Temperatures of gases inside the combustion chamber of a four-stroke automobile

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 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 each cylinder. Assume you have a 5-cylinder engine, and each cylinder has a diameter of 8.75 cm and height of 11.0 cm and is 2.50 mm thick. The temperature on the inside of the cylinders is 191.8°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 95.0°C. What volume flow rate of coolant 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.

Equate the heat transferred per unit time to the liquid coolant to the power transferred by conduction through the cylinder walls. 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.

I have tried this problem and these are INCORRECT answers, 2660 and 48.709

Explanation / Answer

n = 5 cylinder engine
final temperature of the outside of cylinder, Tf = 130 C
initial temperature of the coolant, Ti = 95 C
temperature inside the cylinder, T = 191.8 C
radius of cylinder , r = 0.0875/2 m
height of cylinder, H = 0.11 m
thickness of cylinder , t = 2.5*10^-3 m

so, rate of heat transfer from walls of one cylinder, q = 2*k*pi*L(T - Tf)/ln((r + t)/r)
where k is conductivity of the material of cylinder
given k = 110 W/m C

so, q = 2*110*pi*0.11(191.8-130)/ln(1 + 2.5*2/87.5) = 84550.169 w

so in 1 second, heat gained by the coolant = m*C(Tf - Ti)
where m is mass of coolant needed in 1 second and C is heat capacity of coolant, C = 3.75 J/g C
so, considering all cylinders
84550.169*5 = m*3.75(130 - 95)
m = 3220.9588 grams / sec
now mass flow rate is rho*A*V
rho is density of the coolant
Av is volume flow rate
so, 1070*Av = 3220.9588/1000
Av = 3010.2418 cm^3/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.