Question

Temperatures of gases inside the combustion chamber of a four‑stroke automobile engine can reach up to 1000 ∘C. To remove this enormous amount of heat, the engine utilizes a closed liquid‑cooled system that relies on conduction to transfer heat from the engine block into the liquid and then into the atmosphere by flowing coolant around the outside surface of each cylinder.

Suppose that, in a particular 6‑cylinder engine, each cylinder has a diameter of 8.25 cm, a height of 11.3 cm, and a thickness of 3.94 mm. The temperature on the inside of the cylinders is 188.4 ∘C, and the temperature outside, where the coolant passes, is 130.0 ∘C. The temperature of the incoming liquid (a mixture of water and antifreeze) is maintained at 103.4 ∘C.

What volume flow rate of coolant Vt would be required to cool this engine? Assume that the coolant reaches thermal equilibrium with the outer cylinder walls before exiting the engine. The specific heat of the coolant is 3.75 J/g⋅∘C and its density is 1.070×103 kg/m3. The cylinder walls have thermal conductivity of 1.10×102 W/m⋅∘C. Assume that no heat passes through the ends of the cylinders.

V/t = ________ cm³/s

Answer 1

The mode of heat transfer here is conduction

Law of conduction stats :

Q' = -kAdT/dx

A= Conduction AREA = 2πrh = 2x3.141x0.04125x0.113 = 0.02927 m2

dT = temperature difference = 188.4 - 130 = 58.4

dx = length along which heat flows = 3.94x10^-3 m

Heat transfer from cylinder:

Q = 1.10x10^2 x 0.02927 x 58.4 / 3.94x10^-3

= 0.47723 x 10^5 Joules/sec

Ths same heat(provided by 6 cylinders) is absorbed by the coolant leading to its temperature increase:

Qt = 2.86x10^5 Joules

Q = mCdT ......(i)

here dt = (130 - 103.4)

=26.6

C=3.75 j/g/C

density = 1.070 x10^3 kg/m3

Putting the values in equation-(i)

2.86x10^5 = m'x3.75x26.6

m' =2867.16 g/sec

m' = 2.867 kg/sec

V' = 2.867 / 1.070 x10^3

     = 2.679 x10^-3 m3

     = 2.679 x10^3 cm3/s ---> Answer