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krok68 [10]
3 years ago
7

Is it possible to maintain a pressure of 10 kpa in a condenser that is being cooled by river water entering at 20 C.?

Engineering
1 answer:
SIZIF [17.4K]3 years ago
7 0

Answer:

Yes,If we use river water which is entering at 20⁰ C in the condenser then it is possible to maintain the pressure of 10 KPa in condenser.

Explanation:

Yes,If we use river water which is entering at 20⁰ C in the condenser then it is possible to maintain the pressure of 10 KPa in condenser.

The saturation temperature of steam is 45.81⁰ C at the pressure of 10 KPa which is higher than 20⁰C of river water. So river water at 20⁰C can be used to maintain the condenser pressure to 10 KPa.

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Explanation:

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Why do we study National studies
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Answer:

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2 years ago
Given a square matrix [A], write a single line MATLAB command that will create a new matrix [Aug] that consists of the original
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Above command is tested in MATLAB command window and is attached in figure below

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3 years ago
Which of the following is true about ideal standards? a.Ideal standards provide allowance for normal breakdowns and interruption
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3 years ago
How to find the voltage(B Aab) in series parallel circuit? ​
Sindrei [870]

Answer:

  Vab ≈ 3.426 V

Explanation:

First of all, it is convenient to find the equivalent parallel resistance of R5 and R6. That will be ...

  R56 = (R5)(R6)/(R5 +R6) = (1000)(1500)/(1000 +1500) = 600

Then we can call V1 the voltage at the top of R2. The voltage at Va is a divider from V1:

  Va = V1·(R4/(R3+R4)) = V1(560/1030) ≈ 0.543689V1

The voltage at Vb is also a divider from V1:

  Vb = V1·(R7+R8)/(R2 +R56 +R7 +R8) = V1(780/1710) ≈ 0.456140V1

The parallel branches containing Va and Vb have an effective resistance of ...

  (1030)(1710)/(1030+1710) = 642.81

That forms a divider with R1 to give V1:

  V1 = (100 V)642.81/(1000 +642.81) ≈ 39.1287 V

The difference Va-Vb is ...

  Vab = (39.1287 V)(0.543689 -0.456140) ≈ 3.426 V

_____

We have done this using parallel resistance and voltage divider calculations. You can also do it using node voltage equations. Using the same definition for V1 as above, we have ...

  (Vs -V1)/R1 +(Vb -V1)/(R56+R2) +(Va-V1)/R3 = 0

  (V1 -Vb)/(R56 +R2) -Vb/(R7+R8) = 0

  (V1 -Va)/R3 -Va/R4 = 0

The solution of interest is the value of Vab, shown in the attachment. It computes as 154200/45013 V ≈ 3.42568 V.

4 0
3 years ago
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