Answer:
Hydraulic gradient=0.0173
Explanation:
The hydraulic head at the points A, B and C will be as follows
A=95-5=90 m
B=110-30=80 m
C=135-35= 100 m
By drawing the equipotential lines, the direction of ground water can be seen as in the triangle
The length of flow will be
The hydraulic gradient will be given by
Answer:
6.5 × 10¹⁵/ cm³
Explanation:
Thinking process:
The relation
With the expression Ef - Ei = 0.36 × 1.6 × 10⁻¹⁹
and ni = 1.5 × 10¹⁰
Temperature, T = 300 K
K = 1.38 × 10⁻²³
This generates N₀ = 1.654 × 10¹⁶ per cube
Now, there are 10¹⁶ per cubic centimeter
Hence,
Answer:
Explanation:
= 113.27 g - 49.31 g = 63.96 g
= 100.06 g - 49.31 g = 50.75
Moisture /water content, w = =
= 0.26 = 26%
Void ratio = water content X specific gravity of solid
= 0.26 X 2.80 =0.728
Answer:
Hello some parts of your question is missing below is the missing part
Convection coefficient = 11 w/m^2. °c
answer : 44.83 watts
Explanation:
Given data :
surface emissivity ( ε )= 0.95
head ( sphere) diameter( D ) = 0.25 m
Temperature of sphere( T ) = 35° C
Temperature of surrounding ( T∞ ) = 25°C
Temperature of surrounding surface ( Ts ) = 15°C
б = ( 5.67 * 10^-8 )
Determine the total rate of heat loss
First we calculate the surface area of the sphere
As =
= = 0.2 m^2
next we calculate heat loss due to radiation
Qrad = ε * б * As( ) ---- ( 1 )
where ;
ε = 0.95
б = ( 5.67 * 10^-8 )
As = 0.2 m^2
T = 35 + 273 = 308 k
Ts = 15 + 273 = 288 k
input values into equation 1
Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )
= 22.83 watts
Qrad ( heat loss due to radiation ) = 22.83 watts
calculate the heat loss due to convection
Qconv = h* As ( ΔT )
= 11*0.2 ( 35 -25 ) = 22 watts
Hence total rate of heat loss
= 22 + 22.83
= 44.83 watts