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
Answer:
the heat transfer from the pipe will decrease when the insulation is taken off for r₂< 
where;
r₂ = outer radius
= critical radius
Explanation:
Note that the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h .
The rate of heat transfer from the cylinder increases with the addition of insulation for outer radius less than critical radius (r₂< ) 0, and reaches a maximum when r₂ = , and starts to decrease for r₂< . Thus, insulating the pipe may actually increase the rate of heat transfer from the pipe instead of decreasing it when r₂< .
Answer:
38 kJ
Explanation:
The solution is obtained using the energy balance:
ΔE=E_in-E_out
U_2-U_1=Q_in+W_in-Q_out
U_2=U_1+Q_in+W_in-Q_out
=38 kJ
Answer:
Hi there...
Explanation:
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