3 Hududni kengaytirish bo'yicha savollarga misollar
1. 20 o C haroratdagi po'lat plitaning uzunligi 50 sm va kengligi 30 sm. Agar po'latning chiziqli kengayish koeffitsienti 10 -5 o C -1 bo'lsa, 60 o C haroratda maydon va umumiy maydonning oshishi … ga teng.
Munozara
Ma'lumki:
Boshlang'ich harorat (T1 ) = 20 o C
Yakuniy harorat (T2 ) = 60 o C
Harorat o'zgarishi (Δ T) = 60 o C – 20 o C = 40 o C
Boshlang'ich maydon (A1 ) = Uzunlik x kenglik = 50 sm x 30 sm = 1500 sm 2
Po'latning chiziqli kengayish koeffitsienti (α ) = 10 -5 o C -1
Po'latning kengayish koeffitsienti (β ) = 2a = 2 x 10 -5 o C -1
Savol: Maydonning oshishi (Δ A )
Maydonning oshishi (Δ A ):
ΔA = β A 1Δ T
ΔA = ( 2 x 10 -5 ° C -1 )(1500 sm 2 )(40 ° C)
DA = (80 x 10 -5 )(1500 sm 2 )
ΔA = 1 20.000 x 10 -5 sm 2
ΔA = 1,2 x 10 5 x 10 -5 sm 2
ΔA = 1,2 sm2
Umumiy maydoni ( A2 ) :
A 2 = A 1 + Δ A
A 2 = 1500 sm 2 + 1,2 sm 2
A 2 = 1501 , 2 sm 2
2. Chiziqli kengayish koeffitsienti 24 x 10 -6 / o C bo'lgan alyuminiy plastinkaning 30 o C haroratda 40 sm2 maydoni bor. Agar alyuminiy plastinkaning maydoni 40,2 sm2 ga oshsa, yakuniy haroratni aniqlang.
Munozara
Ma'lumki:
Boshlang'ich harorat (T1 ) = 30 o C
Alyuminiyning chiziqli kengayish koeffitsienti (α ) = 24 x 10 -6 o C -1
Alyuminiyning maydon kengayish koeffitsienti (β ) = 2α = 2 x 24 x 10 -6 o C -1 = 48 x 10 -6 o C -1
Boshlang'ich maydon (A1 ) = 40 sm2
Yakuniy maydon (A2 ) = 40,2 sm2
Maydonning o'zgarishi (Δ A) = 40,2 sm 2 – 40 sm 2 = 0,2 sm 2
Savol: Yakuniy haroratni aniqlang ( T2 )
Javob:
Maydon o'zgarishi formulasi (Δ A) :
ΔA = β A 1 Δ T
Yakuniy harorat (T2 ) :
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 ) (T 2 – 30)
0,2 = (2 x 10 -3 ) (T 2 – 30)
0,2 / (2 x 10 -3 ) = T 2 – 30
0,1 x 10 3 = T 2 – 30
1 x 10 2 = T 2 – 30
100 = T 2 – 30
100 + 30 = T2
T2 = 130
Yakuniy harorat = 130 o C
3. 20 o C haroratdagi dumaloq jismning radiusi 20 sm ga teng. Agar 100 o C haroratda radius 20,5 sm ga oshsa, u holda jismning chiziqli kengayish koeffitsienti… ga teng.
Munozara
Ma'lumki:
Boshlang'ich harorat (T1 ) = 30 o C
Yakuniy harorat (T2 ) = 100 o C
Harorat o'zgarishi (Δ T) = 100 o C – 30 o C = 70 o C
Boshlang'ich radius (r1 ) = 20 sm
Yakuniy radius (r2 ) = 20,5 sm
Savol: Ob'ektning maydonini kengaytirish koeffitsienti ( b )
Javob:
Aylananing boshlang'ich maydoni (A1 ) = π r12 = ( 3,14 )(20 sm) 2 = (3,14)(400 sm2 ) = 1256 sm2
Aylananing yakuniy maydoni (A2 ) = π r22 = ( 3,14 )(20,5 sm) 2 = (3,14)(420,25 sm2 ) = 1319,585 sm2
Doira maydonining oshishi (Δ A) = 1319,585 sm 2 – 1256 sm 2 = 63,585 sm 2
Maydon o'zgarishi formulasi (Δ A) :
ΔA = β A 1 ΔT
Maydon kengayish koeffitsienti:
ΔA = β A 1 ΔT
63,585 sm 2 = b (1 256 sm 2 )(70 o C)
63,585 = β ( 87920 ° C )
β = 63,585 / 87920 o C
β = 0,00072 / o C
β = 7,2 x 10 -4 / o C
β = 7,2 x 10 -4 o C -1
Chiziqli kengayish koeffitsienti (α ):
β = 2 α
α = β / 2
α = (7,2 x 10 -4 ) / 2
α = 3,6 x 10 -4 o C -1
