Question
Really stuck on this! Thanks!
Temperatures of gases inside the combustion chamber of a four stroke automobile engine can reach up to 1000 degree C. To remove this enourmous 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. Let's assume we have a 5 cylinder engine and each cylinder has a diameter of 8.75 cm and height of 10.1 cm and is 2.82 mm thick. The temperature on the inside of the cylinders is 188.4 degree C whereas the temperature where the coolant passes is 131.4 degree C. The temperature of the liquid (a mixture of water and antifreeze) is to be maintained at 103.4 degree 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.) uate 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 degree C and has density of 1.070 times 103 kg/m3 and the cylinder wall has thermal conductivity of 1.10 times 102 W/m degree C. Ignore the ends of the cylinders in the area calculation since these are not cooled.
Explanation / Answer
Heat transfer rate per cyclinder = (188.4-131.4)/R_thermal
R_thermal for cyclinder =L/k*Amean= (R2-R1)/(k*A_mean)
A_mean = ln(A2/A1)
R_thermal = 1/2*pi*K*H * ln(R2/R1)
= 1/(2*pi*110*0.101)*ln(8.75/3.11)
= 0.0148
tota heat transfer Q_dot = 3846.51*5 = 19232.55 watts.
this is equal to heat lost to coolant
rho*Vol_rate*cp(103.4- T_amb) = 3846.51*5
assuming T_ambient = 25 C
1070*Vol_rate*3.75*1000*78.4 = 3846.51*5
Vol_rate = 61.14 cm^3/s
