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3 years ago

7

Liquid water at 300 kPa and 20°C is heated in a chamber by mixing it with superheated steam at 300 kPa and 300°C. Cold water ent

ers the chamber at a rate of 2.6 kg/s. If the mixture leaves the mixing chamber at 60°C.
Required:
Determine the mass flow rate of the superheated steam required.

1 answer:

lesya692 [45]3 years ago

3 0

Answer:

0.154kg/s

Explanation:

From this question we have the following information:

P1 = 300kpa

T1 = 20⁰c

M1 = 2.6kg/s

For superheated system

P2 = 300kpa

T2 = 300⁰c

M2 = ??

T2 = 60⁰c

From saturated water table

h1 = 83.91kj/kg

h3 = 251.18kj/kg

From superheated water,

h2 = 3069.6kj/kg

The equation of energy balance

m1h1 + m2h2 = m3h3

When we input all the corresponding values:

We get

m2 = -434.902/-2818.42

m2 = 0.15430

m2 = 0.154kg/s

This is the mass flow rate of the superheated steam

Please check attachment for more detailed explanation.

thank you!

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LLE equations:

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A12 = 2.148; A21 = 2.781

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LLE equations:

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4 years ago

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Vera_Pavlovna [14]

Answer:

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Also Maximum power can be carried by wave guide only in dominant mode

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input all the given values back to ZTE equation

ZTE = 285 ohms

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B) The dielectric materials given data/parameters

∈ = 2.5 ∈o ∪ = ∪o

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free space impedance n = $\sqrt{\frac{u}{e} } = \sqrt{\frac{UoUr}{EoEr} }$

therefore for the given dielectric n = $\sqrt{\frac{Uo}{Eo} } \sqrt{\frac{1}{2.5} } = \frac{377}{\sqrt{2.5} }$ n = 238.43

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therefore ZTE = 180.56 ohms

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To calculate the number of time power can be transmitted by the waveguide = power carried 2 / power carried 1

= 58*10^11 / 1.4*10^11 = 41.42 ≈ 41

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