while performing a running compression test how should running compression compare to static compression
1 answer:
You might be interested in
This question is incomplete, the complete question is;
A centrifugal pump is used to extract water from a reservoir at 14,000 gal/min. The pipe connecting the pump inlet to the reservoir is 12 inches in diameter and is 65 ft long.
The average friction factor in this pipe is 0.018. The pump performance curves indicate that, at this flow rate, the head rise across the pump is 320 ft, the efficiency is 81% and the required NPSH is 25 ft.
Please estimate: The required brake horsepower.
Answer:
The required brake horsepower is 1400.08
Explanation:
Given the data in the question;
Power required to drive the pump can be determined using the formula;
P = rQH / η₀(0.745)
given that; centrifugal pump is used to extract water from a reservoir at 14,000 gal/min.
Q = 14,000 gal/min = ( 14,000 × 0.00006309 )m³/sec = 0.883 m³/sec
the head rise across the pump is 320 ft,
H = 320 ft = ( 320 × 0.3048 )m = 97.536 m
the efficiency η₀ = 81% = 0.81
r = 9.81 kN/m³
so we substitute our values into the formula
P = [ 9.81 × 0.883 × 97.536 ] / 0.81(0.745)
P = 844.87926528 / 0.60345
P = 1400.08 HP
Therefore, The required brake horsepower is 1400.08
Answer:
- C. 1/4"
- B. 3/16"
Explanation:
1. For hex bolts, lag bolts, and square bolts, the wrench size is 1/4" larger than the bolt size for 1/2" and 9/16" bolts. For 5/8" bolts and larger, the wrench size is <em>50% larger than the bolt size</em>.
__
2. For 7/16" bolts, the wrench size is 5/8", so is 3/16" larger than the bolt. This holds down to 1/4" bolts, where the wrench size may be 3/8" or 7/16".
