Answer:
(a) 561.12 W/ m² (b) 196.39 MW
Explanation:
Solution
(a) Determine the energy and power of the wave per unit area
The energy per unit are of the wave is defined as:
E = 1 /16ρgH²
= 1/16 * 1025 kg/ m3* 9.81 m/s² * (2.5 m )²
=3927. 83 J/m²
Thus,
The power of the wave per unit area is,
P = E/ t
= 3927. 83 J/m² / 7 s = 561.12 W/ m²
(b) The average and work power output of a wave power plant
W = E * л * A
= 3927. 83 J/m² * 0.35 * 1 *10^6 m²
= 1374.74 MJ
Then,
The power produced by the wave for one km²
P = P * л * A
= 5612.12 W/m² * 0.35 * 1* 10^6 m²
=196.39 MW
Answer:
1.887 m
Explanation:
(15 *pi)/180
= 0.2618 rad
Polar moment
= Pi*d⁴/32
= (22/7*20⁴)/32
= 15707.96
Torque on shaft
= ((22/7)*20³*110)/16
= 172857.14
= 172.8nm
Shear modulus
G = 79.3
L = Gjθ/T
= 79.3x10⁹x(1.571*10^-8)x0.2618/172.8
= 1.887 m
The length of the bar is therefore 1.887 meters
Answer:The move from hubs (shared networks) to switched networks was a big improvement. Control over collisions, increased throughput, and the additional features offered by switches all provide ample incentive to upgrade infrastructure. But Layer 2 switched topologies are not without their difficulties. Extensive flat topologies can create congested broadcast domains and can involve compromises with security, redundancy, and load balancing. These issues can be mitigated through the use of virtual local area networks, or VLANs. This chapter provides the structure and operation of VLANs as standardized in IEEE 802.1Q. This discussion will include trunking methods used for interconnecting devices on VLANs.
Problem: Big Broadcast Domains
With any single shared media LAN segment, transmissions propagate through the entire segment. As traffic activity increases, more collisions occur and transmitting nodes must back off and wait before attempting the transmission again. While the collision is cleared, other nodes must also wait, further increasing congestion on the LAN segment.
The left side of Figure 4-1 depicts a small network in which PC 2 and PC 4 attempt transmissions at the same time. The frames propagate away from the computers, eventually colliding with each other somewhere in between the two nodes as shown on the right. The increased voltage and power then propagate away from the scene of the collision. Note that the collision does not continue past the switches on either end. These are the boundaries of the collision domain. This is one of the primary reasons for switches replacing hubs. Hubs (and access points) simply do not scale well as network traffic increases.
Answer:
a) 75%
b) 82%
Explanation:
Assumptions:
Properties: The density of water 
Conversions:
Analysis:
Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:
Then;
gh = 0.491 kJ/kg
= 1559 kW
Therefore; the overall efficiency is:
= 0.75
= 75%
b) mechanical efficiency of the turbine:
thus;
![\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%](https://tex.z-dn.net/?f=%5Ceta_%7Bturbine%7D%20%3D%20%5Cdfrac%7B%5Ceta_%7B%5Bturbine-%20generator%5D%7D%20%7D%7B%5Ceta_%7Bgenerator%7D%7D%20%5C%5C%20%5C%5C%20%5Ceta_%7Bturbine%7D%20%3D%20%5Cdfrac%7B0.75%7D%7B0.92%7D%20%5C%5C%20%5C%5C%20%5Ceta_%7Bturbine%7D%20%3D%200.82%20%5C%5C%20%5C%5C%20%5Ceta_%7Bturbine%7D%20%3D%2082%5C%25)
