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Tcecarenko [31]
3 years ago
10

Air is to be heated steadily by an 8-kW electric resistance heater as it flows through an insulated duct. If the air enters at 5

5°C at a rate of 2 kg/s, determine the exit temperature of air. Solve using appropriate software.
Engineering
1 answer:
Furkat [3]3 years ago
6 0

To solve this problem it is necessary to apply the concepts related to the heat exchange of a body.

By definition heat exchange in terms of mass flow can be expressed as

W = \dot{m}c_p \Delta T

Where

C_p = Specific heat

\dot{m}= Mass flow rate

\Delta T = Change in Temperature

Our values are given as

C_p = 1.005kJ/kgK \rightarrow Specific heat of air

T_1 = 50\°C

\dot{m} = 2kg/s

W = 8kW

From our equation we have that

W = \dot{m}c_p \Delta T

W = \dot{m}c_p (T_2-T_1)

Rearrange to find T_2

T_2 = \frac{W}{\dot{m}c_p}+T_1

Replacing

T_2 = \frac{8}{2*1.005}+(50+273)

T_2 = 326.98K \approx 53.98\°C

Therefore the exit temperature of air is 53.98°C

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Ivahew [28]

The weight of the specimen in SSD condition is 373.3 cc

<u>Explanation</u>:

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apparent specific gravity = \frac{A}{A-C}

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Apparent specific gravity = 2.88

b) Bulk specific gravity G_{B}^{O D}=\frac{A}{B-C}

G_{B}^{O D}=\frac{1034}{1048.9-675 \cdot 6}

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c) Bulk specific gravity (SSD):

G_{B}^{S S D}=\frac{B}{B-C}

=\frac{1048 \cdot 9}{1048 \cdot 9-675 \cdot 6}

G_{B}^{S S D} = 2.80

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=\frac{1048 \cdot 9-1034}{1034} \times 100

Absorption = 1.44 %

e) Bulk Volume :

v_{b}=\frac{\text { weight of dispaced water }}{P \omega t}

=\frac{1048 \cdot 9-675 \cdot 6}{1}

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Explanation:

Hope it helps you

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Answer:

Alice is correct.

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Explanation:

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This is a nested for-loop.

After the first for-loop opening, there is no block of statement to be executed rather a for-loop is called again. And the second for-loop uses the value of i from the first for-loop. The value of N is both called from outside the loop.

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