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valentinak56 [21]

4 years ago

12

Turpentine flows through a 12-nominal schedule 40 pipe. What is the flow rate that corresponds to a Reynolds number of 2000?

1 answer:

r-ruslan [8.4K]4 years ago

4 0

Answer:

flow rate is 8.0385 × $x^{-4}$ m³/s or 12.741 gpm

Explanation:

given data

12-nominal schedule 40 pipe

Reynolds number = 2000

to find out

What is the flow rate

solution

we know the diameter of 12-nominal schedule 40 pipe is

Diameter = 12.75 inch

D = 0.32385 m

and

dynamic viscosity of Turpentine is = 0.001375 Pa-s

and Density of Turpentine is 870 kg/m³

so

Reynolds number is express as

Re = $\frac{\rho*V*D}{\mu}$

here ρ is density and D is diameter and V is velocity and µ is viscosity

so put here all value

2000 = $\frac{870*V*0.3238}{0.001375}$

V = 9.7619 × $x^{-3}$ m/s

and

flow rate is

Q = V × A

here A is area and Q is flow rate

Q = 9.7619 × $x^{-3}$ × $\frac{\pi }{4} * 0.3238^2$

Q = 8.0385 × $x^{-4}$ m³/s

so flow rate is 8.0385 × $x^{-4}$ m³/s or 12.741 gpm

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I think it’s D ?? I’m not completely sure tho

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3 years ago

A closely wound, circular coil with radius 2.80 cm has 720 turns. Part APart complete What must the current in the coil be if th

lara [203]

Answer:

I = 4.642 Ampere.

x = 2.145 cm

Explanation:

a) As we know, the magnetic field on the axis of the loop is given as

$B = \frac{ \mu NIa^{2} }{2(x^{2} + a^{2})^{\frac{3}{2} } }$

where a = radius of loop

x = point on the axis of loop

N = No of turns of coil

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Therefore, the above equation can be rewritten as

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I = $\frac{2aB}{μN}$

by putting values

I = $\frac{2 X 0.0750 T X 0.028 m}{4\pi X 10^{-7} T.\frac{m}{A} X 720}$

Current in the loop for which the magnetic field at the center is 0.0750 Tesla = I = 4.642 Ampere

b)

Now for part b the magnetic field at a distance x from the center is given as

$B = \frac{ \mu N I a^{2} }{2(x^{2} + a^{2})^{\frac{3}{2} } }$

multiply and divide by a on both sides we get

$B = \frac{ \mu NI}{2a} X \frac{a^{3} }{(x^{2} + a^{2})^{\frac{3}{2} } }$

As we know, according to Biot sivorts law,the Magnetic Field at the Center of a circular loop is given as

$B = \frac{ \mu NI}{2a}$ ( Magnetic field at center) = $B_{c}$

So we got magnetic field at any point x as

$B_{x}$ = $B_{c}$ X $\frac{a^{3} }{(x^{2} + a^{2})^{\frac{3}{2} } }$

For magnetic field at x is half of the B at center

$B_{x}$ = $\frac{1}{2}$ $B_{c}$

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$\frac{a^{3} }{(x^{2} + a^{2})^{\frac{3}{2} } }$ = $\frac{1}{2}$

$(x^{2} + a^{2})^{3} = 4a^{6}$

$x = \sqrt{4^{\frac{1}{3}} -1 } X a^{}$

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6 0

3 years ago

Plateau Creek carries 5.0 m^3 /s of water with a selenium (Se) concentration of 0.0015 mg/L. A farmer withdraws water at a certa

Bond [772]

Answer:

The correct answer is "4.8137 m³". The further explanation is given below.

Explanation:

Firstly we have to calculate the concentration of Se:

$C = 0.0015 \ mg/L\times \frac{1g}{1000 mg}\times \frac{1 \ mol}{79 \ g}$

$=1.9\times 10^{-8} \ mol/L$

Concentration the fish can take:

$=0.04 \ mg/L\times \frac{1 \ g}{1000mg}\times \frac{1 \ mol}{79 \ g}$

According to the general dilution principle will be:

⇒ $M_1V_1 = M_2V_2$

The volume that can take the farmer will be:

$V_2 = 1.9\times 10^{-8} M\times \frac{5\times 10^3 \ L}{5.1\times 10-7 M}$

$=186.27 \ L$

On converting this into m³, we get

= $0.18627 \ m^3$

Finally the volume the farmer can remove would be:

$V = 5-0.18627$

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7 0

3 years ago

Convert 850 nm wavelength into frequency, eV, wavenumber, joules and ergs.

Sholpan [36]

Answer:

Frequency = $3.5294\times 10^{14}s^{-1}$

Wavenumber = $1.1765\times 10^6m^{-1}$

Energy = $2.3365\times 10^{-19}J$

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Energy = $2.3365\times 10^{-12}erg$

Explanation:

As we are given the wavelength = 850 nm

conversion used : $(1nm=10^{-9}m)$

So, wavelength is $850\times 10^{-9}m$

The relation between frequency and wavelength is shown below as:

$Frequency=\frac{c}{Wavelength}$

Where, c is the speed of light having value = $3\times 10^8m/s$

So, Frequency is:

$Frequency=\frac{3\times 10^8m/s}{850\times 10^{-9}m}$

$Frequency=3.5294\times 10^{14}s^{-1}$

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So,

$Wavenumber=\frac{1}{Wavelength}=\frac{1}{850\times 10^{-9}m}$

$Wavenumber=1.1765\times 10^6m^{-1}$

Also,

$Energy=h\times frequency$

where, h is Plank's constant having value as $6.62\times 10^{-34}J.s$

So,

$Energy=(6.62\times 10^{-34}J.s)\times (3.5294\times 10^{14}s^{-1})$

$Energy=2.3365\times 10^{-19}J$

Also,

$1J=6.24\times 10^{18}eV$

So,

$Energy=(2.3365\times 10^{-19})\times (6.24\times 10^{18}eV)$

$Energy=1.4579eV$

Also,

$1J=10^7erg$

So,

$Energy=(2.3365\times 10^{-19})\times 10^7erg$

$Energy=2.3365\times 10^{-12}erg$

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3 years ago

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Answer:

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