A compressor delivers 130 scfm of air through a 1-in schedule 40 pipe at a receiver pressure of 105 psig. Find the pressure loss
1 answer:
Answer:
The pressure loss in the pipe is 7.792 psig
Explanation:
Given;
flow rate Q = 130 scfm
length of pipeline =170ft
atmospheric pressure = 14.7 psia
pressure in pipe = 105 psig, to psia = 105 + 14.7 = 119.7 psia
Compression ratio (CR) = pressure in pipe/atmospheric pressure
Compression ratio (CR) = 119.7/14.7 = 8.143
Pressure loss in the pipe can be calculated as follows:
from schedule table, 1-in schedule 40 pipe has a diameter d.
Substituting these values, pressure loss in the pipe becomes
Therefore, the pressure loss in the pipe is 7.792 psig
You might be interested in
Answer:
linear density of the string = 4.46 × 10⁻⁴ kg/m
Explanation:
given,
mass of the string = 31.2 g
length of string = 0.7 m
linear density of the string =
linear density of the string =
linear density of the string = 44.57 × 10⁻³ kg/m
linear density of the string = 4.46 × 10⁻⁴ kg/m
Answer:
(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz
(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz
Explanation:
(a) The fundamental, f₁, frequency is given as follows;
Where;
T = The tension in the string
μ = The linear density of the string
L = The length of the string
f₁ = The fundamental frequency = 560 Hz
If the tension in the string is increased by 15%, we will have;
Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz
(b) When the string length is decreased by one-third, we have;
The new length of the string, = 2/3·L
The value of the fundamental frequency will then be given as follows;
When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.