Answer:
Condition A
Heat flux is 1400 W/M^2
Condition B
Heat flux is 12800 w/m^2
Explanation:
Given that:
is given as 30 degree celcius
condition A
Air temperature = - 5 degree c
convection coefficient h = 40 w/m^2. k
condition A
water temperature = 10 degree c
convection coefficient = 800 w/m^2.k
Answer:
a. 9947 m
b. 99476 times
c. 2*10^11 molecules
Explanation:
a) To find the mean free path of the air molecules you use the following formula:
R: ideal gas constant = 8.3144 Pam^3/mol K
P: pressure = 1.5*10^{-6} Pa
T: temperature = 300K
N_A: Avogadros' constant = 2.022*10^{23}molecules/mol
d: diameter of the particle = 0.25nm=0.25*10^-9m
By replacing all these values you obtain:
b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:
c) By using the equation of the ideal gases you obtain:
Answer:
investment 10 years from now is $1,238,000 .
Explanation:
given data
sum = $500,000
rate = 12% =0.12
total time = 10 year
solution
as present value After 2 years from now is $500,000
so time period is now = 8 year ( 10 - 2 )
so we apply future value formula that is
Future value = present value × ............1
put here value we get
Future value = $500,000 ×
Future value = $500,000 × 2.476
Future value = $1,238,000
so investment 10 years from now is $1,238,000 .
Answer:
S = 5.7209 M
Explanation:
Given data:
B = 20.1 m
conductivity ( K ) = 14.9 m/day
Storativity ( s ) = 0.0051
1 gpm = 5.451 m^3/day
calculate the Transmissibility ( T ) = K * B
= 14.9 * 20.1 = 299.5 m^2/day
Note :
t = 1
U = ( r^2* S ) / (4*T*<em> t </em>)
= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4
Applying the thesis method
W(u) = -0.5772 - In(U)
= 7.9
next we calculate the pumping rate from well ( Q ) in m^3/day
= 500 * 5.451 m^3 /day
= 2725.5 m^3 /day
Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping
S =
where : Q = 2725.5
T = 299.5
W(u) = 7.9
substitute the given values into equation above
S = 5.7209 M