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:
From the main bearings, the oil passes through feed-holes into drilled passages in the crankshaft and on to the big-end bearings of the connecting rod.
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.