pengantar rekayasa dan desain
DESCRIPTION
Tugas Mata kuliah Pengantar Rekayasa dan Desain Institut Teknologi Bandung. KU1101 Kelas 23Dinda Ayu Rahmi NIM 16513212TRANSCRIPT
Pengantar Rekayasa dan Desain Dinda Ayu Rahmi
SOAL LATIHAN 1. Tekanan bejana, p = 350 psi = 24,608 kgf/cm2 = 2,413 MPa
Solution: 1psi = 6,895 KPa. 350 Psi = 2431,25 KPa = 2,413 MPa 1 kgf/cm2 = 98,0665 KPa. 2431,25 KPa : 98,0665 KPa = 24,608 kgf/cm2
2. Bahan bejana, σallowable = σboleh = 15000 psi = 15 ksi = 10,546 kgf/mm2 = 103,425 MPa
Solution: 15 ksi x 6,895 = 103,425 MPa 103425 KPa : 98,0665 = 1054,6 kgf/cm2 = 10,546 kgf/mm2
3. Spesifikasi bahan, St.37, Su = 37 kgf/mm2 = 52,626 ksi Solution: 1 kgf/mm2 = 9806,65 KPa. 1 psi = 6,895 KPa. 9806,65 : 6,895 = 1422,28426. 37 x 1422,28426 = 52624 psi = 52,624 ksi
4. Berat jenis baja, γ = 7,8 kgf/dm3 = 0,2817 lbf/in3 5. Modulus elastis baja, E = 200 GPa = 29.007.547,5 psi
Solution: 1 GPa = 1 000 000 KPa 200 000 000 : 6,895 = 29 007 547,5
6. Nilai kalor (HHV: higher heating value) LNG 10000 kcal/m3 = 11,6222222 kWh/m3 = 0,00112370329 MM Btu/ft3 Exercise (Exploring Engineering page. 32) If a US gallon has a volume of 0.134 ft3 and a human mouth has a volume of 0.900 in3, then how many mouthfuls of water are required to fill a 5.00 US gallon can? Solution: 5 US gallon = 5 x 0,134 = 0,67 ft3 0,67 ft3 = 1157.76 in3 1157.76 in3 / 0.900 in3 = 1286,4 mouthfuls ≈ 1,29 x 103 mouthfuls