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AveGali [126]
3 years ago
10

The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through alumi

num radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm .
The term deltaT/dis called the temperature gradient which is the temperature difference deltaT between x coolant inside and the air outside per unit thickness of tube.
Cooling temperature gradient?
A) twice that of the old unit
B) three halves that of the old unit
C) same as that of the old unit
D) two third that of the old unit
E) one-half that of the old unit
Part B
If the old radiator is replaced with anew one made of a material that has twice the thermal conductivity of the material of the old radiator what should the total surface area available for heat exchange in the new radiator be to achieve the desired temperature gradient ?
Assume that the tubes of the new radiator are as thick as tubes in older unit.
A) twice that of the old unit
B) three halves that of the old unit
C) same as that of the old unit
D) two third that of the old unit
E) one-half that of the old unit
Physics
1 answer:
Alex777 [14]3 years ago
8 0

Answer:

(A) The correct answer is option (B) three halves that of the old unit.

(B) The answer is three fourth that of old unit

Explanation:

from the relation;

(A) Fromthe expression;

K = Qd/AΔT

Anew = 3/2 A(old)

(B)

K¹ = 2K(old), so we get

A(old) = A(old)/2

Combining with part A, we have;

Anew = 3/2 *A(old)/2

          = 3/4A(old)

The answer is three fourth that of old unit

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The magnitude and direction (inward or outward) of the net flux through the cell boundary is calculated as follows;

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A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
atroni [7]

(a) 2.79 rev/s^2

The angular acceleration can be calculated by using the following equation:

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

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\omega_i = 11.0 rev/s is the initial angular speed

\alpha is the angular acceleration

\theta=50.0 rev is the number of revolutions made by the disk while accelerating

Solving the equation for \alpha, we find

\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

\theta=? is the number of revolutions made by the disk while accelerating

Solving the equation for \theta, we find

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev

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Answer:

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