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2 years ago

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A preheater involves the use of condensing steam at 100o C on the inside of a bank of tubes to heat air that enters at I atm and

25o C. The air moves at 5 m/s in cross flow over the tubes. Each tube is 1 m long and has an outside diameter of 10 mm. The bank consists of 196 tubes in a square, aligned array for which ST = SL = 15 mm. What is the total rate of heat transfer to the air? What is the pressure drop associated with the airflow?
(b) Repeat the analysis of part (a), but assume that the tubes have a staggered arrangement with ST = 15 mm and SL = 10 mm.

1 answer:

Anna35 [415]2 years ago

8 0

Answer:

Please see the attached file for the complete answer.

Explanation:

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A fire hose nozzle has a diameter of 1.125 in. According to some fire codes, the nozzle must be capable of delivering at least 2

Furkat [3]

Answer:

$P_{1} = 403,708\,kPa\,(58.553\,psi)$

Explanation:

Let assume that changes in gravitational potential energy can be neglected. The fire hose nozzle is modelled by the Bernoulli's Principle:

$\frac{P_{1}}{\rho\cdot g} = \frac{P_{2}}{\rho \cdot g} + \frac{v^{2}}{2\cdot g}$

The initial pressure is:

$P_{1} = P_{2}+ \frac{1}{2}\cdot \rho v^{2}$

The speed at outlet is:

$v=\frac{\dot Q}{\frac{\pi}{4}\cdot D^{2}}$

$v=\frac{(250\,\frac{gal}{min} )\cdot (\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot [(1.125\,in)\cdot(\frac{0.0254\,m}{1\,in} )]^{2} }$

$v\approx 24.592\,\frac{m}{s}\,(80.682\,\frac{ft}{s} )$

The initial pressure is:

$P_{1} = 101.325\times 10^{3}\,Pa+\frac{1}{2}\cdot (1000\,\frac{kg}{m^{3}} )\cdot (24.592\,\frac{m}{s} )^{2}$

$P_{1} = 403,708\,kPa\,(58.553\,psi)$

7 0

2 years ago

Read 2 more answers

Write using about 10-15 lines for each of the six materials (metals, ceramics, glasses, polymers, composites, and semiconductors

Svetradugi [14.3K]

Answer:

See Answer below- Explanation is the entire answer

Explanation:

Metals:

Properties: Ductile, good heat conductivity, good electrical conductivity, high strength;

Drawbacks: Relatively high weight, reactive with oxygen to create oxides- corrosion is presented;

Examples: steel, aluminum alloys, brass, copper, titanium

Applications: Body of the vehicles, structures in the skyscrapers, cooking pots.

Ceramics:

Properties: Brittle, poor heat conductors, poor electrical conductors, high wear resistance, corrosion resistance;

Drawbacks: Deforms by fracturing, shock resistance is low, no conductivity of electricity;

Examples: concrete, tungsten carbide, diamond

Applications: bricks for constructions, clay pots to keep heat, cutting tools for metals;

Glasses:

Properties: amorphous, transparent, high weight

Drawbacks: poor conductors of heat and electricity; brittle; low shock resistance;

Examples: Silica, lead glass, glaze;

Applications: windows, protection screens;

Polymers:

Properties: low density, recyclable, poor heat and electrical conductors, plastic deformation;

Drawbacks: low strength, low operating temperatures;

Examples: polyethylene, nylon, ABS-plastic, rubber;

Applications: toys, tires, insulation covers for the wires.

Composites:

Properties: high strength to weight ratio, can get combination of properties from the used materials, rarely conductive, good shock resistance;

Drawbacks: high cost, hard to recycle, expensive;

Examples: steel-reinforced concrete, carbon fiber, fiber glass, Nomex, sandwich roof panels;

Applications: buildings, bullet proof vests, body of the Formula 1 cars, rockets, roof panels.

Semiconductors:

Properties: brittle, change conductive behavior under certain scenario, poor heat conductors;

Drawbacks: hard to manufacture, expensive;

Examples: Silicon-based semiconductors, Germanium-based semiconductors, Ga-based semiconductors;

Applications: chips, LED, diodes, transistors, op-amps, microprocessors.

8 0

2 years ago

A wire of diameter d is stretched along the centerline of a pipe of diameter D. For a given pressure drop per unit length of pip

JulsSmile [24]

Answer:

Part A: (d/D=0.1)

DeltaV percent=42.6%

Part B:(d/D=0.01)

DeltaV percent=21.7%

Explanation:

We are going to use the following volume flow rate equation:

$DeltaV=\frac{\pi * DeltaP}{8*u*l}(R^{4}-r^{4} -\frac{(R^{2}-r^{2})}{ln\frac{R}{r}}^{2})$

Above equation can be written as:

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{r}{R} )^{4}+\frac{(1-(\frac{r}{R} )^{2})}{ln\frac{r}{R}}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{d}{D} )^{4}+\frac{(1-(\frac{d}{D})^{2})}{ln\frac{d}{D}}^{2})$

First Consider no wire i.e d/D=0

Above expression will become:

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0)^{4}+\frac{(1-(0)^{2})}{ln0}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}$

Part A: (d/D=0.1)

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.1)^{4}+\frac{(1-(0.1)^{2})}{ln0.1}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.574$

$DeltaV percent=\frac{(\frac{\pi*R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.574}{\frac{\pi*R^{4}*DeltaP}{8*u*l} }*100$

$DeltaV percent=\frac{1-0.574}{1}*100$

DeltaV percent=42.6%

Part B:(d/D=0.01)

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.01)^{4}+\frac{(1-(0.01 )^{2})}{ln0.01}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.783$

$DeltaV percent=\frac{(\frac{\pi *R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.783}{\frac{\pi *R^{4}*DeltaP}{8*u*l} }*100$

$DeltaV percent=\frac{1-0.783}{1}*100$

DeltaV percent=21.7%

5 0

2 years ago

The extruder head in a fused- deposition modeling setup has a diameter of 1.25 mm (0.05 in) and produces layers that are 0.25mm

Angelina_Jolie [31]

Answer:

The time taken will be "1 hour 51 min". The further explanation is given below.

Explanation:

The given values are:

Number of required layers:

= $\frac{38}{0.25}$

= $152 \ layers$

Diameter (d):

= 1.25 mm

Velocity (v):

= 40 mm/s

Now,

The area of one layer will be:

= $38\times 38 \ mm^2$

= $1444 \ mm^2$

The area covered every \second will be:

= $d\times v$

= $1.25\times 40$

= $50 \ mm^2$

The time required to deposit one layer will be:

= $\frac{1444}{50}$

= $28.88 \ sec$

The time required for one layer will be:

= $15 \ sec$

∴ Total times required for one layer will be:

= $15+28.88$

= $43.88 \ sec$

So,

Number of layers = 152

Therefore,

Total time will be:

= $152\times 43.88$

= $6669.76 \ sec$

= $1 \ hour \ 51 \ min$

6 0

2 years ago

Determine the speed of sound in air at 400 K. Also determine the Mach number of an aircraft moving in the air at a velocity of 3

Reika [66]

Answer:

$\alpha = \sqrt{1.4 *0.287 \frac{KJ}{Kg K}*\frac{1000J}{1KJ} *400 K}= 400.899 m/s$

$Ma= \frac{310 m/s}{400.899 m/s}= 0.773$

Explanation:

For this case we have given the following data:

$T= 400 K$ represent the temperature for the air

$v = 310 m/s$ represent the velocity of the air

$k = 1.4$ represent the specific heat ratio at the room

$R = 0.287 KJ/ Kg K$ represent the gas constant for the air

And we want to find the velocity of the air under these conditions.

We can calculate the spped of the sound with the Newton-Laplace Equation given by this equation:

$\alpha = \sqrt{\frac{K}{\rho}}=\sqrt{k RT}$

Where K = is the Bulk Modulus of air, k is the adiabatic index of air= 1.4, R = the gas constant for the air, $\rho$ the density of the air and T the temperature in K

So on this case we can replace and we got:

$\alpha = \sqrt{1.4 *0.287 \frac{KJ}{Kg K}*\frac{1000J}{1KJ} *400 K}= 400.899 m/s$

The Mach number by definition is "a dimensionless quantity representing the ratio of flow velocity past a boundary to the local speed of sound" and is defined as:

$Ma=\frac{v}{\alpha}$

Where v is the flow velocity and $\alpha$ the volocity of the sound in the medium and if we replace we got:

$Ma= \frac{310 m/s}{400.899 m/s}= 0.773$

And since the Ma<0.8 we can classify the regime as subsonic.

7 0

2 years ago