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Rudiy27
2 years ago
7

The two major forces opposing the motion of a vehicle moving on a level road are the rolling resistance of the tires, Fr, and th

e aerodynamic drag force of the air flowing around the vehicle, Fd, given respectively by Fr, = fW, Fd= CdA1/2 rhoV2 where f and Cd are constants known as the rolling resistance coefficient and drag coefficient, respectively, W and A are the vehicle weight and projected frontal area, respectively, V is the vehicle velocity, and rho is the air density. For a passenger car with W = 3,550 lbf, A = 23.3 ft^2, and Cd = 0.34, and where f = 0.02 and rho = 0.08 lbm/ft^3.
Required:
Determine the power required, in HP, to overcome rolling resistance and aerodynamic drag when V is 55 mph.
Engineering
1 answer:
anzhelika [568]2 years ago
7 0

Answer:

The power required to overcome rolling resistance and aerodynamic drag is 19.623 h.p.

Explanation:

Let suppose that vehicle is moving at constant velocity. By Newton's Law of Motion, the force given by engine must be equal to the sum of the rolling resistance and the aerodynamic drag force of the air. And by definition of power, we have the following formula:

\dot W = \left(f\cdot W +\frac{\rho\cdot C_{D}\cdot A\cdot v^{2}}{2\cdot g_{c}} \right)\cdot v (1)

Where:

\dot W- Power, in pounds-force-feet per second.

f - Rolling resistance coefficient, no unit.

W - Weight of the passanger car, in pounds-force.

\rho - Density of air, in pounds-mass per cubic feet.

C_{D} - Drag coefficient, no unit.

A - Projected frontal area, in square feet.

v - Vehicle speed, in feet per second.

g_{c} - Pound-mass to pound-force ratio, in pounds-mass to pound-force.

If we know that f = 0.02, W = 3,550\,lbf, \rho = 0.08\,\frac{lbm}{ft^{3}}, C_{D} = 0.34, A = 23.3\,ft^{2}, v = 80.685\,\frac{ft}{s} and g_{c} = 32.174\,\frac{lbm}{lbf}, then the power required by the car is:

\dot W = \left(f\cdot W +\frac{\rho\cdot C_{D}\cdot A\cdot v^{2}}{2\cdot g_{c}} \right)\cdot v

\dot W = 10901.941\,\frac{lbf\cdot ft}{s}

\dot W = 19.623\,h.p.

The power required to overcome rolling resistance and aerodynamic drag is 19.623 h.p.

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The current in a 20 mH inductor is known to be: 푖푖=40푚푚푚푚푡푡≤0푖푖=푚푚1푒푒−10,000푡푡+푚푚2푒푒−40,000푡푡푚푚푡푡≥0The voltage across the induct
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Answer:

a) The expression for electrical current: i = -0.134*e^(-10,000*t) + 0.174*e^(-40,000*t) A

The expression for voltage: v = 26.8*e^(-10,000*t) - 139.2*e^(-40,000*t) V

b) For t<=0 the inductor is storing energy and for t > 0 the inductor is delivering energy.

Explanation:

The question text is corrupted. I found the complete question on the web and it goes as follow:

The current in a 20 mH inductor is known to be: i = 40 mA at t<=0 and i = A1*e^(-10,000*t) + A2*e^(-40,000*t) A at t>0. The voltage across the inductor (passive sign convention) is -68 V at t = 0.

a. Find the numerical expressions for i and v for t>0.

b. Specify the time intervals when the inductor is storing energy and is delivering energy.

A inductor stores energy in the form of a magnetic field, it behaves in a way that oposes sudden changes in the electric current that flows through it, therefore at moment just after t = 0, that for convenience we'll call t = 0+, the current should be the same as t=0, so:

i = A1*e^(-10,000*(0)) + A2*e^(-40,000*(0))

40*10^(-3) = A1*e^(-10,000*0) + A2*e^(-40,000*0)

40*10^(-3) = (A1)*1 + (A2)*1

40*10^(-3) = A1 + A2

A1 + A2 = 40*10^(-3)

Since we have two variables (A1 and A2) we need another equation to be able to solve for both. For that reason we will use the voltage expression for a inductor, that is:

V = L*di/dt

We have the voltage drop across the inductor at t=0 and we know that the current at t=0 and the following moments after that should be equal, so we can use the current equation for t > 0 to find the derivative on that point, so:

di/dt = d(A1*e^(-10,000*t) + A2*e^(-40,000*t))/dt

di/dt = [d(-10,000*t)/dt]*A1*e^(-10,000*t) + [d(-40,000*t)/dt]*A2*e^(-40,000*t)

di/dt = -10,000*A1*e^(-10,000*t) -40,000*A2*e^(-40,000*t)

By applying t = 0 to this expression we have:

di/dt (at t = 0) = -10,000*A1*e^(-10,000*0) - 40,000*A2*e^(-40,000*0)

di/dt (at t = 0) = -10,000*A1*e^0 - 40,000*A2*e^0

di/dt (at t = 0) = -10,000*A1- 40,000*A2

We can now use the voltage equation for the inductor at t=0, that is:

v = L di/dt (at t=0)

68 = [20*10^(-3)]*(-10,000*A1 - 40,000*A2)

68 = -400*A1 -800*A2

-400*A1 - 800*A2 = 68

We now have a system with two equations and two variable, therefore we can solve it for both:

A1 + A2 = 40*10^(-3)

-400*A1 - 800*A2 = 68

Using the first equation we have:

A1 = 40*10^(-3) - A2

We can apply this to the second equation to solve for A2:

-400*[40*10^(-3) - A2] - 800*A2 = 68

-1.6 + 400*A2 - 800*A2 = 68

-1.6 -400*A2 = 68

-400*A2 = 68 + 1.6

A2 = 69.6/400 = 0.174

We use this value of A2 to calculate A1:

A1 = 40*10^(-3) - 0.174 = -0.134

Applying these values on the expression we have the equations for both the current and tension on the inductor:

i = -0.134*e^(-10,000*t) + 0.174*e^(-40,000*t) A

v = [20*10^(-3)]*[-10,000*(-0.134)*e^(-10,000*t) -40,000*(0.174)*e^(-40,000*t)]

v = [20*10^(-3)]*[1340*e^(-10,000*t) - 6960*e^(-40,000*t)]

v = 26.8*e^(-10,000*t) - 139.2*e^(-40,000*t) V

b) The question states that the current for the inductor at t > 0 is a exponential powered by negative numbers it is expected that its current will reach 0 at t = infinity. So, from t =0 to t = infinity the inductor is delivering energy. Since at time t = 0 the inductor already has a current flow of 40 mA and a voltage, we can assume it already had energy stored, therefore for t<0 it is storing energy.

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The inlet and exhaust flow processes are not included in the analysis of the Otto cycle. How do these processes affect the Otto
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Answer:

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Explanation:

Step1

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Step2

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Step3

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Answer:

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Answer:

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