Answer:
(a) E = 0 N/C
(b) E = 0 N/C
(c) E = 7.78 x10^5 N/C
Explanation:
We are given a hollow sphere with following parameters:
Q = total charge on its surface = 23.6 μC = 23.6 x 10^-6 C
R = radius of sphere = 26.1 cm = 0.261 m
Permittivity of free space = ε0 = 8.85419 X 10−12 C²/Nm²
The formula for the electric field intensity is:
E = (1/4πεo)(Q/r²)
where, r = the distance from center of sphere where the intensity is to be found.
(a)
At the center of the sphere r = 0. Also, there is no charge inside the sphere to produce an electric field. Thus the electric field at center is zero.
<u>E = 0 N/C</u>
(b)
Since, the distance R/2 from center lies inside the sphere. Therefore, the intensity at that point will be zero, due to absence of charge inside the sphere (q = 0 C).
<u>E = 0 N/C</u>
(c)
Since, the distance of 52.2 cm is outside the circle. So, now we use the formula to calculate the Electric Field:
E = (1/4πεo)[(23.6 x 10^-6 C)/(0.522m)²]
<u>E = 7.78 x10^5 N/C</u>
Answer:
flow ( m ) = 4.852 kg/s
Explanation:
Given:
- Inlet of Turbine
P_1 = 10 MPa
T_1 = 500 C
- Outlet of Turbine
P_2 = 10 KPa
x = 0.9
- Power output of Turbine W_out = 5 MW
Find:
Determine the mass ow rate required
Solution:
- Use steam Table A.4 to determine specific enthalpy for inlet conditions:
P_1 = 10 MPa
T_1 = 500 C ---------- > h_1 = 3375.1 KJ/kg
- Use steam Table A.6 to determine specific enthalpy for outlet conditions:
P_2 = 10 KPa -------------> h_f = 191.81 KJ/kg
x = 0.9 -------------> h_fg = 2392.1 KJ/kg
h_2 = h_f + x*h_fg
h_2 = 191.81 + 0.9*2392.1 = 2344.7 KJ/kg
- The work produced by the turbine W_out is given by first Law of thermodynamics:
W_out = flow(m) * ( h_1 - h_2 )
flow ( m ) = W_out / ( h_1 - h_2 )
- Plug in values:
flow ( m ) = 5*10^3 / ( 3375.1 - 2344.7 )
flow ( m ) = 4.852 kg/s
Answer:
1709.07 ft^3/s
Explanation:
Annual peak streamflow = Log10(Q [ft^3/s] )
mean = 1.835
standard deviation = 0.65
Probability of levee been overtopped in the next 15 years = 1/5
<u>Determine the design flow ins ft^3/s </u>
P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2
∴ T = 67.72 years
Q₁₅ = 1 - 0.2 = 0.8
Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )
K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )
= 2.1504
back to equation 1
Zt = 1.835 + ( 2.1504 * 0.65 ) = 3.23276
hence:
Log₁₀ ( Qt(ft^3/s) ) = Zt = 3.23276
hence ; Qt = 10^3.23276
= 1709.07 ft^3/s
