Answer:
b)false
Explanation:
Rolling is a process in which work piece passes through rolls to produce desired out put of the work piece.Rolling is a metal forming process.
We know that friction force is responsible for motion of work piece between rolls.If friction force is so small at the entrance side then work piece will not enter in the forming zone and forming process will not occurs.So the friction force should be high at the entrance side and low at the exit side.
So given statement is wrong.
Answer:
a. ε₁=-0.000317
ε₂=0.000017
θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain =3.335 *10^-4
Associated average normal strain ε(avg) =150 *10^-6
θ = 31.71 or -58.29
Explanation:

ε₁=-0.000317
ε₂=0.000017
To determine the orientation of ε₁ and ε₂

θ= -13.28° and 76.72°
To determine the direction of ε₁ and ε₂

=-0.000284 -0.0000335 = -0.000317 =ε₁
Therefore θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain

=3.335 *10^-4

ε(avg) =150 *10^-6
orientation of γmax

θ = 31.71 or -58.29
To determine the direction of γmax

= 1.67 *10^-4
Answer:
/* C Program to rotate matrix by 90 degrees */
#include<stdio.h>
int main()
{
int matrix[100][100];
int m,n,i,j;
printf("Enter row and columns of matrix: ");
scanf("%d%d",&m,&n);
/* Enter m*n array elements */
printf("Enter matrix elements: \n");
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
scanf("%d",&matrix[i][j]);
}
}
/* matrix after the 90 degrees rotation */
printf("Matrix after 90 degrees roration \n");
for(i=0;i<n;i++)
{
for(j=m-1;j>=0;j--)
{
printf("%d ",matrix[j][i]);
}
printf("\n");
}
return 0;
}
Scuze , am incercat , dat chiar nu stiu. Prefer sa nu raspund si sa scriu asta decat sa raspund gresit.❤️
Answer:
1) The exergy of destruction is approximately 456.93 kW
2) The reversible power output is approximately 5456.93 kW
Explanation:
1) The given parameters are;
P₁ = 8 MPa
T₁ = 500°C
From which we have;
s₁ = 6.727 kJ/(kg·K)
h₁ = 3399 kJ/kg
P₂ = 2 MPa
T₂ = 350°C
From which we have;
s₂ = 6.958 kJ/(kg·K)
h₂ = 3138 kJ/kg
P₃ = 2 MPa
T₃ = 500°C
From which we have;
s₃ = 7.434 kJ/(kg·K)
h₃ = 3468 kJ/kg
P₄ = 30 KPa
T₄ = 69.09 C (saturation temperature)
From which we have;
h₄ =
+ x₄×
= 289.229 + 0.97*2335.32 = 2554.49 kJ/kg
s₄ =
+ x₄×
= 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)
The exergy of destruction,
, is given as follows;
= T₀ ×
= T₀ ×
× (s₄ + s₂ - s₁ - s₃)
= T₀ ×
×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)
∴
= 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138 - 2554.49) ≈ 456.93 kW
The exergy of destruction ≈ 456.93 kW
2) The reversible power output,
=
+
≈ 5000 + 456.93 kW = 5456.93 kW
The reversible power output ≈ 5456.93 kW.