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Snezhnost [94]
3 years ago
8

You will be observing laminar-turbulent transition for room temperature (about 20°C) water flowing in a 0.602"" ID pipe (Schedul

e 40 1/2"" pipe). Estimate the flow rate in gallons/minute (gpm) at which transition should begin (ReD= 2300) and the flow rate for fully turbulent flow (ReD> 4000). Repeat calculations for a 0.804"" ID pipe (Schedule 40 3/4"" pipe).

Engineering
1 answer:
igomit [66]3 years ago
8 0

Answer:

b) 0.7667gallons/minutes

Explanation:

check the attachment below for other answers

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A flywheel performs each of these functions except: A. Contains a gear used for engine starting B. Smoothes engine operation C.
torisob [31]

Answer:

C. Provides lubrication to parts

Explanation:

Flywheel :

 Flywheel is a device which stored the mechanical energy.This energy can be use when more energy required during any operation.Due to high moment of inertia of the flywheel it resist the change in the speed .The flywheel is attached to the crank shaft at the rear side of the engine.

The flywheel perform following function:

1. It connects the crankshaft and the transmission system.

2.It makes the engine operation smooth.

3.It contains gear and other parts of the engine.

But it can not provide lubrication.

C. Provides lubrication to parts

4 0
3 years ago
Which is a better hydraulic cross section for an open channel: one with a small or a large hydraulic radius?
Cerrena [4.2K]
Hydraulic radius is caused by pressurized hydrogen air so that should mean the answer is hydraulic radius
6 0
3 years ago
a metal rod 24mm diameter and 2m long is subjected to an axial pull of 40 kN. If the rod is 0.5mm, then find the stress-induced
ozzi

Answer:

i dont know but i will take the points tho hahah

Explanation:

8 0
3 years ago
An electrical current of 700 A flows through a stainlesssteel cable having a diameter of 5 mm and an electricalresistance of 610
KatRina [158]

Answer:

778.4°C

Explanation:

I = 700

R = 6x10⁻⁴

we first calculate the rate of heat that is being transferred by the current

q = I²R

q = 700²(6x10⁻⁴)

= 490000x0.0006

= 294 W/M

we calculate the surface temperature

Ts = T∞ + \frac{q}{h\pi Di}

Ts = 30+\frac{294}{25*\frac{22}{7}*\frac{5}{1000}  }

Ts=30+\frac{294}{0.3928} \\

Ts =30+748.4\\Ts = 778.4

The surface temperature is therefore 778.4°C if the cable is bare

6 0
3 years ago
For each function , sketch the Bode asymptotic magnitude and asymptotic phase plots.
horrorfan [7]

Answer:

attached below

Explanation:

a) G(s) = 1 / s( s+2)(s + 4 )

Bode asymptotic magnitude and asymptotic phase plots

attached below

b) G(s) = (s+5)/(s+2)(s+4)

phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4

attached below

c) G(s)= (s+3)(s+5)/s(s+2)(s+4)

solution attached below

5 0
3 years ago
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