Question

I have to design an insulated tank which can keep the grains for 65^oC for 1 hour. outside temp 25^oC, volume of the tank 5000L cp=4.1 kj/kgK, p=1060kg/m³. optimize the height and radius of the tank.

Answer 1

	Inputs
1)	Inside Temp -Tin = 65 C
2)	Outside Temp - T out = 25 C
3)	Volume of tank - V= 5000 lit
4)	cp =4.1 Kj/kg K
5)	density - p = 1060 kg/m^3
	Solution
	In this problem,
	ther are two constaints
	a) volume of tank - 5000 lit = 5 m3
	b) for minimum heat loss - surface area should be minimum
	If r is radius of tank and h is height ,
	volume V is given by
	V = $\tiny \prod$ r2h
	And
	surface area is given by
	A = 2 ∏ r2 + 2∏ rh
	Here A is objective that is to be minimized and V = 5 m^3
	represents the constraint
	From constaint equation h = V/( ∏ r2)
	substituting this relationship into expression for Area will give
	A = 2 ∏ r2 + 2∏ r V/( ∏ r2)
	A = 2 ∏ r2 + 2 V/ r
	Differentaiting A with respect to r and setting derivative equal to zero
	to obtain the radius to optimum value ,
	We get
	dA/ dr = 4 ∏ r - 2 V/ r2 = 0
	4 ∏ r = 2 V/ r2
	4 ∏ r3 = 2 V
	r3 = V /2∏
	r = ( V /2∏)1/3 &
	h =V/ ∏r2 = (4V/∏ )1/3
	Putting V =5 m3
	r = (5/2 x ∏)1/3 = 0.9266 m
	h = (4V/∏ )1/3 =(4*5/∏)1/3 = 1.853 m
	A = 2 ∏ r2 + 2∏ rh =16.189 m2
	Second derivative is calculated to determine nature of the optimum.
	Thus
	$\tiny \frac{\partial^2 }{\partial x^2}$ A = 4 x ∏ + 4 V/r3
	As r is positive, second derivative is also positive, it indicates that surface area is a
	minimum as required for mimimum heat loss.
	Thus minimum Radius - r = 0.9266 m
	and minimum height - h = 1.853 m
	for maintaing condition of minimum heat loss that grain temp is kept at 65 C for 1 hour