Temperatures of gases inside the combustion chamber of a fou

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 which 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 the cylinder. Assume you have a 5-cylinder engine, and each cylinder has a diameter of 9.75 cm and height of 11.3 cm and is 2.50 mm thick. The temperature on the inside of the cylinders is 185.0°C whereas the temperature where the coolant passes is 131.4°C. The temperature of the liquid (a mixture of water and antifreeze) is to be maintained at 101.3°C. Using the given parameters, what flow rate would need to be supplied by the water pump to cool this engine? (See the hint panel for other needed constants.)

hint : Equate the heat equation per unit time for the liquid coolant to the power transferred by conduction through the cylinder walls to the coolant. You can arrive at an equation which is in terms of volume per unit time. The specific heat of the water/antifreeze mixture is 3.75 J/g·°C and has density of 1.070 × 103 kg/m3 and the cylinder wall has thermal conductivity of 1.10 × 102 W/m·°C. Ignore the ends of the cylinders in the area calculation since these are not cooled.

Solution

temp. inside the chamber = 185 deg.C

temp. outside = 131.4 deg.C

conductivity K = 1.1e+2 W/m-C

area of the surface A= pi*0.0975*0.113 = 3.46e-2 sq.m

length l = 2.5e-3 m

heat conduction rate through the cylinder surface

= 1.1e+2*3.46e-2*(183-131.4)/2.5e-3 = 81600 w/s

this heat need to be rmoved by the coolant whose tempearture need to be maintained at 101.3 deg.C

if M is the mass of the coolant flows per unit time then heat removed per unit time

= M*C*delT

specifric heat of coolant C = 3.75 J/g-C

del T = 131.4 - 101.3 = 30.1 deg.c

M*3.75*30.1 = 81600

M = 722.92 gms/s flow fate of the coolant

density of the coolant d = 1.070e+3 kg/cu.m   = 1.070 gm/cc

flow rate = 722.92/1.070 = 675.63 cc/s

Note: inlet temperature of the coolant is not given. the flowrate depends on howmuch heat the coolant can take. In the above computation we have assumed the inlet temperature as 101.3 C.

Instead if the inlet temp =0 C

then the coolant can be heated up to 101.3 C

heat absorbed by the coolant.

M*3.75*101.3 = 81600

M = 218.4 gms/s flow fate of the coolant

density of the coolant d = 1.070e+3 kg/cu.m   = 1.070 gm/cc

flow rate = 218.4/1.070 = 200.75 cc/s