Answer:
(a) Effectiveness of the regenerator= 0.433
(b) The rate of heat removal=21.38 kW
Explanation:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Answer:
diameter is 14 mm
Explanation:
given data
power = 15 kW
rotation N = 1750 rpm
factor of safety = 3
to find out
minimum diameter
solution
we will apply here power formula to find T that is
power = 2π×N×T / 60 .................1
put here value
15 × = 2π×1750×T / 60
so
T = 81.84 Nm
and
torsion = T / Z ..........2
here Z is section modulus i.e = πd³/ 16
so from equation 2
torsion = 81.84 / πd³/ 16
so torsion = 416.75 / / d³ .................3
so from shear stress theory
torsion = σy / factor of safety
so here σy = 530 for 1020 steel
so
torsion = σy / factor of safety
416.75 / d³ = 530 × / 3
so d = 0.0133 m
so diameter is 14 mm
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