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3 years ago

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Determine the total condensation rate of water vapor onto the front surface of a vertical plate that is 10 mm high and 1 m in th

e horizontal direction. The plate is exposed to saturated steam at atmospheric pressure and is maintained at 75oC.

2 answers:

castortr0y [4]3 years ago

8 0

Answer:

Q = 63,827.5 W

Explanation:

Given:-

- The dimensions of plate A = ( 10 mm x 1 m )

- The fluid comes at T_sat , 1 atm.

- The surface temperature, T_s = 75°C

Find:-

Determine the total condensation rate of water vapor onto the front surface of a vertical plate

Solution:-

- Assuming drop-wise condensation the heat transfer coefficient for water is given by Griffith's empirical relation for T_sat = 100°C.

h = 255,310 W /m^2.K

- The rate of condensation (Q) is given by Newton's cooling law:

Q = h*As*( T_sat - Ts )

Q = (255,310)*( 0.01*1)*( 100 - 75 )

Q = 63,827.5 W

Hatshy [7]3 years ago

7 0

Answer:

<em>Q = 63,827.5 W</em>

Explanation:

<em>From the questions given, we recall the following,</em>

<em>The dimensions of vertical plate A = ( 10 mm x 1 m )</em>

<em>The fluid comes at T_sat, 1 atm.</em>

<em>The surface temperature pressure maintained T_s = 75°C </em>

<em>Thus we determine,</em>

<em>The total rate of condensation of water vapor onto the front surface of a vertical plate</em>

<em> Assuming drop-wise condensation the heat transfer coefficient for water is given by </em>

<em>Griffith's empirical relation for T_sat = 100°C.</em>

<em>Then</em>

<em> h = 255,310 W /m^2.K</em>

<em>The rate of condensation (Q) is stated by Newton's cooling law:</em>

<em> Q = h x As x ( T_sat - Ts )</em>

<em> Q = (255,310)x ( 0.01 x 1)x ( 100 - 75 )</em>

<em>Therefore</em>

<em> Q = 63,827.5 W</em>

<em />

<em />

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