Answer:
<em>Never</em>
Explanation:
In a reversible adiabatic process, there is not transfer of heat or matter between the system and its environment. An adiabatic reversible process is a process with constant entropy, i.e ΔQ=0. The internal energy is solely dependent on the work done either due to compression or expansion. So the entropy of the gas will never increase.
Answer:
Is there supposed to be an attachment?
Explanation:
Answer:
The three dimensions shown in an isometric drawing are the height, H, the length, L, and the depth, D
Explanation:
An isometric drawing of an object in presents a pictorial projection of the object in which the three dimension, views of the object's height, length, and depth, are combined in one view such that the dimensions of the isometric projection drawing are accurate and can be measured (by proportion of scale) to draw the different views of the object or by scaling, for actual construction of the object.
Answer:
class TestCode {
public static void main(String args[]) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int arr[] = new int[n];
int min = Integer.MAX_VALUE;
for(int i = 0;i<n;i++){
arr[i] = scanner.nextInt();
if(min > arr[i]){
min = arr[i];
}
}
for(int i = 0;i<n;i++){
System.out.print((arr[i]-min) + " ");
}
}
}
Answer:
-Differential equation: d²T/dx² = 0
-The boundary conditions are;
1) Heat flux at bottom;
-KAdT(0)/dx = ηq_e
2) Heat flux at top surface;
-KdT(L)/dx = h(T(L) - T(water))
Explanation:
To solve this question, let's work with the following assumptions that we are given;
- Heat transfer is steady and one dimensional
- Thermal conductivity is constant.
- No heat generation exists in the medium
- The top surface which is at x = L will be subjected to convection while the bottom surface which is at x = 0 will be subjected to uniform heat flux.
Will all those assumptions given, the differential equation can be expressed as; d²T/dx² = 0
Now the boundary conditions are;
1) Heat flux at bottom;
q(at x = 0) is;
-KAdT(0)/dx = ηq_e
2) Heat flux at top surface;
q(at x = L):
-KdT(L)/dx = h(T(L) - T(water))