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4 years ago

(3) What makes the Boxer engine balance so well?

Engineering

1 answer:

lord [1]4 years ago

5 0

Answer:

Just like two boxing athletes competing in a match, the boxer engine works similarly. The horizontally positioned pistons inside the engine create a better balance with their side-to-side motion as compared to V-shaped engines.

Explanation:

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3 years ago

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A two-dimensional flow field described by

Oduvanchick [21]

Answer:

the answer is

Explanation:

<h2> We now focus on purely two-dimensional flows, in which the velocity takes the form </h2><h2>u(x, y, t) = u(x, y, t)i + v(x, y, t)j. (2.1) </h2><h2>With the velocity given by (2.1), the vorticity takes the form </h2><h2>ω = ∇ × u = </h2><h2> </h2><h2>∂v </h2><h2>∂x − </h2><h2>∂u </h2><h2>∂y </h2><h2>k. (2.2) </h2><h2>We assume throughout that the flow is irrotational, i.e. that ∇ × u ≡ 0 and hence </h2><h2>∂v </h2><h2>∂x − </h2><h2>∂u </h2><h2>∂y = 0. (2.3) </h2><h2>We have already shown in Section 1 that this condition implies the existence of a velocity </h2><h2>potential φ such that u ≡ ∇φ, that is </h2><h2>u = </h2><h2>∂φ </h2><h2>∂x, v = </h2><h2>∂φ </h2><h2>∂y . (2.4) </h2><h2>We also recall the definition of φ as </h2><h2>φ(x, y, t) = φ0(t) + Z x </h2><h2>0 </h2><h2>u · dx = φ0(t) + Z x </h2><h2>0 </h2><h2>(u dx + v dy), (2.5) </h2><h2>where the scalar function φ0(t) is arbitrary, and the value of φ(x, y, t) is independent </h2><h2>of the integration path chosen to join the origin 0 to the point x = (x, y). This fact is </h2><h2>even easier to establish when we restrict our attention to two dimensions. If we consider </h2><h2>two alternative paths, whose union forms a simple closed contour C in the (x, y)-plane, </h2><h2>Green’s Theorem implies that </h2><h2> </h2><h2> </h2><h2> </h2><h2> </h2><h2> </h2><h2></h2><h2></h2>

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3 years ago

A smooth concrete pipe (1.5-ft diameter) carries water from a reservoir to an industrial treatment plant 1 mile away and dischar

Kamila [148]

ANSWER:

Q = 0.17ft3/s

EXPLANATION: since the water runs downhill on a 1:100 slope, that means the flow is laminar.

Using poiseuille equation:

Q = (π × D^4 × ∆P) ÷ (128 × U × ∆X)

Q is the volume flow rate.

π is pie constant value at 3.142

D is the diameter of the pipe

∆P is the pressure drop

U is the viscosity

∆X is the length of the pipe or distance of flow.

Form the question, we are to determine U then Find Q

Therefore;

D = 1.5ft

∆P = 1pa since the minor losses are negligible.

∆X = 1mile = 5280ft.

STEP1: FIND U

Viscosity is a function of the temperature of the liquid. An increase in temperature increases the viscosity of the liquid.

We know that at room temperature, which is 25°C the viscosity of water is 8.9×10^-4pa.s . We can find the viscosity of water at 4°C by cross multiplying.

Therefore;

25°C = 8.9×10^-4pa.s

4°C = U

Cross multiply

U25°C = 4°C × 8.9×10^-4pa.s

U25°C = 0.00356°C.pa.s

Therefore;

U = 0.00356°C.pa.s ÷ 25°C

U = 1.424×10^-4pa.s

Therefore at 4°C the viscosity of water in the pipe is 1.424×10^-4pa.s

STEP2: FIND Q

Imputing the values into poiseuille equation above.