Answer:
Enthalpy at outlet=284.44 KJ
Explanation:


We need to Find enthalpy of outlet.
Lets take the outlet mass m and outlet enthalpy h.
So from mass conservation

m=1+1.5+2 Kg/s
m=4.5 Kg/s
Now from energy conservation

By putting the values

So h=284.44 KJ
Answer:
Exit velocity
m/s.
Explanation:
Given:
At inlet:

Properties of steam at 100 bar and 600°C

At exit:Lets take exit velocity 
We know that if we know only one property inside the dome then we will find the other property by using steam property table.
Given that dryness or quality of steam at the exit of nozzle is 0.85 and pressure P=80 bar.So from steam table we can find the other properties.
Properties of saturated steam at 80 bar

So the enthalpy of steam at the exit of turbine



Now from first law for open system

In the case of adiabatic nozzle Q=0,W=0

m/s
So Exit velocity
m/s.
Answer:
The required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm²
.
Explanation:
We are given current density of 459 A/cm² and we want to limit the current to 0.53 A in a fuse wire. We are asked to find the corresponding diameter of the fuse wire.
Recall that current density is given by
j = I/A
where I is the current flowing through the wire and A is the area of the wire
A = πr²
but r = d/2 so
A = π(d/2)²
A = πd²/4
so the equation of current density becomes
j = I/πd²/4
j = 4I/πd²
Re-arrange the equation for d
d² = 4I/jπ
d = √4I/jπ
d = √(4*0.53)/(459π)
d = 0.0383 cm
Therefore, the required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm²
.