Answer:
Engineering aims to allow technical parts, structures and/or systems, such as those of a machine, to fulfill their function. To this end, occurring and/or desired processes (the operation thereof) are investigated and elaborated. Appropriate parts are designed individually or in a team, occurring stresses are calculated and the parts to be produced are modeled and/or specified on technical drawings. In some fields, maintenance is a standard part of the work to be performed. For a large part of engineering, standards and agreements have been drawn up with a view to the desired safety and practical applicability.
Answer:
The maximum water pressure at the discharge of the pump (exit) = 496 kPa
Explanation:
The equation expressing the relationship of the power input of a pump can be computed as:
![E _{pump,u} = \dfrac{m(P_2-P_1)}{\rho}](https://tex.z-dn.net/?f=E%20_%7Bpump%2Cu%7D%20%3D%20%5Cdfrac%7Bm%28P_2-P_1%29%7D%7B%5Crho%7D)
where;
m = mass flow rate = 120 kg/min
the pressure at the inlet
= 96 kPa
the pressure at the exit
= ???
the pressure
= 1000 kg/m³
∴
![0.8 \times 10^{3} \ W = \dfrac{(120 \ kg/min * 1min/60 s)(P_2-96000)}{1000}](https://tex.z-dn.net/?f=0.8%20%5Ctimes%2010%5E%7B3%7D%20%5C%20W%20%3D%20%5Cdfrac%7B%28120%20%5C%20kg%2Fmin%20%2A%201min%2F60%20s%29%28P_2-96000%29%7D%7B1000%7D)
![0.8 \times 10^{3}\times 1000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}](https://tex.z-dn.net/?f=0.8%20%5Ctimes%2010%5E%7B3%7D%5Ctimes%201000%20%3D%20%7B%28120%20%5C%20kg%2Fmin%20%2A%201min%2F60%20s%29%28P_2-96000%29%7D)
![800000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}](https://tex.z-dn.net/?f=800000%20%3D%20%7B%28120%20%5C%20kg%2Fmin%20%2A%201min%2F60%20s%29%28P_2-96000%29%7D)
![\dfrac{800000}{2} = P_2-96000](https://tex.z-dn.net/?f=%5Cdfrac%7B800000%7D%7B2%7D%20%3D%20P_2-96000)
400000 = P₂ - 96000
400000 + 96000 = P₂
P₂ = 496000 Pa
P₂ = 496 kPa
Thus, the maximum water pressure at the discharge of the pump (exit) = 496 kPa
Solution :
Given :
![$V_1 = 7 \ m/s$](https://tex.z-dn.net/?f=%24V_1%20%3D%207%20%5C%20m%2Fs%24)
Operation time,
= 3000 hours per year
![$V_2 = 10 \ m/s$](https://tex.z-dn.net/?f=%24V_2%20%3D%2010%20%5C%20m%2Fs%24)
Operation time,
= 2000 hours per year
The density, ρ = ![$1.25 \ kg/m^3$](https://tex.z-dn.net/?f=%241.25%20%5C%20kg%2Fm%5E3%24)
The wind blows steadily. So, the K.E. = ![$(0.5 \dot{m} V^2)$](https://tex.z-dn.net/?f=%24%280.5%20%5Cdot%7Bm%7D%20V%5E2%29%24)
![$= \dot{m} \times 0.5 V^2$](https://tex.z-dn.net/?f=%24%3D%20%5Cdot%7Bm%7D%20%5Ctimes%200.5%20V%5E2%24)
The power generation is the time rate of the kinetic energy which can be calculated as follows:
Power = ![$\Delta \ \dot{K.E.} = \dot{m} \frac{V^2}{2}$](https://tex.z-dn.net/?f=%24%5CDelta%20%5C%20%5Cdot%7BK.E.%7D%20%3D%20%5Cdot%7Bm%7D%20%5Cfrac%7BV%5E2%7D%7B2%7D%24)
Regarding that
. Then,
Power
→ Power = constant x ![$V^3$](https://tex.z-dn.net/?f=%24V%5E3%24)
Since,
is constant for both the sites and the area is the same as same winf turbine is used.
For the first site,
Power, ![$P_1= \text{const.} \times V_1^3$](https://tex.z-dn.net/?f=%24P_1%3D%20%5Ctext%7Bconst.%7D%20%5Ctimes%20V_1%5E3%24)
![$P_1 = \text{const.} \times 343 \ W$](https://tex.z-dn.net/?f=%24P_1%20%3D%20%5Ctext%7Bconst.%7D%20%5Ctimes%20343%20%5C%20W%24)
For the second site,
Power, ![$P_2 = \text{const.} \times V_2^3 \ W$](https://tex.z-dn.net/?f=%24P_2%20%3D%20%5Ctext%7Bconst.%7D%20%5Ctimes%20V_2%5E3%20%5C%20W%24)
Answer: Solid state welding
Explanation: Solid state welding is the welding procedure which is based on the temperatures and pressure but without any liquid or vapor as aid for welding.This process is carried mainly cohesive forces and considering forces as less important. Brazing,soldering and adhesive is the process in which material are joint which the help of solid welding agent.Thus solid state welding is the correct option.