    T

    4

    =

    P

    u

    2 4 : 5

    R

    N o w W

    0

    = 7 5 R + R (

    P

    u

    2 4 5

    R

    ; 3 7 3 P

    u

    ) = 7 5 R + R P

    u

    (

    2 4 5

    R

    ; 3 7 3 ) .

  • 8/12/2019 exweek3

    4/4

    T h e r e s t r i c t i o n t h a t t h e w o r k d o n e o n t h e g a s i n t h e r s t c y c l e e q u a l s t h e w o r k d o n e

    b y t h e g a s i n t h e s e c o n d c y c l e i s e q u i v a l e n t t o ; W = W

    0

    .

    W + W

    0

    = 0 = ; 7 5 R l n 2 + 7 5 R + R P

    u

    (

    2 4 : 5

    R

    ; 3 7 3 )

    P

    u

    =

    7 5 ( l n 2 ; 1 )

    2 4 5

    R

    ; 3 7 3

    I n t h e a b o v e R = 8 : 3 1 4

    J

    m o l e K

    1 a t m

    1 0 1 3 1 0

    5

    P a

    1 0

    3

    L

    m

    3

    = 0 : 0 8 2 0 7

    L a t m

    m o l e K

    . S o P

    u

    = 0 : 3 a t m .