3 Tusaalooyin su'aalo ku saabsan ballaarinta aagga
1. Xaashi bir ah oo heerkulkeedu yahay 20 o C waxay leedahay dherer dhan 50 cm iyo ballac dhan 30 cm. Haddii isku-dhafka ballaarinta toosan ee birta uu yahay 10 -5 o C -1 markaa kororka aagga iyo guud ahaan bedka heerkulka 60 o C waa….
Dood
Waa la ogyahay in:
Heerkulka bilowga ah (T 1 ) = 20 o C
Heerkulka kama dambaysta ah (T2 ) = 60 o C
Isbeddelka heerkulka (Δ T) = 60 o C – 20 o C = 40 o C
Bedka bilowga ah (A 1 ) = Dhererka x ballaca = 50 cm x 30 cm = 1500 cm 2
Isugeynta ballaarinta toosan ee birta (α ) = 10 -5 o C -1
Isugeynta ballaarinta birta (β)) = 2a = 2 x 10-5 oC-1
La weydiiyay: Kordhinta aagga (ΔA)
Kordhinta aagga (ΔA):
ΔA = β A1ΔT
ΔA = ( 2 x 10 -5 o C -1 )(1500 cm 2 )(40 o C)
ΔA = (80 x 10 -5 ) (1500 cm 2 )
ΔA = 1 20.000 x 10 -5 cm 2
ΔA = 1,2 x 105 x 10-5 cm2
ΔA = 1,2 cm2
Wadarta bedka (A2):
A2 =A1 +ΔA
A2 = 1500cm2 + 1,2 cm2
A2 = 1501,2 cm2
2. Saxanka aluminiumka oo leh isku-darka ballaarinta toosan ee 24 x 10-6 /oC waxay leedahay bedka 40 cm2 heerkul ah 30oC. Go'aami heerkulka kama dambaysta ah haddii bedka saxanka aluminiumku uu kordho 40,2 cm2.
Dood
Waa la ogyahay in:
Heerkulka bilowga ah (T 1 ) = 30 o C
Isugeynta ballaarinta toosan ee aluminiumka (α)) = 24 x 10-6 oC-1
Isugeynta ballaarinta aluminiumka (β)) = 2α = 2 x 24 x 10-6 oC-1 = 48x 10-6 oC-1
Aagga bilowga ah (A)1) = 40cm2
Aagga kama dambaysta ah (A)2) = 40,2cm2
Isbeddelka aagga (ΔA) = 40,2 cm2 - 40 cm2 = 0,2cm2
La weydiiyay: Go'aami heerkulka kama dambaysta ah (T)2)
Jawaab:
Qaacidada isbeddelka aagga (Δ)A) :
ΔA = β A1 ΔT
Heerkulka kama dambaysta ah (T)2):
ΔA = β A1 (T2 - T1)
0,2 cm2 = (48 x) 10-6 oC-1(40 cm)2)(T2 - 30oC)
0,2 = (1920 x 10 -6 ) (T 2 – 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 = T 2
T 2 = 130
Heerkulka kama dambaysta ah = 130oC
3. Shay wareegsan oo heerkulkiisu yahay 20oC wuxuu leeyahay gacan 20 cm ah. Haddii heerkulku yahay 100oC gacanku wuxuu kordhaa ilaa 20,5 cm ka dibna isku-dhafka ballaarinta toosan ee shayga waa….
Dood
Waa la ogyahay in:
Heerkulka bilowga ah (T 1 ) = 30 o C
Heerkulka kama dambaysta ah (T2 ) = 100 o C
Isbeddelka heerkulka (Δ T) = 100 o C – 30 o C = 70 o C
Gacanka bilowga ah (r)1) = 20cm
Gacanka kama dambaysta ah (r)2) = 20,5cm
La weydiiyay: Isugeynta ballaarinta aagga ee shayga (b)
Jawaab:
Aagga bilowga ah ee goobada (A)1) = π r12 = (3,14)(20 cm)2 = (3,14) (400 cm)2) = 1256cm2
Aagga ugu dambeeya ee goobada (A)2) = π r22 = (3,14)(20,5 cm)2 = (3,14) (420,25 cm)2) = 1319,585cm2
Kordhinta aagga goobada (ΔA) = 1319,585 cm2 - 1256 cm2 = 63,585cm2
Qaacidada isbeddelka aagga (Δ)A) :
ΔA = β A1 ΔT
Isugeynta ballaarinta aagga:
ΔA = β A1 ΔT
63,585 cm2 = β (1256 cm2(70) oC)
63,585 = β (87920 oC)
β = 63,585 / 87920 o C
β = 0,00072 / o C
β = 7,2 x 10-4 /oC
Isugeynta ballaarinta toosan (α)):
β = 2 α
α = β / 2
α = (7,2 x 1 0 -4 ) / 2
α = 3,6 x 10-4 oC-1
