Answer:
Building the Canadian hydroelectric plant reduced the rate of erosion after 1906. This change in the rate of erosion is represented by the change in the slope on the graph. Hydroelectric dams convert kinetic energy from moving water to electrical energy.
Answer:
23.06262m^3/s
Explanation:
The volume flow rate of the water is determined from the needed power output and the elevation difference:
Where, height (h) =221m, power(w)=50MW=50*10^6w
Density of water (ρ)=1000kg/m^3
Efficiency of turbine(η)=100%=1
V=W/ρηgh
=50*10^6m^3/(1)*(1000)*(9.81)*(221)s=23.06262m^3/s
Answer:
(a) Effectiveness of the regenerator= 0.433
(b) The rate of heat removal=21.38 kW
Explanation:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Answer:
Explanation:
Before: PT= 0.10, PB= 0.03 (given) ET = 2.5 ER = 2.0 (Table 6.5)
fHV= 0.847 (Eq. 6.5) PHF = 0.95, fp= 0.90, N=2, V = 2200 (given)
vp= V/[PHF⋅fHVTB⋅fp⋅N] = 1518.9 (Eq. 6.3)
BFFS = 50+5, BFFS =55 (given) fLW= 6.6
TLC=6+3=9 fLC= 0.65
fM= 0.0
fA= 1.0
FFS = BFFS −fLW−fLC–fM–fA= 46.75 (Eq. 6.7)
Use FFS=45 D= vp/S = 33.75pc/mi/ln Eq (6.6)
After: fA= 3.0
FFS = BFFS −fLW−fLC–fM–fA= 44.75 (Eq. 6.7)
Use FFS=45 Vnew= 2600 Vp= Va/[PHF⋅fHB⋅fp⋅N] = 1795 (Eq. 6.3) D= vp/S = 39.89pc/mi/ln