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3 years ago

Is cross flow more efficient or counter flow

Engineering

1 answer:

pochemuha3 years ago

7 0

Answer:

Cross flow

Explanation:

Preferred for several reasons that include overall efficiency and serviceability.

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A damped harmonic oscillator consists of a mass on a spring, with a damping force proportional to the speed of the block. If the

Murrr4er [49]

Answer:

TOTAL ENERGY = 0.74 j

Explanation:

Given data:

spring constant is 350 N/m

m = 0.24 kg

b = 0.41 kg/s

A = 0.075 M

$\omega = \sqrt{\frac{k}{m}}$

$= \sqrt{\frac{350}{0.24}}$

$y = e^{\frac{-b}{2m} t} A cos(\omega t)$

$=e^{\frac{-0.41}{2*0.24} t} cos (\sqrt{\frac{350}{0.24}} t) *0.075$

after one full cycle, mass will be at extreme point, hence K,E = 0

But total energy remain same

y after one full cycle is\

$y =e^{\frac{-0.41}{2*0.24} 2\pi \sqrt{\frac{0.24}{350}}} cos (2\pi*0.075)$

y = 0.06517 m

y = 6.517 cm

total energy $= \frac{1}{2} k y^2$

$= \frac{1}{2}* 350* 0.06517^2 = 0.742 J$

8 0

4 years ago

Compare the output of full-wave rectifier with and without filter

lara31 [8.8K]

Answer:

Full wave rectification flips the negative half cycle of the sine wave to positive so the result is two positive half cycles.

Explanation:

hope it helps a lil

4 0

2 years ago

A rigid insulated tank is divided into 2 equal compartments by a thin rigid partition. One of the compartments contains air, ass

Illusion [34]

Https://www.slader.com/discussion/question/an-insulated-rigid-tank-is-divided-into-two-equal-parts-by-a-partition-initially-one-part-contains-4/

there will be the answer

6 0

3 years ago

As a general rule of thumb, in-line engines are easier to work on than the other cylinder arrangements.

siniylev [52]

Answer:

The general rule of thumb is that the SMALLER a substance's atoms and the STRONGER the bonds, the harder the substance. Two of the strongest forms of chemical bonds are the ionic and covalent bonds.

Explanation:

5 0

3 years ago

Air at 1 atm enters a thin-walled ( 5-mm diameter) long tube ( 2 m) at an inlet temperature of 100°C. A constant heat flux is ap

alexdok [17]

Answer:

heat rate = 7.38 W

Explanation:

Given Data:

Pressure = 1atm

diameter (D) = 5mm = 0.005m

length = 2

mass flow rate (m) = 140*10^-6 kg/s

Exit temperature = 160°C,

At 400K,

Dynamic viscosity (μ) = 22.87 *10^-6

Prandtl number (pr) = 0.688

Thermal conductivity (k) = 33.65 *10^-3 W/m-k

Specific heat (Cp) = 1.013kj/kg.K

Step 1: Calculating Reynolds number using the formula;

Re = 4m/πDμ

= (4*140*10^-6)/(π* 0.005*22.87 *10^-6)

= 5.6*10^-4/3.59*10^-7

= 1559.

Step 2: Calculating the thermal entry length using the formula

Le = 0.05*Re*Pr*D

Substituting, we have

Le = 0.05 * 1559 * 0.688 *0.005

Le = 0.268

Step 3: Calculate the heat transfer coefficient using the formula;

Nu = hD/k

h = Nu*k/D

Since Le is less than given length, Nusselt number (Nu) for fully developed flow and uniform surface heat flux is 4.36.

h = 4.36 * 33.65 *10^-3/0.005

h = 0.1467/0.005

h = 29.34 W/m²-k

Step 4: Calculating the surface area using the formula;

A = πDl

=π * 0.005 * 2

=0.0314 m²

Step 5: Calculating the temperature Tm

For energy balance,

Qc = Qh

Therefore,

H*A(Te-Tm) = MCp(Tm - Ti)

29.34* 0.0314(160-Tm) = 140 × 10-6* 1.013*10^3 (Tm-100)

0.921(160-Tm) = 0.14182(Tm-100)

147.36 -0.921Tm = 0.14182Tm - 14.182

1.06282Tm = 161.542

Tm = 161.542/1.06282

Tm = 151.99 K

Step 6: Calculate the rate of heat transferred using the formula

Q = H*A(Te-Tm)

= 29.34* 0.0314(160-151.99)

= 7.38 W

the Prandtl number using the formula

5 0

3 years ago