Temperatures of gases inside the combustion chamber of a four‑stroke automobile engine can reach up to 1000°C. To remove this en
ormous 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 4‑cylinder engine, each cylinder has a diameter of 8.75 cm, a height of 11.3 cm, and a thickness of 3.14 mm. The temperature on the inside of the cylinders is 195.2°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.
Suppose that, in a particular 4‑cylinder engine, each cylinder has a diameter of 8.75 cm, a height of 11.3 cm, and a thickness of 3.14 mm. The temperature on the inside of the cylinders is 195.2°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
1 answer:
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From the given information in the question, the correct option is Option 1: 14 cm.
A non-stretched elastic spring has a conserved potential energy which gives it the ability to perform work. The elastic potential energy can be expressed as:
PE = k
Where PE is the energy, k is the spring constant and x is extension.
i. Given that: PE = 10 J and x = 10 cm, then;
PE = k
10 = k
20 = 100k
k = 0.2 J/cm
ii. To determine how far the spring is needed to be stretched, given that PE = 20 J.
PE = k
20 = (0.2)
40 = 0.2
= 200
x =
= 14.1421
x = 14.14 cm
So that;
x is approximately 14.00 cm.
Thus, the spring need to be stretched to 14.00 cm to give the spring 20 J of elastic potential energy.
For more information, check at: brainly.com/question/1352053.
Answer:
I₁ > I₃ > I₂
Explanation:
Taking the pic shown, we have
m₁ = 10m₀
m₂ = 2m₀
m₃ = m₀
r₁ = r₀
r₂ = 2r₀
r₃ = 3r₀
We apply the formula
I = mr²
then
I₁ = m₁r₁² = (10m₀)(r₀)² = 10m₀r₀²
I₂ = m₂r₂² = (2m₀)(2r₀)² = 8m₀r₀²
I₃ = m₃r₃² = (m₀)(3r₀)² = 9m₀r₀²
finally we have
I₁ > I₃ > I₂
