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dlinn [17]
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.
Engineering
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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Who is/are the founder/founders of transistor? ​
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William Shockley, Walter Houser Brattain and John Bardeen.

Explanation:

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A(n) is a detailed, structured diagram or drawing.
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Answer:

Schematics

Explanation:

A schematic is a detailed structured diagram or drawing. It employs illustrations to help the viewer understand detailed information on the machine or object being described. Its main aim is not to help the observer know what the object looks like physically. It is rather aimed at helping the viewer know how the machine works. This is achieved by only including key and important details to the drawing.

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3 years ago
A lake has a carrying capacity of 10,000 fish. At the current level of fishing, 2,000 fish per year are taken with the catch uni
arlik [135]

Answer:

The population size would be p' = 5000

The yield would be    MaxYield = 2082 \ fishes \ per \ year

Explanation:

So in this problem we are going to be examining the application of a  population dynamics a fishing pond and stock fishing and objective would be to obtain the maximum sustainable yield and and the population of the fish at the obtained maximum sustainable yield,  so basically we would be applying an engineering solution to fishing

 

    So the current  yield which is mathematically represented as

                               \frac{dN}{dt} =   \frac{2000}{1 \ year }

 Where dN is the change in the number of fish

            and dt is the change in time

So in order to obtain the solution we need to obtain the  rate of growth

    For this we would be making use of the growth rate equation which is

                                      r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}

  Where N is the population of the fish which is given as 4,000 fishes

          and  K is the carrying capacity which is given as 10,000 fishes

             r is the growth rate

        Substituting these values into the equation

                              r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]}  =0.833

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                      Max Yield  = \frac{rK}{4} = \frac{(10,000)(0.833)}{4} = 2082 \ fishes \ per \ year

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                          p' = \frac{K}{2}  = \frac{10,000}{2}  = 5000\ fishes          

7 0
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Answer:

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