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.
Answer:
heat transfer for the process is - 643.3 kJ
Explanation:
given data
mass m = 2 kg
pressure p1 = 500 kPa
temperature t1 = 400°C = 673.15 K
temperature t2 = 40°C = 313.15 K
pressure p2 = 300 kPa
to find out
heat transfer for the process
solution
we know here mass is constant so
m1 = m2
so by energy equation
m ( u2 - u1 ) = Q - W
Q is heat transfer
and in process P = A+ N that is linear spring
so
W = ∫PdV
= 0.5 ( P1+P2) ( V1 - V2)
so for case 1
P1V1 = mRT
put here value
500 V1 = 2 (0.18892) (673.15)
V1 = 0.5087 m³
and
for case 2
P2V2 = nRT
300 V2 = 2 (0.18892) (313.15)
V2 = 0.3944 m³
and
here W will be
W = 0.5 ( 500 + 300 ) ( 0.3944 - 0.5087 )
W = -45.72 kJ
and
Q is here for Cv = 0.83 from ideal gas table
Q = mCv ( T2-T1 ) + W
Q = 2 × 0.83 ( 40 - 400 ) - 45.72
Q = - 643.3 kJ
heat transfer for the process is - 643.3 kJ
Answer:
Explanation:
Hello!
In this case, we compute the heat output from coal, given its heating value and the mass flow:
Next, since the work done by the power plant is 230 MW, we compute the efficiency as shown below:
Best regards!
Answer:
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Explanation:
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