1. Sewage-treatment plant, a large concrete tank initially contains 440,000 liters liquid and 10,000 kg fine suspended solids. To flush this material out of the tank, water is pumped into the vessel at a rate of 40,000 liter/h. Liquid containing solids leaves at the same rate. Estimate the concentration of suspended solids in the tank at the end of 5 h.
Answers
Answer:
Concentration = 10.33 kg/m³
Explanation:
We are given;
Mass of solids; 10,000 kg
Volume; V = 440,000 L = 440 m³
Rate at which water is pumped out = 40,000 liter/h
Thus, at the end of 5 hours we amount of water that has been replaced with fresh water is = 40,000 liter/h x 5 hours = 200,000 L = 200 m³
Now, since the tank is perfectly mixed, therefore we can calculate a ratio of fresh water to sewage water as;
200m³/440m³ = 5/11
Thus, the amount left will be calculated by multiplying that ratio by the amount of solids;
Thus,
Amount left; = 10000 x (5/11) = 4545 kg
The concentration would be calculated by:
Concentration = amount left/initial volume
Thus,
Concentration = 4545/440 = 10.3 kg/m³
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b) Thermal barrier coating. Cement, aluminum, engineering ceramic, super alloy, steel and glass
Answers
Answer:
Answer explained below
Explanation:
3.] a] A turbine blade is the individual component which makes up the turbine section of a gas turbine. The blades are responsible for extracting energy from the high temperature, high pressure gas produced by the combustor.
The turbine blades are often the limiting component of gas turbines. To survive in this difficult environment, turbine blades often use exotic materials like superalloys and many different methods of cooling, such as internal air channels, boundary layer cooling, and thermal barrier coatings. The blade fatigue failure is one of the major source of outages in any steam turbines and gas turbines which is due to high dynamic stresses caused by blade vibration and resonance within the operating range of machinery.
To protect blades from these high dynamic stresses, friction dampers are used.
b] Thermal barrier coatings (TBC) are highly advanced materials systems usually applied to metallic surfaces, such as on gas turbine or aero-engine parts, operating at elevated temperatures, as a form ofexhaust heat management.
These 100μm to 2mm coatings serve to insulate components from large and prolonged heat loads by utilizing thermally insulating materials which can sustain an appreciable temperature difference between the load-bearing alloys and the coating surface.
In doing so, these coatings can allow for higher operating temperatures while limiting the thermal exposure of structural components, extending part life by reducing oxidation and thermal fatigue.
In conjunction with active film cooling, TBCs permit working fluid temperatures higher than the melting point of the metal airfoil in some turbine applications.
Due to increasing demand for higher engine operation (efficiency increases at higher temperatures), better durability/lifetime, and thinner coatings to reduce parasitic weight for rotating/moving components, there is great motivation to develop new and advanced TBCs.
Answers
Answer:
18.722 ft
Explanation:
The elevation difference between the PVT and a point on the curve at station
111 + 00
attached below is a detailed solution to the problem
Δelevation ( elevation difference )
= Yt - Y
= 19.355 - 0.632 = 18.722 ft
Answers
Answer:
Cu= 60
Mg= 67
Mn= 6
Si= 76
Zn= 101
Explanation:
Solution steps :
1) Creating matrix A holds the composition of each raw material.
2) Submission of composition of each raw material.
3) Multiplying each amount by the total amount needed to be produced.
note
find the attached code
Answers
Find the given attachments for complete explanation
Answers
Answer:
(a) ΔU = 125 kJ
(b) ΔU = -110 kJ
Explanation:
(a) Suppose that 150 kJ of work are used to compress a spring, and that 25 kJ of heat are given off by the spring during this compression. What is the change in internal energy of the spring during the process?
The work is done to the system so w = 150 kJ.
The heat is released by the system so q = -25 kJ.
The change in internal energy (ΔU) is:
ΔU = q + w
ΔU = -25 kJ + 150 kJ = 125 kJ
(b) Suppose that 100 kJ of work is done by a motor, but it also gives off 10 kJ of heat while carrying out this work. What is the change in internal energy of the motor during the process?
The work is done by the system so w = -100 kJ.
The heat is released by the system so q = -10 kJ.
The change in internal energy (ΔU) is:
ΔU = q + w
ΔU = -10 kJ - 100 kJ = -110 kJ
Answers
Answer:
hello your question has a missing part below is the missing part
Consider the string length equal to
answer : 2cos(2t) sin(2x) - 10cos(10t)sin(10x)
Explanation:
Given string length =
distorted function f(x) = 2sin(2x) - 10sin(10x)
Determine the wave formed in the string
attached below is a detailed solution of the problem