Answer:
Most hydraulic systems develops pressure surges that may surpass settings valve. by exposing the hose surge to pressure above the maximum operating pressure will shorten the hose life.
Explanation:
Solution
Almost all hydraulic systems creates pressure surges that may exceed relief valve settings. exposing the hose surge to pressure above the maximum operating pressure shortens the hose life.
In systems where pressure peaks are severe, select or pick a hose with higher maximum operating pressure or choose a spiral reinforced hose specifically designed for severe pulsing applications.
Generally, hoses are designed or created to accommodate pressure surges and have operating pressures that is equal to 25% of the hose minimum pressure burst.
Answer:Science is the body of knowledge that explores the physical and natural world. Engineering is the application of knowledge in order to design, build and maintain a product or a process
Explanation:
Answer:
401.3 kg/s
Explanation:
The power plant has an efficiency of 36%. This means 64% of the heat form the source (q1) will become waste heat. Of the waste heat, 85% will be taken away by water (qw).
qw = 0.85 * q2
q2 = 0.64 * q1
p = 0.36 * q1
q1 = p /0.36
q2 = 0.64/0.36 * p
qw = 0.85 *0.64/0.36 * p
qw = 0.85 *0.64/0.36 * 600 = 907 MW
In evaporation water becomes vapor absorbing heat without going to the boiling point (similar to how sweating takes heat from the human body)
The latent heat for the vaporization of water is:
SLH = 2.26 MJ/kg
So, to dissipate 907 MW
G = qw * SLH = 907 / 2.26 = 401.3 kg/s
Answer:
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Explanation:
Answer:
The surface temperature of the ground is = 296.946K
Explanation:
Solution
Given
r₁= 0.05m
r₂= 0.08m
Tn =Ti = 77K
Ki = 0.0035 Wm-1K-1
Kg = 1 Wm-1K-1
Z= 2m
Now,
The outer type temperature (Skin temperature pipe)
Q = T₀ -T₁/ln (r2/r1)/2πKi = 2πKi T0 -T1/ln (r2/r1)
Thus,
10 w/m = 2π * 0.0035 = T0 -77/ln 0.08/0.05
⇒ T₀ -77 = 231.72
T₀= 290.72K
The shape factor between the cylinder and he ground
S = 2πL/ln 4z/D
where L = length of pipe
D = outer layer of pipe
S = 2π * 1/4 *2/ 2 * 0.08 = 1.606m
The heat gained in the pipe is = S * Kg * (Tg- T₀)
(10* 1) = 1.606 * 1* (Tg- 290.72)
Tg - 290.72 = 6.2266
Tg = 296.946K
Therefore the surface temperature to the ground is 296.946K
