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Natasha2012 [34]
3 years ago
6

What is a Flame Front Generator?

Engineering
1 answer:
Inessa [10]3 years ago
7 0

Answer and Explanation:

Flame Front Generator: It is a ignition system which is very useful in flaring system .In this system the air and gases are mixed together and make a combustible air gas mixture. There is a flame front region where the combustion reaction takes place , it is the region where gases as like hydrogen and air mixed with each other and form combustible gases.

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true or false incident reports, such as situation reports and status reports enhance situational awareness and ensure that perso
Degger [83]
True becauseidbajqkfkmxzmbakwlgof
4 0
4 years ago
Why or why not the following materials will make good candidates for the construction of
zvonat [6]

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.

3 0
4 years ago
1. True/False The Pressure Relief valve maintains the minimum pressure in the hydraulic circuit​
elena55 [62]
Yeah it’s true. Good luck!!
3 0
3 years ago
3. A 4-m × 5-m × 7-m room is heated by the radiator of a steam-heating system. The steam radiator transfers heat at a rate of 10
Natali [406]

Answer:

14.52 minutes

<u>OR</u>

14 minutes and 31 seconds

Explanation:

Let's first start by mentioning the specific heat of air at constant volume. We consider constant volume and NOT constant pressure because the volume of the room remains constant while pressure may vary.

Specific heat at constant volume at 27°C = 0.718 kJ/kg*K

Initial temperature of room (in kelvin) = 283.15 K

Final temperature (required) of room = 293.15 K

Mass of air in room= volume * density= (4 * 5 * 7) * (1.204 kg/m3) = 168.56kg

Heat required at constant volume: 0.718 * (change in temp) * (mass of air)

Heat required = 0.718 * (293.15 - 283.15) * (168.56) = 1,210.26 kJ

Time taken for temperature rise: heat required / (rate of heat change)

Where rate of heat change = 10000 - 5000 = 5000 kJ/hr

Time taken = 1210.26 / 5000 = 0.24205 hours

Converted to minutes = 0.24205 * 60 = 14.52 minutes

4 0
4 years ago
The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet
alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

T_{h_i}=200^{\circ}C

T_{h_o}=120^{\circ}C

T_{c_i}=100^{\circ}C

T_{c_o}=125^{\circ}C

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]

\frac{m_hc_{ph}}{m_cc_{pc}}=\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )

we can see that heat capacity of hot fluid is minimum

Also from energy balance

Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )

NTU=\frac{UA}{\left ( mc_p\right )_{h}}=\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}

T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}

T_m=41.63^{\circ}C

NTU=1.921

And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}

\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}

\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}

\epsilon =\frac{1-0.2669}{1-0.0834}

\epsilon =0.799

5 0
4 years ago
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