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:
a. Rotational speed of the drill = 375.96 rev/min
b. Feed rate = 75 mm/min
c. Approach allowance = 3.815 mm
d. Cutting time = 0.67 minutes
e. Metal removal rate after the drill bit reaches full diameter. = 9525 mm³/min
Explanation:
Here we have
a. N = v/(πD) = 15/(0.0127·π) = 375.96 rev/min
b. Feed rate = fr = Nf = 375.96 × 0.2 = 75 mm/min
c. Approach allowance = tan 118/2 = (12.7/2)/tan 118/2 = 3.815 mm
d. Approach allowance T∞ =L/fr = 50/75 = 0.67 minutes
e. R = 0.25πD²fr = 9525 mm³/min.
Explanation:
Label and group products. One would think that a general cleanup would be the first step, but no, it's not. ...Clean up the area. ...Put up demarcation lines. ...Stack properly. ...Keep the aisles, paths and ramps clear. ...Have all the safety signs in place.
Answer:
a) 8kW
b) $128
Explanation:
Given the coefficient of performance of the heat pump cycle to be 2.5
Energy delivered by the heat pump = 20kW
a) net power required to operate the heat pump = Energy delivered / coefficient of performance
Net power required = 20/2.5
= 8kW
b) Given the cost of electricity is $0.08 for 1kWhour
Since net power required to operate heat pump = 8kW
If the heat pump operate for 200hours, total power required for a month = 8kW×200hours = 1600kWhour
since 1kWh of electricity costs $0.08, cost of electricity used in a month when the pump operates for 200hour will be 1600kWh×$0.08 which is equivalent to $128
