Sabtu, 14 Maret 2009

tugas anorganik3

The Railways

The influence of thermal strains/stresses caused by braking on the fatigue life of cast nodular iron disc brakes for railway vehicles. If not restrained, rails would lengthen in hot weather and shrink in cold weather. To provide this restraint, the rail is prevented from moving in relation to the sleeper by use of clips or anchors. Anchors are more common for wooden sleepers, whereas most concrete or steel sleepers are fastened to the rail by special clips which resist longitudinal movement of the rail. There is no theoretical limit to how long a welded rail can be. However, if longitudinal and lateral restraint are insufficient, the track could become distorted in hot weather and cause a derailment. Distortion due to heat expansion is known in North America as sun kink, and elsewhere as buckling. In North America a rail broken due to cold-related contraction is known as a pull-apart. Attention needs to be paid to compacting the ballast effectively, including under, between, and at the ends of the sleepers, to prevent the sleepers from moving. In extreme hot weather special inspections are required to monitor sections of track known to be problematic.
After new segments of rail are laid, or defective rails replaced (welded-in), the rails are artificially stressed. The stressing process involves either heating the rails causing them to expand, or stretching the rails with hydraulic equipment. They are then fastened (clipped) to the sleepers in their expanded form. This process ensures that the rail will not expand much further in subsequent hot weather. In cold weather the rails try to contract, but because they are firmly fastened, cannot do so. In effect, stressed rails are a bit like a piece of stretched elastic firmly fastened down.
Engineers try to heat the rail to a temperature roughly midway between the average extremes of hot and cold (this is known as the 'rail neutral temperature'). If temperatures reach outside normal ranges however, welded rail can buckle in a hotter than usual summer or can actually break in a colder than anticipated winter. In North America, because broken rails are typically detected by the signaling system; they are seen as less of a problem than heat kinks which are not detected. For this reason, and because it is harder to break a rail than displace the trackbed, CWR is usually installed at a temperature of 90 °F (32 °C), to cope with rail temperature extremes of nearly 120 °F (50 °C) in the summer sun.
Referensi:
http://en.wikipedia.org/wiki/Rail_tracks
http://www.revue-metallurgie.org/index.php?option=article&access=standard&Itemid=129&url=/articles/metal/pdf/2007/10/metal07562.pdf

tugas anorganik3

1.12. panjang satuan sel intan terukur 0,3567 nm, hitung volum satuan sel kubus intan dan hitung rapatan teoritis intan jika massa satu atom karbon adalah 12,01 g/mol; bandingkan hasilnya dengan rapatan intan terukur pada 25 C yaitu 3.513 g/ml?
Jawab :
Diket : a = 0,3567 nm
= 0,3567x10^-9 m
= 0,3567x10^-7 cm
Banyaknya atom intan adalah 4 atom utama dari bangun fcc dan 4 atom terikat secara tetrahedral di dalamnya; oleh karena itu setiap satuan sel intan terdapat : (8 x 1/8) + (6 x 1/2 atom pusat muka) + 4 atom interior = 8 atom
Volume = a ^3
= (0,3567x10^-7)^3
= 4.538x10^-23 ml
Rapatan pada suhu 25 C adalah 3,513 g/ml,
Rapatan = banyaknya atom x massa atom relative : (bil.avogadro x volum sel)
= 8 x 12 : ( 6,02x10^23 x 4,538x10^-23)
= 3,514 g/ml
jika dibandingkan antara rapatan teoritik dengan rapatan intan terukur terdapat selisih yang sangat kecil, hal ini disebabkan adanya pengotor (impurity).

tugas anorganik3

Jawaban soal latihan ikatan metalik
1.10. satuan sel emas adalah kubus pusat muka (fcc). Berapa jumlah atom yang menempati satu satuan sel emas, dan berapa massa satu satuan sel emas ini?
Jawab:
Untuk kubus bangun pusat muka terdapat enam atom tambahan yang menempati keenam muka kubus,sehingga tiap satuan sel kubus pusat muka terdapat:
6(1/2) + (8(1/8))=4 atom
Massa satuan sel = Massa molar x banyaknya atom : bil. avogadro
= 196,97 x 4 : (6,02x10^23)
= 1,3088x10^-21
1.11. panjang satuan sel emas adalah 0,4079 nm, hitung volum satu satuan sel kubus emas dengan informasi dari soal 1.10, hitung pula rapatan teoritis emas ini?
Jawab :
Diket: a = 0,4079 nm
= 0,4079x10^-9 m
= 0,4079x10^-7 cm
Volume= a^3
= (0,4079x10^-7)^3
= 6,787x10^-23 ml
Rapatan= banyaknya atom x massa atom relative : (bil. Avogadro x volume sel)
= 4 x 196,97 : ( 6,02x10^23 x 6,787x10^-23)
= 19,28 g/ml