You need to choose a thermos that can keep 3 litters of water at temperature of
ID: 1596776 • Letter: Y
Question
You need to choose a thermos that can keep 3 litters of water at temperature of 360K (assume constant) as long as possible while you are skiing on a cold day with outside temperature of 273K. You have six thermoses to choose from: The first three are made of the material with the mean thermal conductivity of 0.8W/(m*K), wall thickness of 6mm and available in three shapes: i) cubic, ii) hollow cylinder with inner diameter of 14cm and iii) hollow cylinder with inner diameter of 5cm Other three thermoses are made of a material with mean thermal conductivity of 0.5 W/(m*K), thickness of 4mm and the available shapes are: iv) spherical, v) hollow cylinder with inner radius 16cm and vi) hollow cylinder with inner radius of 3cm. Assuming that the fluid inside thermoses occupies the whole volume and ignoring the heat transfer due to convection, determine which container will be more efficient preserving the fluid temperature (i.e. has the lowest rate of heat loss). Support your answer with numerical evidence.Explanation / Answer
Rate of heat transfer through a cubical container = 6K (x1) (x2) ( delta T) / ( x2 - x1)
x1 and x2 are inner and outer side of cube.
(x1)3 = 3000 cm3
x1 = 10 x 31/2 cm x2 = x1 + 0.6 cm
Hence H1 = 6x 0.008 x 10 x 31/2 x ( 10(3)1/2 + 0.6 ) (delta T ) / 0.6 = 24.83 (delta T) W
Rate of heat transfer through a cylindrical container = K 2 pi L (delta T) / Ln(r2/r1)
where r1 and r2 are inner and outer radius of cylinder and l is it's height.
For thermose 2 , r1 = 7, r2 = 7.6
3000 = pi r12 L
L = 19.5 cm
H2 = 0.008 x 2 x 3.14 x 19.5 (delta T) / Ln( 7.6/7) = 11.9 (delta T) W
For thermose 3 , r1 = 2.5, r2 = 3.1
L = 152.9 cm
H3 = 0.008 x2 x3.14 x 152.9 (delta T) / Ln ( 3.1/2.5) = 35.7 (delta T) W
For thermose 5 , r1 = 16, r2 = 16.4
L = 3.73 cm
H5 = 0.005 x 2 x 3.14 x 3.73 (deltaT) / Ln( 16.4/16) = 4.74 (delta T) W
For thermose 6 , r1 = 3, r2 = 3.4
L = 106.1
H6 = 0.005 x 2x 3.14 106.1 (delta T) / Ln(3.4/3) = 26.6 (delta T) W
Rate of heat transfer through spherical contaner = 4 k pi r1 r2 (delta T) / ( r2 -r1)
3000 = 4 pi r13 /3
r1 = 8.95 cm
H4 = 4 x0.005 x 8.95 x 9.35 x3.14 9delta T) / 0.4 = 13.14 (delta T) W
Thermose 5 is most efficient as, rate of heat transfer through this per unit tempreture difference across walls is minimum.
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