3 Hududni kengaytirish bo'yicha savollarga misollar
1. Po'lat list suhu 20oC ning uzunligi 50 sm va kengligi 30 sm. Agar po'latning chiziqli kengayish koeffitsienti 10 ga teng bo'lsa-5 oC-1 keyin 60 daraja haroratda maydon va umumiy maydonning oshishioC bu….
Munozara
Ma'lumki:
Boshlang'ich harorat (T)1) = 20oC
Yakuniy harorat (T)2) = 60oC
Harorat o'zgarishi (Δ)T) = 60oFZR - 20oC = 40oC
Boshlang'ich maydon (A)1) = Uzunlik x kenglik = 50 sm x 30 sm = 1500 sm2
Po'latning chiziqli kengayish koeffitsienti (α)) = 10-5 oC-1
Po'latning kengayish koeffitsienti (β)) = 2a = 2 x 10-5 oC-1
So'ralgan: Maydonning oshishi (Δ)A)
Maydonning oshishi (Δ)A):
ΔA = β A1DT
ΔA = (2 x 10-5 oC-1)(1500 sm2)(40)oC)
ΔA = (80) x 10-5)(1500 sm2)
ΔA = 120.000 x 10-5 cm2
ΔA = 1,2 x 105 x 10-5 cm2
ΔA = 1,2 sm2
Umumiy maydon (A2):
A2 = A1 +ΔA
A2 = 1500 sm2 + 1,2 sm2
A2 = 1501,2 sm2
2. Chiziqli kengayish koeffitsienti 24 x 10 bo'lgan alyuminiy plastinka-6 /oC ning maydoni 40 sm ga teng2 30 daraja haroratdaoC. Agar alyuminiy plastinkaning maydoni 40,2 sm ga oshsa, yakuniy haroratni aniqlang2.
Munozara
Ma'lumki:
Boshlang'ich harorat (T)1) = 30oC
Alyuminiyning chiziqli kengayish koeffitsienti (α)) = 24 x 10-6 oC-1
Alyuminiyning kengayish koeffitsienti (β)) = 2α = 2 x 24 x 10-6 oC-1 = 48 x 10-6 oC-1
Boshlang'ich maydon (A)1) = 40 sm2
Yakuniy maydon (A)2) = 40,2 sm2
Maydondagi o'zgarish (Δ)A) = 40,2 sm2 - 40 sm2 = 0,2 sm2
So'ralgan: Yakuniy haroratni aniqlang (T2)
Javob:
Maydon o'zgarishi formulasi (Δ)A) :
ΔA = β A1 ΔT
Yakuniy harorat (T)2):
ΔA = β A1 (T2 - T1)
0,2 sm2 = (48 x 10-6 oC-1)(40 sm2)(T2 - 30oC)
0,2 = (1920 x 10-6)(T2 - 30)
0,2 = (1,920 x 10-3)(T2 - 30)
0,2 = (2 x 10-3)(T2 - 30)
0,2 / (2 x 10-3) = T2 - 30
0,1 x 103 = T2 - 30
1 x 102 =T2 - 30
100 = T2 - 30
100 + 30 = T2
T2 = 130
Yakuniy harorat = 130oC
3. 20 daraja haroratdagi dumaloq jismoC ning radiusi 20 sm ga teng. Agar 100 haroratda bo'lsaoC radiusi 20,5 sm ga oshsa, obyektning chiziqli kengayish koeffitsienti… ga teng.
Munozara
Ma'lumki:
Boshlang'ich harorat (T)1) = 30oC
Yakuniy harorat (T)2) = 100oC
Harorat o'zgarishi (Δ)T) = 100oFZR - 30oC = 70oC
Boshlang'ich radius (r)1) = 20 sm
Yakuniy radius (r)2) = 20,5 sm
So'ralgan: Ob'ekt maydonining kengayish koeffitsienti (b)
Javob:
Doira boshlang'ich maydoni (A1) = π r12 = (3,14)(20 sm)2 = (3,14)(400 sm)2) = 1256 sm2
Doiraning oxirgi maydoni (A)2) = π r22 = (3,14)(20,5 sm)2 = (3,14)(420,25 sm)2) = 1319,585 sm2
Doira maydonining oshishi (ΔA) = 1319,585 sm2 - 1256 sm2 = 63,585 sm2
Maydon o'zgarishi formulasi (Δ)A) :
ΔA = β A1 DT
Maydon kengayish koeffitsienti:
ΔA = β A1 DT
63,585 sm2 = β (1256 sm2)(70) oC)
63,585 = β (87920 oC)
b = 63,585 / 87920 oC
β = 0,00072 /oC
β = 7,2 x 10-4 /oC
Chiziqli kengayish koeffitsienti (α)):
β = 2 α
α = β / 2
α = (7,2) x 10-4) / 2
α = 3,6 x 10-4 oC-1
