Answer:
used for ordinary combustibles, such as wood, paper, some plastics, and textiles. This class of fire requires the heat-absorbing effects of water or the coating effects of certain dry chemicals.
Explanation:
Answer:
The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero
Explanation:
The expression for the maximum shear stress is given:

Where
σx = stress in vertical plane = 20 ksi
σy = stress in horizontal plane = -30 ksi
τM = 32 ksi
Replacing:

Solving for τxy:
τxy = ±19.98 ksi
The principal stress is:

Where
σp1 = 20 ksi
σp2 = -30 ksi
(equation 1)
equation 2
Solving both equations:
σp1 = 27 ksi
σp2 = -37 ksi
The shear stress on the vertical plane is zero
Answer:
See explanation
Explanation:
Solution:-
- The shell and tube heat exchanger are designated by the order of tube and shell passes.
- A single tube pass: The fluid enters from inlet, exchange of heat, the fluid exits.
- A multiple tube pass: The fluid enters from inlet, exchange of heat, U bend of the fluid, exchange of heat, .... ( nth order of pass ), and then exits.
- By increasing the number of passes we have increased the "retention time" of a specific volume of tube fluid; hence, providing sufficient time for the fluid to exchange heat with the shell fluid.
- By making more U-turns we are allowing greater length for the fluid flow to develop with " constriction and turns " into turbulence. This turbulence usually at the final passes allows mixing of fluid and increases the heat transfer coefficient by:
U ∝ v^( 0.8 ) .... ( turbulence )
- The higher the velocity of the fluids the greater the heat transfer coefficient. The increase in the heat transfer coefficient will allow less heat energy carried by either of the fluids to be wasted ; hence, reduced losses.
Thereby, increases the thermal efficiency of the heat exchanger ( higher NTU units ).
Answer:
MRR = 1.984
Explanation:
Given that
Depth of cut ,d=0.105 in
Diameter D= 1 in
Speed V= 105 sfpm
feed f= 0.015 ipr
Now the metal removal rate given as
MRR= 12 f V d
d= depth of cut
V= Speed
f=Feed
MRR= Metal removal rate
By putting the values
MRR= 12 f V d
MRR = 12 x 0.015 x 105 x 0.105
MRR = 1.984
Therefore answer is -
1.944