Answer:Topographic map. Contour line. Learning Objectives. After completing this chapter, you will be able to: □ Define civil engineering and civil drafting.
Explanation:
Answer:
Power = 371.28 kW
Explanation:
Initial pressure, P1 = 5 bar
Final pressure, P2 = 1 bar
Initial temperature, T1 = 320°C
Final temperature, T2 = 160°C
Volume flow rate, V = 0.65m³/s
From steam tables at state 1,
h1 = 3105.6 kJ/kg, s1 = 7.5308 kJ/kgK
v1 = 0.5416 m³/kg
Mass flow rate, m = V/v1
m = 1.2 kg/s
From steam tables, at state 2
h2 = 2796.2 kJ/kg, s2 = 7.6597 kJ/kgK
Power developed, P = m(h1 - h2)
P = 1.2(3105.6-2796.2)
P = 371.28 kW
Answer:
≅ 111 KN
Explanation:
Given that;
A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8
mass = 85,000 kg
drag co-efficient (C) = 0.37
(velocity (v)= 230 m/s
density (ρ) = 1.0 kg/m³
To calculate the thrust; we need to determine the relation of the drag force; which is given as:
=
× CρAv²
where;
ρ = density of air wind.
C = drag co-efficient
A = Area of the jet
v = velocity of the jet
From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0
SO, ![F_{drag}-F_{thrust} = 0](https://tex.z-dn.net/?f=F_%7Bdrag%7D-F_%7Bthrust%7D%20%3D%200)
We can as well say:
![F_{drag}= F_{thrust}](https://tex.z-dn.net/?f=F_%7Bdrag%7D%3D%20F_%7Bthrust%7D)
We can now replace
in the above equation.
Therefore,
=
× CρAv²
The A which stands as the area of the jet is given by the formula:
![A=\frac{\pi d^2}{4}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Cpi%20d%5E2%7D%7B4%7D)
We can now have a new equation after substituting our A into the previous equation as:
=
× Cρ ![(\frac{\pi d^2}{4})v^2](https://tex.z-dn.net/?f=%28%5Cfrac%7B%5Cpi%20d%5E2%7D%7B4%7D%29v%5E2)
Substituting our data from above; we have:
=
× ![(0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2](https://tex.z-dn.net/?f=%280.37%29%281.0kg%2Fm%5E3%29%28%5Cfrac%7B%5Cpi%283.8m%29%5E2%20%7D%7B4%7D%29%28230m%2Fs%29%5E2)
= ![\frac{1}{8} (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B8%7D%20%20%20%280.37%29%281.0kg%2Fm%5E3%29%28%7B%5Cpi%283.8m%29%5E2%20%7D%29%28230m%2Fs%29%5E2)
= 110,990N
in N (newton) to KN (kilo-newton) will be:
= ![(110,990N)*\frac{1KN}{1,000N}](https://tex.z-dn.net/?f=%28110%2C990N%29%2A%5Cfrac%7B1KN%7D%7B1%2C000N%7D)
= 110.990 KN
≅ 111 KN
In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.
Answer:
Greywater.
Explanation:
Greywater is also known as sullage and it can be defined as any form of gently used wastewater derived from sources within a residential or office building such as showers, washing machines, bathroom sinks, bathroom tub, etc.
Generally, greywater or sullage is completely free of fecal materials (faeces) because it is independent from all toilet activities. However, greywater is not clean for direct use because it usually contains food particles, dirt, oil from dishes, hair, etc.
In this scenario, Anna makes arrangements to reuse waste water that has been used in sinks and showers. Greywater is a term which refers to the waste water that Anna reuses to conserve resources.
Therefore, Anna reuses greywater to conserve resources.