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
The efficiency of a transformer is mainly dependent on a)- Core losses
Hope this helps! :)
Answer:
η=0.19=19% for p=14.7psi
η=0.3=30% for p=1psi
Explanation:
enthalpy before the turbine, state: superheated steam
h1(p=200psi,t=500F)=2951.9KJ/kg
s1=6.8kJ/kgK
Entalpy after the turbine
h2(p=14.7psia, s=6.8)=2469KJ/Kg
Entalpy before the boiler
h3=(p=14.7psia,x=0)=419KJ/Kg
Learn to pronounce
the efficiency for a simple rankine cycle is
η=
η=(2951.9KJ/kg-2469KJ/Kg)/(2951.9KJ/kg-419KJ/Kg)
η=0.19=19%
second part
h2(p=1psia, s=6.8)=2110
h3(p=1psia, x=0)=162.1
η=(2951.9KJ/kg-2110KJ/Kg)/(2951.9KJ/kg-162.1KJ/Kg)
η=0.3=30%
Answer:
Heat transfer = 2.617 Kw
Explanation:
Given:
T1 = 300 k
T2 = 440 k
h1 = 300.19 KJ/kg
h2 = 441.61 KJ/kg
Density = 1.225 kg/m²
Find:
Mass flow rate = 1.225 x [1.3/60]
Mass flow rate = 0.02654 kg/s
mh1 + mw = mh2 + Q
0.02654(300.19 + 240) = 0.02654(441.61) + Q
Q = 2.617 Kw
Heat transfer = 2.617 Kw
