All categories

3 years ago

Which is the required type of fire extinguisher for standard naval vessels

Engineering

1 answer:

Bess [88]3 years ago

3 0

Answer:

Mentioned below are the required types of fire extinguishers for standard naval vessels:

Soda Acid Fire Extinguisher
Water Extinguisher
Foam Extinguisher – Chemical and Mechanical
Carbon Dioxide Extinguisher
Dry Powder Extinguisher

Explanation:

A fire extinguisher is a functioning fire insurance gadget used to douse or control little fires, regularly in crisis circumstances. It isn't planned for use on a wild fire, for example, one which has arrived at the roof, jeopardizes the client (i.e., no way out course, smoke, blast danger, and so on.), or in any case requires the mastery of a fire unit. Ordinarily, a fire extinguisher comprises of a hand-held barrel shaped weight vessel containing an operator that can be released to stifle a fire. Fire extinguishers made with non-round and hollow weight vessels likewise exist however are less normal.

A naval vessel is a military boat (or in some cases pontoon, contingent upon arrangement) utilized by a naval force. Naval boats are separated from non military personnel delivers by development and reason. By and large, naval boats are harm versatile and furnished with weapon frameworks, however combat hardware on troop transports is light or non-existent. Naval vessel is planned fundamentally for naval fighting are named warships, rather than help (assistant boats) or shipyard activities.

You might be interested in

Does anybody know how to take a screenshot on a HP pavilion computer?

Setler79 [48]

Answer:

I do i do it everyday

Explanation:

Press windows and prt sc at the same time

6 0

2 years ago

Tensile testing provides engineers with the ability to verify and establish material properties related to a specific material.

Sedbober [7]

Answer:

True

Explanation:

Tensile testing which is also referred to as tension testing is a process which materials are subjected to so as to know how well it can be stretched before it reaches breaking point. Hence, the statement in the question is true

7 0

2 years ago

Steam enters an adiabatic turbine at 10 MPa and 500°C and leaves at 10 kPa with a quality of 90 percent. Neglecting the changes

Anna35 [415]

Answer:

The mass flow rate of steam m=5.4 Kg/s

Explanation:

Given:

At the inlet of turbine P=10 MPa ,T=500 C

AT the exit of turbine P=10 KPa ,x=0.9

Required power=5 MW

From steam table

h=3374 KJ/Kg ,s=6.59 KJ/Kg-K (Super heated steam table)

$h_g$ =2675.1 KJ/Kg, $h_f$ =417.51 KJ/Kg

$s_g$ = 7.3 KJ/Kg-K , $s_f$ =1.3 KJ/Kg-K

So enthalpy of steam at the exit of turbine

h= $h_f+x(h_g- h_f)$

Now by putting the values

h= 417.51+0.9(2675.1- 417.51) KJ/Kg

h=2449.34 KJ/Kg

Lets take m is the mass flow rate of steam

So $5\times 10^3=m\times (3374-2449.34)$

m=5.4 Kg/s

So the mass flow rate of steam m=5.4 Kg/s

8 0

3 years ago

The importance of reading a circuit diagram to interpret a wiring diagram?

Nataly [62]

Answer:

The ability to read electrical schematics is a really useful skill to have. To start developing your schematic reading abilities, it's important to memorize the most common schematic symbols. ... You should also be able to get a rough idea of how the circuit works, just by looking at the schematic.

Explanation:

7 0

2 years ago

Give the approximate temperature (in K) at which creep deformation becomes an important consideration for each of the following

andrezito [222]

Answer:

691K, 543K, 725K, 1473K, 240K, 373K

Explanation:

Creep deformation of any metal is the transformational tendency of a metal to distort rapidly or slowly when attacked by any form of mechanical stress. The temperature significant for a metal to deform is gotten by the division of the actual temperature of the metal by its melting point. This is termed homologous temperature which is 0.4 or higher. It is calculated by the equation:

0.4Tm

Therefore for the listed metals...

For Nickel, 0.4Tm = 0.4 ×(1455 + 273) = 691 K

For Copper, 0.4Tm = 0.4 ×(1085 + 273) = 543 K

For Iron, 0.4Tm = 0.4 ×(1538 + 273) = 725 K

For Tungsten, 0.4Tm = 0.4 ×(3410 + 273) = 1473 K

For Lead, 0.4Tm = 0.4 × (327 + 273) = 240 K

For Aluminium, 0.4Tm = 0.4 ×(660 + 273) = 373 K

5 0

3 years ago