Water is pumped from one large reservoir to another at a higher elevation. If the flow rate is 2.5 ft3 /s and the pump delivers
1 answer:
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Answer:
a) 49.95 watts
b) The self locking condition is satisfied
Explanation:
Given data
weight of the square-thread power screw ( w ) = 100 kg = 1000 N
diameter (d) = 20 mm ,
pitch (p) = 2 mm
friction coefficient of steel parts ( f ) = 0.1
Gravity constant ( g ) = 10 N/kg
Rotation of electric power screwdrivers = 300 rpm
A ) Determine the power needed to raise to the basket board
first we have to calculate T
T = Wtan (∝ + Ф ) * ------------- equation 1
Dm = d - 0.5 ( 2) = 19mm
Tan ∝ = where L = 2*2 = 4
hence ∝ = 3.83⁰
given f = 0.1 , Tan Ф = 0.1. hence Ф = 5.71⁰
insert all the values into equation 1
T = 1.59 Nm
Determine the power needed using this equation
=
= 49.95 watts
B) checking if the self-locking condition of the power screw is satisfied
Ф > ∝ hence it is self locking condition is satisfied
This question is not complete, the complete question is;
The stagnation chamber of a wind tunnel is connected to a high-pressure air bottle farm which is outside the laboratory building. The two are connected by a long pipe of 4-in inside diameter. If the static pressure ratio between the bottle farm and the stagnation chamber is 10, and the bottle-farm static pressure is 100 atm, how long can the pipe be without choking? Assume adiabatic, subsonic, one-dimensional flow with a friction coefficient of 0.005
Answer:
the length of the pipe is 11583 in or 965.25 ft
Explanation:
Given the data in the question;
Static pressure ratio; p1/p2 = 10
friction coefficient f = 0.005
diameter of pipe, D =4 inch
first we obtain the value from FANN0 FLOW TABLE for pressure ratio of ( p1/p2 = 10 )so
4fL / D = 57.915
we substitute
(4×0.005×L) / 4 = 57.915
0.005L = 57.915
L = 57.915 / 0.005
L = 11583 in
Therefore, the length of the pipe is 11583 in or 965.25 ft
