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irakobra [83]
3 years ago
11

A rigid vessel with a volume of 10 m3 contains a water-vapor mixture at 400 kPa. If the quality is 60 percent, find the mass. Th

e pressure is lowered to 300 kPa by cooling the vessel; find mg and mf.
Engineering
1 answer:
trapecia [35]3 years ago
7 0
A) The pressure (p)=400kPa=0.400MPa
Mass of mixture =M
quality (X)= 0.60
Volume of mixture (V)=10 m3
From steam table at P=0.400MPa
Specific volume of saturated water (vf)=0.00108355 m3/kg
Specific volume of saturated steam (vg)=0.46238 m3/kg
Therefore, the volume of steam having X=0.6 is given by
V=M[vf+X(vg-vf)]
10= M[0.00108355+0.6(0.46238-0.00108355)]
M=36.04748 kg
If pressure is lowered to 300kPa
p=0.300MPa
From steam table we get,
Specific volume of saturated water (vf)=0.00107317 m3/kg
Specific volume of saturated steam (vg)=0.60576 m3/kg
Therefore, the volume of steam having X=0.6 is given by
V=M[vf+X(vg-vf)]
10= M[0.001007317+0.6(0.60576-0.00107317)]
M= 26.4825 kg
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2. The following segment of carotid artery has an inlet velocity of 50 cm/s (diameter of 15 mm). The outlet has a diameter of 11
ahrayia [7]

This question is incomplete, the missing diagram is uploaded along this answer below.

Answer:

the forces required to keep the artery in place is 1.65 N

Explanation:

Given the data in the question;

Inlet velocity V₁ = 50 cm/s = 0.5 m/s

diameter d₁ = 15 mm = 0.015 m

radius r₁ = 0.0075 m

diameter d₂ = 11 mm = 0.011 m

radius r₂ = 0.0055 m

A₁ = πr² = 3.14( 0.0075 )² =  1.76625 × 10⁻⁴ m²

A₂ = πr² = 3.14( 0.0055 )² =  9.4985 × 10⁻⁵ m²

pressure at inlet P₁ = 110 mm of Hg = 14665.5 pascal

pressure at outlet P₂ = 95 mm of Hg = 12665.6 pascal

Inlet volumetric flowrate = A₁V₁ = 1.76625 × 10⁻⁴ × 0.5 = 8.83125 × 10⁻⁵ m³/s

given that; blood density is 1050 kg/m³

mass going in m' = 8.83125 × 10⁻⁵ m³/s × 1050 kg/m³ = 0.092728 kg/s

Now, using continuity equation

A₁V₁ = A₂V₂

V₂ = A₁V₁ / A₂ = (d₁/d₂)² × V₁

we substitute

V₂ =  (0.015 / 0.011 )² × 0.5

V₂ = 0.92975 m/s

from the diagram, force balance in x-direction;

0 - P₂A₂ × cos(60°) + Rₓ = m'( V₂cos(60°) - 0 )    

so we substitute in our values

0 - (12665.6 × 9.4985 × 10⁻⁵)  × cos(60°) + Rₓ = 0.092728( 0.92975 cos(60°) - 0 )    

0 - 0.6014925 + Rₓ =  0.043106929 - 0

Rₓ = 0.043106929 + 0.6014925

Rₓ = 0.6446 N

Also, we do the same force balance in y-direction;

P₁A₁ - P₂A₂ × sin(60°) + R_y = m'( V₂sin(60°) - 0.5 )  

we substitute

⇒ (14665.5 × 1.76625 × 10⁻⁴) - (12665.6 × 9.4985 × 10⁻⁵) × sin(60°) + R_y = 0.092728( 0.92975sin(60°) - 0.5 )

⇒ 1.5484 + R_y = 0.092728( 0.305187 )

⇒ 1.5484 + R_y = 0.028299    

R_y = 0.028299 - 1.5484

R_y = -1.52 N

Hence reaction force required will be;

R = √( Rₓ² + R_y² )

we substitute

R = √( (0.6446)² + (-1.52)² )

R = √( 0.41550916 + 2.3104 )

R = √( 2.72590916 )

R = 1.65 N

Therefore, the forces required to keep the artery in place is 1.65 N

 

7 0
2 years ago
Match each titration term with its definition.
Illusion [34]

Answer:

1) titration

2) titrand

3) equivalence point

4) titrant

5) Burette

6) Indicator

Explanation:

The process in which a known volume of a standard solution is added to another solution so that the standard solution can react with the solution of unknown concentration such that its concentration is determined  can be referred to as titration.

The solution which is added to another solution  is called the titrant.  The titrand is the solution of unknown concentration

A burette is a glassware used to slowly add a known volume of the titrant to the titrand.

The indicator used signals the point when the reaction is complete by a color change. At this point, a stoichiometric amount of titrant has been added to the titrand. This is also referred to as the equivalence point.

3 0
3 years ago
4.116 The lid of a roof scuttle weighs 75 lb. It is hinged at corners A and B and maintained in the desired position by a rod CD
babunello [35]

Answer:

(a) The magnitude of force is 116.6 lb, as exerted by the rod CD

(b) The reaction at A is (-72.7j-38.1k) lb and at B it is (37.5j) lb.

Explanation:

Step by step working is shown in the images attached herewith.

For this given system, the coordinates are the following:

A(0, 0, 0)

B(26, 0, 0)

And the value of angle alpha is 20.95°

Hope that answers the question, have a great day!

5 0
3 years ago
Assign numMatches with the number of elements in userValues that equal matchValue. userValues has NUM_VALS elements. Ex: If user
Thepotemich [5.8K]

Answer:

import java.util.Scanner;

public class FindMatchValue {

  public static void main (String [] args) {

     Scanner scnr = new Scanner(System.in);

     final int NUM_VALS = 4;

     int[] userValues = new int[NUM_VALS];

     int i;

     int matchValue;

     int numMatches = -99; // Assign numMatches with 0 before your for loop

     matchValue = scnr.nextInt();

     for (i = 0; i < userValues.length; ++i) {

        userValues[i] = scnr.nextInt();

     }

     /* Your solution goes here */

         numMatches = 0;

     for (i = 0; i < userValues.length; ++i) {

        if(userValues[i] == matchValue) {

                       numMatches++;

                }

     }

     System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);

  }

}

8 0
3 years ago
Consider a multiprocessor system and a multithreaded program written using the many-to-many threading model. Let the number of u
Montano1993 [528]

Answer:

At the point when the quantity of bit strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just part strings to processors and not client level strings to processors. At the point when the quantity of part strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used all the while. Be that as it may, when a part string obstructs inside the portion (because of a page flaw or while summoning framework calls), the comparing processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, in this way expanding the use of the multiprocessor system.When the quantity of part strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just bit strings to processors and not client level strings to processors. At the point when the quantity of bit strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used at the same time. Be that as it may, when a part string hinders inside the piece (because of a page flaw or while summoning framework calls), the relating processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, along these lines expanding the usage of the multiprocessor framework.

4 0
3 years ago
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