Answer:
Rate of heat transfer to river=1200MW
So the actual amount of heat rejected ti the river will be less as there will some heat loss to surrounding and in pipes
Explanation:
In order to find the actual heat transfer rate is lower or higher than its value we will first find the rate of heat transfer to power plant:
From First law of thermodynamics:
Rate of heat transfer to river=heat transfer to power plant-work done
Rate of heat transfer to river=2000-800
Rate of heat transfer to river=1200MW
So the actual amount of heat rejected ti the river will be less as there will some heat loss to surrounding and in pipes.
Answer:
, 
Explanation:
The drag force is equal to:
Where 
is the drag coefficient and 
is the frontal area, respectively. The work loss due to drag forces is:
The reduction on amount of fuel is associated with the reduction in work loss:
Where 
and 
are the original and the reduced frontal areas, respectively.
The change is work loss in a year is:
![\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)](https://tex.z-dn.net/?f=%5CDelta%20W%20%3D%20%280.3%29%5Ccdot%20%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Ccdot%20%281.20%5C%2C%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%7D%29%5Ccdot%20%2827.778%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%29%5E%7B2%7D%5Ccdot%20%5B%281.85%5C%2Cm%29%5Ccdot%20%281.75%5C%2Cm%29%20-%20%281.50%5C%2Cm%29%5Ccdot%20%281.75%5C%2Cm%29%5D%5Ccdot%20%2825%5Ctimes%2010%5E%7B6%7D%5C%2Cm%29)
The change in chemical energy from gasoline is:
The changes in gasoline consumption is:
Lastly, the money saved is:
Answer:
The answer is "
".
Explanation:
Please find the correct question in the attachment file.
using formula:
Answer:
R = 31.9 x 10^(6) At/Wb
So option A is correct
Explanation:
Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A
Thus; R = L/μA,
Now from the question,
L = 4m
r_1 = 1.75cm = 0.0175m
r_2 = 2.2cm = 0.022m
So Area will be A_2 - A_1
Thus = π(r_2)² - π(r_1)²
A = π(0.0225)² - π(0.0175)²
A = π[0.0002]
A = 6.28 x 10^(-4) m²
We are given that;
L = 4m
μ_steel = 2 x 10^(-4) Wb/At - m
Thus, reluctance is calculated as;
R = 4/(2 x 10^(-4) x 6.28x 10^(-4))
R = 0.319 x 10^(8) At/Wb
R = 31.9 x 10^(6) At/Wb
