Answer: The overhead percentage is 7.7%.
Explanation:
We call overhead, to all those bytes that are delivered to the physical layer, that don't carry real data.
We are told that we have 700 bytes of application data, so all the other bytes are simply overhead, i.e. , 58 bytes composed by the transport layer header, the network layer header, the 14 byte header at the data link layer and the 4 byte trailer at the data link layer.
So, in order to assess the overhead percentage, we divide the overhead bytes between the total quantity of bytes sent to the physical layer, as follows:
OH % = (58 / 758) * 100 = 7.7 %
Answer:
The solution for the given problem is done below.
Explanation:
M1 = 2.0
= 0.3636
= 0.5289
= 0.7934
Isentropic Flow Chart: M1 = 2.0 ,
= 1.8
T1 =
(1.8)(288K) = 653.4 K.
In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.
At the inlet,
T02=
= (1.8)(288K) = 518.4 K.
Q= Cp(T02-T01) =
= 135.7*
J/Kg.
Explanation:
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