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3 years ago

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Carbon dioxide steadily flows into a constant pressure heater at 300 K and 100 kPa witha mass flow rate of 9.2 kg/s. Heat transf

er in the rate of 175 kW is supplied to the carbondioxide as it flows through the heater. Determinea) the carbon dioxide temperature at the exit andb) carbon dioxide volume flow rate at the heater exit.Assume constant specific heat, cp, of 1075 J/(kg K).

1 answer:

docker41 [41]3 years ago

8 0

Answer:

Carbon dioxide temperature at exit is 317.69 K

Carbon dioxide flow rate at heater exit is 20.25 m³/s

Explanation:

Detailed steps are attached below.

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PolarNik [594]

Answer:

a= - 2.6 m/s².

Explanation:

u = 32 m/s

The speed after 6 s is half of u

$v= \dfrac{32}{2}=16\ m/s$

t= 6 s

The average acceleration = a

We know v = u +at

v=final velocity

u=initial velocity

Now by putting the values in the above equation

16= 32 + a x 6

$a=\dfrac{16-32}{6}\ m/s^2$

$a=-2.6\ m/s^2$

Therefore the acceleration will be - 2.6 m/s².

a= - 2.6 m/s².

Negative indicates that velocity and acceleration is is opposite direction.

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3 years ago

The disk of radius 0.4 m is originally rotating at ωo=4 rad/sec. If it is subjected to a constant angular acceleration of α=5 ra

Lina20 [59]

Answer:32.4m/ $s^2$

Explanation:

Given data

$radius\left ( r\right )$ =0.4m

Intial angular velocity $\left ( \omega_0\right )$ =4rad/s

angular acceleration $\left ( \alpha\right )$ =5rad/ $s^2$

angular velocity after 1 sec

$\omega$ = $\omega_0$ + $\alpha\times\t$

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$\omega$ =9rad/s

Velocity of point on the outer surface of disc $\left ( v\right )$ = $\omega_0\timesr$

v= $9\times0.4$ m/s=3.6m/s

Normal component of acceleration $\left ( a_c\right )$ = $\frac{v^2}{r}$

$a_c$ = $\frac{3.6\times3.6}{0.4}$ =32.4m/ $s^2$

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4 years ago

Consider steady heat transfer through the wall of a room in winter. The convection heat transfer coefficient at the outer surfac

AveGali [126]

The heat transfer which is in steady state, the heat transfer rate to the wall is equal to the wall.

<u>Explanation:</u>

The convection transfer of heat to the wall is

$Q=h A\left(T_{s}-T_{f}\right)$

Here, $T_{S}$ is the temperature of solid surface, $T_{f}$ is the temperature of moving fluid stream which is adjacent of solid surface, h is the heat transfer coefficient.
The coefficient of convection heat transfers outer surface contains 3 times to the inner surface which experience smaller drop of temperature for 3 times that compares to inner surface.
Hence, the temperatures outer surface get close to the surroundings of air temperature.

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3 years ago

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Ainat [17]

Answer

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Explanation:

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4 years ago

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xenn [34]

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Explanation:

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