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FinnZ [79.3K]
4 years ago
15

To cool a summer home without using a vapor compression refrigeration cycle, air is routed through a plastic pipe (k=0.15 W/m*K,

D_inner=0.15m, D_outer=0.17m) that is submerged in an adjoining body of water. the water temperature is nominally at T_infinity=17 degrees Celsius and has a convection coefficient of h=1500W/m^2*K is maintained at the outer surface of the pipe. If air from the home enters at T_m,i=29 degrees Celsius and a volumetric flow rate of V_i=0.025 m^3/s, what pipe length, L, is needed to provide a discharge temperature of T_m,o=21 degrees Celsius?

Engineering
1 answer:
Shalnov [3]4 years ago
8 0

Answer:

See attachment for complete solving

Explanation:

Given that: Brainly.com

What is your question?

mkasblog

College Engineering 5+3 pts

To cool a summer home without using a vapor compression refrigeration cycle, air is routed through a plastic pipe (k=0.15 W/m*K, D_inner=0.15m, D_outer=0.17m) that is submerged in an adjoining body of water. the water temperature is nominally at T_infinity=17 degrees Celsius and has a convection coefficient of h=1500W/m^2*K is maintained at the outer surface of the pipe. If air from the home enters at T_m,i=29 degrees Celsius and a volumetric flow rate of V_i=0.025 m^3/s, what pipe length, L, is needed to provide a discharge temperature of T_m,o=21 degrees Celsius

See attachment fir completed solving

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Answer:

<u>Option-(A)</u>

Explanation:

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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
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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

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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

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8 0
4 years ago
A heated long cylindrical rod is placed in a cross flow of air at 20°C (1 atm) with velocity of 10 m/s. The rod has a diameter o
postnew [5]

Answer:

Ts = 413.66 K

Explanation:

given data

temperature = 20°C

velocity = 10 m/s

diameter = 5 mm

surface emissivity = 0.95

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heat flux dissipated = 17000 W/m²

to find out

surface temperature

solution

we know that here properties of air at 70°C

k = 0.02881 W/m.K

v = 1.995 ×10^{-5} m²/s

Pr = 0.7177

we find here reynolds no for air flow that is

Re = \frac{\rho V D }{\mu } = \frac{VD}{v}    

Re = \frac{10*0.005}{1.99*10^{-5}}

Re = 2506

now we use churchill and bernstein relation for nusselt no

Nu = \frac{hD}{k} = 0.3 + \frac{0.62 Re6{0.5}Pr^{0.33}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\frac{2506}{282000})^{5/8}]^{4/5}

h = \frac{0.02881}{0.005}0.3 + \frac{0.62*2506{0.5}0.7177^{0.33}}{[1+(0.4/0.7177)^{2/3}]^{1/4}} [1+ (\frac{2506}{282000})^{5/8}]^{4/5}

h = 148.3 W/m².K

so

q conv = h∈(Ts- T∞ )

17000 = 148.3 ( 0.95) ( Ts - (20 + 273 ))

Ts = 413.66 K

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4 years ago
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Answer: i got you its d

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3 years ago
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Answer:

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Explanation:

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We'll need to check the diameters under stress and strain.

Now, we know that the formula for stress is;

Stress = Force/Area

Thus,

Area = Force/stress

So for this stress, area required is;

A_req = 24000/60 = 4000 in²

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Area = πd²/4

So, 4000 = πd²/4

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Making A the subject to obtain;

A = 24000/(120 x 0.00278)

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Area = πd²/4

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Required diameter here is;

d = √91599.4

d = 302.65 in

The larger diameter is 302.65 inchesand it's therefore the required one.

3 0
4 years ago
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