TUGAS 3CP gas merupakan fungsi suhu

CP= a + bt + ct2

T0F 0 200 400 600 800 1000CP Btu/lb.mol0F

9 9,53 10,04 10,54 11,01 11,45

Hitunglah panas yang diperlukan untuk menaikkan suhu gas tersebut dari 2000F sampai 8000F dengan cara sbb

1. Carilah dulu persamaan CP diatas dari 3 titik (CP,t) secara simultan lalu diintegralkan 2. Cara integrasi grafis

Jawab :

# CP = a +bt+ct2

10,54 =a +b.600+c.6002 10,54 = a+600b+360000c10,04 =a+b.400+c.4002 10,04 =a+400b+160000c

0,5 = 200b+200000c

# 10,04 = a+400b+160000c x1 10,04= a+400b+160000c

9,53 = a+200b+40000c x2 19,06=2a+400b+80000c

-9.02=-a+80000c

a= 9.02 +80000c

# 10,54 = a+600b+ 360000c x1 10,54 = a+ 600b+ 360000c

9,53 = a+200b+40000c x9 85,77 = 9a + 1800b +360000c

-75,23 = -8a – 1200b

8a = 75,23 – 1200b

8a = 75,23 – 1200b 8(9,02 + 80000c) = 75,23 – 1200b 72,16 + 640000c= 75,23 – 1200b

640000c +1200b= 75,23 – 72,16 640000c +1200b= 3,07

# 3,07 = 1200b + 640000c x1 3,07 = 1200b + 640000c0,5 = 200b + 200000c x6 3 = 1200b + 1200000c

0,07 = - 560000cc = 0,000000125

640000(- 0,000000125) + 1200b = 3,07 - 0,08 +1200b = 3,07

1200b = 3,07 + 0,08 b = 0,002625

# 9,53 = a + 200(0,002625) + 40000 (0,000000125)9,53 = a + 0,525 – 0,005 9,53 = a + 0,52 a = 9,01

# CP =

=

= 9,01 t +1/2 (0,002625)t2 + 1/3 (-0,000000125)t3

= [9,01(800) + ½ (0,002625)(800)2 + 1/3 (-0,000000125)(800)3]- [9,01(200)+1/2(0,002625)(200)2 + 1/3 (-0,000000125)(200)3]

= (7208 + 840 – 21,3 ) – ( 1802 + 52,1 – 1)

=8026,7 – 1853,5

= 6173,2

L1 = ½ (9,53 + 10,04 ) 200

=19,57 x 100

=1957

L2 =½ (10,54 +10,04)200

=20,58 x 100

=2058

L2 =½ (10,54 +11,01)200

=21,55 x 100

=2155

Ltot = 1957 + 2058 + 2155

= 6170