Answer:
-6.326 KJ/K
Explanation:
A) the entropy change is defined as:
In an isobaric process heat (Q) is defined as:
Replacing in the equation for entropy
m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:
Solving the integral we get the expression to estimate the entropy change in the system
The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is 
We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K
The temperature in the final state we can get it from the heat expression, since we know how much heat was lost in the process (-976.71 kJ). By convention when heat is released by the system a negative sign is used to express it.
With 
clearing for T2 we get:
Now we can estimate the entropy change in the system
The entropy change in the system is negative because we are going from a state with a lot of disorder (high temperature) to one more organize (less temperature. This was done increasing the entropy of the surroundings.
b) see picture.
D
Step by step explanation
Answer:
The power produced by the turbine is 23309.1856 kW
Explanation:
h₁ = 3755.39
s₁ = 7.0955
s₂ = sf + x₂sfg =
Interpolating fot the pressure at 3.25 bar gives;
570.935 +(3.25 - 3.2)/(3.3 - 3.2)*(575.500 - 570.935) = 573.2175
2156.92 +(3.25 - 3.2)/(3.3 - 3.2)*(2153.77- 2156.92) = 2155.345
h₂ = 573.2175 + 0.94*2155.345 = 2599.2418 kJ/kg
Power output of the turbine formula =
![Q - \dot{W } = \dot{m}\left [ \left (h_{2}-h_{1} \right )+\dfrac{v_{2}^{2}- v_{1}^{2}}{2} + g(z_{2}-z_{1})\right ]](https://tex.z-dn.net/?f=Q%20-%20%5Cdot%7BW%20%7D%20%3D%20%5Cdot%7Bm%7D%5Cleft%20%5B%20%5Cleft%20%28h_%7B2%7D-h_%7B1%7D%20%20%5Cright%20%29%2B%5Cdfrac%7Bv_%7B2%7D%5E%7B2%7D-%20v_%7B1%7D%5E%7B2%7D%7D%7B2%7D%20%2B%20g%28z_%7B2%7D-z_%7B1%7D%29%5Cright%20%5D)
Which gives;
![560 - \dot{W } = 8\left [ \left (2599.2418-3755.39 \right )+\dfrac{15^{2}- 60^{2}}{2} \right ]](https://tex.z-dn.net/?f=560%20-%20%5Cdot%7BW%20%7D%20%3D%208%5Cleft%20%5B%20%5Cleft%20%282599.2418-3755.39%20%20%5Cright%20%29%2B%5Cdfrac%7B15%5E%7B2%7D-%2060%5E%7B2%7D%7D%7B2%7D%20%5Cright%20%5D)
= -8*((2599.2418 - 3755.39)+(15^2 - 60^2)/2 ) = -22749.1856
= -22749.1856 - 560 = -23309.1856 kJ
= 23309.1856 kJ
Power produced by the turbine = Work done per second = 23309.1856 kW.
Answer:
Explanation:
Assumptions is that
1. The flow is an unsteady one
2. Bubbles diameter is constant
3. The bubble velocity is slow
4. There is no homogenous reaction
5. It has a one dimensional flux model along the radial direction
Answer:
Program that removes all spaces from the given input
Explanation:
// An efficient Java program to remove all spaces
// from a string
class GFG
{
// Function to remove all spaces
// from a given string
static int removeSpaces(char []str)
{
// To keep track of non-space character count
int count = 0;
// Traverse the given string.
// If current character
// is not space, then place
// it at index 'count++'
for (int i = 0; i<str.length; i++)
if (str[i] != ' ')
str[count++] = str[i]; // here count is
// incremented
return count;
}
// Driver code
public static void main(String[] args)
{
char str[] = "g eeks for ge eeks ".toCharArray();
int i = removeSpaces(str);
System.out.println(String.valueOf(str).subSequence(0, i));
}
}
