Solution :


Using the data table for refrigerant-134a at P = 120 psia



∴ 
For pressure, P = 20 psia





Change in temperature, 


Now we find the quality,



The final energy,



Change in internal energy

= 38.09297-40.5485
= -2.4556
Answer:
Three ways that engineers explore possible solutions in their projects are;
1) Prototyping
2) Simulation
3) Calculations
Explanation:
1) Prototyping is the process of experimental testing of samples of design, or model of a product with the possibility of the inclusion of control of parameters in order to determine the workability of a solution.
2) Simulation is the process of creating an imitation of a situation, operation or process which can be used to determine if a particular solution will be able to work as required in the simulated environment of a problem.
3) Calculations are used to find preliminary results of particular situations, their cause and effects based on scientific laws, theories and hypothesis such that the factor of the problem is equated with the available ideas to find the best possible solution.
Answer:
thank you for the free point have a great rest of your day
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