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 5 ‑cylinder engine, each cylinder has a diameter of 8.25 cm, a height of 10.4 cm, and a thickness of 2.66 mm. The temperature on the inside of the cylinders is 195.2 ∘ C, and the temperature outside, where the coolant passes, is 131.4 ∘ C. The temperature of the incoming liquid (a mixture of water and antifreeze) is maintained at 101.3 ∘ C. What volume flow rate of coolant V t 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 × 10 3 kg / m 3 . The cylinder walls have thermal conductivity of 1.10 × 10 2 W / m ⋅ ∘ C. Assume that no heat passes through the ends of the cylinders.

Answer 1

Answer:- Given that Specific Heat -3.75 J/jc density = 1.070 X 10% kg/me Thermal conductivity =lolx16 W/mc Diameter = 8.25cm a height = 10.4cm thickness = 2.66 mm Temparature Inside =195.2'c Temparature outside = 13104'c Temperature for incoming liquid = 101.3°C Total amount of heat Yoss 94 = KA AT KATTA PH = 110X5K (X010 825x0.604) x4 22.66X103 PH = 355,58x163W So the total amount of heat bass is same as that of negative of Total amount : 9c = - PH = -355.58X1oW = -355,58 kW. Total amount of heat gain Pc = mcp (T-T2) - 355,58 = m (3.75) (101.3-131.4) m= 355.58 3.75X30.1 m = 3.15 kgls So the change in valumetric flow rate is va massm. 3.15 density 3 - 7.07X103 V=2.94X163 m/s - 0.00294x10 fem? Is . V= 2.94 4s.