Hot Water Cylinder: Min. Heat Loss Calc.

[Sagesky]

Homework Statement

To reduce heat loss, the surface area of a hot-water tank must be kept to a minimum. If such a tank is 125 litters in capacity, and can be approximated by a cylinder in shape with a hemispherical end cap; calculate the radius and overall height for minimum heat loss.hi can anyone check my ansers please thanks

The Attempt at a Solution

125 litres = 125000 cm^3

The volume is the sum of the volume of a hemisphere and a cylinder.
V= 2/3 πr^3+πr^2 h
The surface area is
S=2πrh+2πr^2
Isolate h in the Volume equation.
V= πr^3+πr^2 h

h=(V- 2/3 πr^2 )/(πr^2 )

Substitute for h into the Surface area equation.
S=2πr^2 (V- 2/3 πr^2 )/(πr^2 )+2πr^2
S=(2V )/r+2/3 πr^2
calculate the derivative.
ds/dr= - (2V )/r^2 +4/3 πr
Solve for r
- 2V + 4/3 πr^3=0

r^3= (3v )/2π

r^3= (3 ×125000 )/2π

r^3= 59683
r= ∛59683
r = 39.1 cm
to calculate the height
h=(125000- 2/3 π〖39.1〗^2 )/(π〖39.1〗^2 )

Height = 25 cmsorry can figuare how to make the equations non linnear

 

[hgfalling]

Your answer looks right to me; I didn't check the last couple calculations, but i get r = 39.1 cm, too.

 

[Sagesky]

thanks did i include the base ok?

 

[hgfalling]

Oh wait, I might have duplicated your mistake. :) The cylinder just has a hemispherical cap in place of one of the ends, but the other end is just a flat "circle" cap, right?

Then:

$$SA = SA_{walls} + SA_{flatcap} + SA_{hemcap} = 2 \pi r h + \pi r^2 + 2 \pi r^2 = 2 \pi r h + 3 \pi r^2$$

And then solve in the same way you did.

 

[Sagesky]

thank s

 

[Sagesky]

do i use the 2pirh + 3pir^2 for making h the subject

am hopeless at this question

 

[Sagesky]

hi i was check over my height equation is it

2pir^2 (v- 0.66 pi r^3/pi r^2) = height?