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2 years ago

9

Consider fully developed flow in a circular pipe with negligible entrance effects. If the length of the pipe is doubled, the pre

ssure drop will _____.

1 answer:

dezoksy [38]2 years ago

5 0

The answer is 6because I just took the test!

You might be interested in

1. A car engine transfers 3000J in 20 seconds. What is the power generated by the engine?

Scorpion4ik [409]

Answer:

150 W

Explanation:

Power is energy per unit time. The power generated by the engine is ...

(3000 J)/(20 s) = 150 J/s = 150 W

The engine generated 150 watts.

3 0

3 years ago

A long aluminum wire of diameter 3 mm is extruded at a temperature of 280°C. The wire is subjected to cross air flow at 20°C at

Musya8 [376]

Answer:

Explanation:

Given:

Diameter of aluminum wire, D = 3mm

Temperature of aluminum wire, $T_{s}=280^{o}C$

Temperature of air, $T_{\infinity}=20^{o}C$

Velocity of air flow $V=5.5m/s$

The film temperature is determined as:

$T_{f}=\frac{T_{s}-T_{\infinity}}{2}\\\\=\frac{280-20}{2}\\\\=150^{o}C$

from the table, properties of air at 1 atm pressure

At $T_{f}=150^{o}C$

Thermal conductivity, $K = 0.03443 W/m^oC$ ; kinematic viscosity $v=2.860 \times 10^{-5} m^2/s$ ; Prandtl number $Pr=0.70275$

The reynolds number for the flow is determined as:

$Re=\frac{VD}{v}\\\\=\frac{5.5 \times(3\times10^{-3})}{2.86\times10^{-5}}\\\\=576.92$

sice the obtained reynolds number is less than $2\times10^5$ , the flow is said to be laminar.

The nusselt number is determined from the relation given by:

$Nu_{cyl}= 0.3 + \frac{0.62Re^{0.5}Pr^{\frac{1}{3}}}{[1+(\frac{0.4}{Pr})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{Re}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}$

$Nu_{cyl}= 0.3 + \frac{0.62(576.92)^{0.5}(0.70275)^{\frac{1}{3}}}{[1+(\frac{0.4}{(0.70275)})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{576.92}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}\\\\=12.11$

The covective heat transfer coefficient is given by:

$Nu_{cyl}=\frac{hD}{k}$

Rewrite and solve for $h$

$h=\frac{Nu_{cyl}\timesk}{D}\\\\=\frac{12.11\times0.03443}{3\times10^{-3}}\\\\=138.98 W/m^{2}.K$

The rate of heat transfer from the wire to the air per meter length is determined from the equation is given by:

$Q=hA_{s}(T_{s}-T{\infin})\\\\=h\times(\pi\timesDL)\times(T_{s}-T{\infinity})\\\\=138.92\times(\pi\times3\times10^{-3}\times1)\times(280-20)\\\\=340.42W/m$

The rate of heat transfer from the wire to the air per meter length is $Q=340.42W/m$

6 0

3 years ago

Two objects labeled K and L have equal mass but densities 0.95Do and Do, respectively. Each of these objects floats after being

Nonamiya [84]

Answer:

C. The buoyant forces are equal on the objects since they have equal mass.

Explanation:

Correct option (C) The buoyant forces are equal on the objects since they have equal mass. For floating objects, the buoyant force equals the weight of the objects. Since each object has the same weight, they must have the same buoyant force to counteract that weight and make them float.

4 0

3 years ago

A jar made of 3/16-inch-thick glass has an inside radius of 3.00 inches and a total height of 6.00 inches (including the bottom

swat32

Answer:

1. Volume of the glass shell (Vg) is simply volume of the empty part of the jar (Ve) subtracted from volume of the entire jar (Vj):

Vg = Vj - Ve

Volume is calculated as base (B) multiplied with height (h). Base of the jar is circle, so its surface is πr^2 (r being the radius).

However radius is different depending on the part of the jar; for empty part of the jar, inner radius is d = 3 in, for the whole jar it is inner radius plus thickness of the glass a = 3 + 3/16 = 3.1875 in.

We are also given height of the whole jar, h = 6 in, but height of the empty part is entire height minus thickness of the jar h' = 6 - 0.1875 = 5.8125 in.

Now, let's calculate:

Vj = πa^2 • h = 191.42 in^3

Ve = πd^2 • h' = 164.26 in^3

So, volume of the glass shell is Vj - Ve which is 27.16 in^3.

2. Mass of the glass jar is density of the glass multiplied with volume:

m = ρ • Vg

Density of the glass is given here in cubic feet so, first, we need to convert it to cubic inches, dividing it by 1728:

ρ = 165 lb/ft^3 / 1728 = 0.095 lb/in^3

So, mass of the jar is:

m = 0.095 lb/in^3 • 27.16 in^3 = 2.59 lb

5. To find weight and volume of the water displaced we first need to find how deep the jar sinks (H), because volume of the displaced water is equal to the volume of the jar submerged. Jar will sink until gravity force (pulling it down) and buoyancy force (pushing it up) become equal. Displaced water is πa^2 • H and the buoyancy is ρw • g • Vd (ρw is density of water which is 62.5 lb/ft^3 / 1728 = 0.036 lb/in^3, and Vd is displaced water).

So, buoyancy is:

B = ρw • g • πa^2 • H

We said that buoyancy must be equal to gravity:

B = m • g (m being mass of the jar). So:

ρw • g πa^2 • H = m • g

ρw • πa^2 • H = m

From this, we can find H:

H = m / ρw•πa^2

H = 2.25 inches

That means that the jar will sink 2.25 inches in the water.

3. Now, it's easy to find volume of displaced water. It's the same as the volume of the jar submerged:

Vd = πa^2 • H

Vd = 71.94 in^3

4. And finally, the weight of water is:

m = ρw • Vd

m = 0.036 lb/in^3 • 71.94 in^3

m = 2.59 lb

Of course, we see that the mass of the jar equals the mass of the displaced water. Taking this as a rule, this question could have been solved easier However I wanted to do it more detailed, to explain it more clearly

6 0

3 years ago

What does the plasma membrane do

aniked [119]

<u>Answer:</u>

The plasma membrane encloses specific structures.

<u>Explanation: </u>

plasma membrane is also called as cell membrane. Plasma membrane only allows the particle to get in and pass out of the cell by osmosis and diffusion method from the outside environment. It is responsible for the molecular traffic inside the cell.

It helps in maintaining the shape of the cell. It has many proteins in it. Therefore cell membrane are responsible for having specific structures.

7 0

3 years ago