Answer:
Q=486.49 KJ/kg
Explanation:
Given that
V= 0.2 m³
At initial condition
P= 2 MPa
T=320 °C
Final condition
P= 2 MPa
T=540°C
From steam table
At P= 2 MPa and T=320 °C
h₁=3070.15 KJ/kg
At P= 2 MPa and T=540°C
h₂=3556.64 KJ/kg
So the heat transfer ,Q=h₂ - h₁
Q= 3556.64 - 3070.15 KJ/kg
Q=486.49 KJ/kg
Answer:
5 microhenries
Explanation:
The effective value of inductors in parallel "add" in the same way that resistors in parallel do. The value is the reciprocal of the sum of the reciprocals of the inductances that are in parallel.
10 uH ║ 10 uH = 5 uH
The effective inductance is 5 uH.
Answer:
False
Explanation:
its self explanatory. a tracter pulls the equipment, so why power the equipment? the tracter is giving it free energy.
The guy above me is right fyi and its not 50 points
Answer:
The value of exit temperature from the nozzle = 719.02 K
Explanation:
Temperature at inlet 
= 450°c = 723 K
Velocity at inlet 
= 55 
velocity at outlet 
= 390
Specific heat at constant pressure for steam 
Apply steady flow energy equation for the nozzle
Put all the values in the above formula we get,
⇒ 18723 × 723 + 
= 
+ 
⇒ = 719.02 K
This is the value of exit temperature from the nozzle.
