Answer:
11.52 hp
Explanation:
<u><em>Givens: </em></u>
p_1 = 15 pisa
p_2 = 70 pisa
V_ol=1.5 ft^3/s
<u><em>Solution: </em></u>
Note: m = p x V_ol (assuming in compressible flow —> p =const)
The total change in the system mechanical energy can be calculated as follows,
Δ
e= (p_2 - p_1 ) /p
The power needed can be calculated as follows
P = W =mΔ
e = p x V_ol x(p_2 - p_1 ) /p
= V_ol x (p_2 - p_1 )
= 44 pisa. ft^3/s
= 44 x (1 btu/5.404pisa. ft^3) x (1 hp/0.7068btu/s)
= 11.52 hp
Answer: 5.36×10-3kg/h
Where 10-3 is 10 exponential 3 or 10 raised to the power of -3.
Explanation:using the formula
M =JAt = -DAt×Dc/Dx
Where D is change in the respective variables. Insulting the values we get,
=5.1 × 10-8 × 0.13 × 3600 × 2.9 × 0.31 / 4×10-3.
=5.36×10-3kg/h
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:
It's true
Explanation:
I took the quiz a few days ago and got it right!
Hope this helps:)