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2 years ago

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Engineering

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romanna [79]2 years ago

8 0

Answer:

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g Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front end is exposed to ambient air th

xxTIMURxx [149]

Answer:

The electrical power is 96.5 W/m^2

Explanation:

The energy balance is:

Ein-Eout=0

$qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0$

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }$

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }$

if Gl≈El(l,5800)

$\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt= 2*5800=11600 um-K, at this value, F=0.941

$\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77$

The hemispherical emissivity is equal to:

$E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt=2*333=666 K, at this value, F=0

$E=0+(1-0.7)(1)=0.3$

The hemispherical absorptivity is equal to:

$qe=EoTs^{4}+h(Ts-T\alpha )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}$

3 0

3 years ago

An air compressor of mass 120 kg is mounted on an elastic foundation. It has been observed that, when a harmonic force of amplit

kupik [55]

Answer:

equivalent stiffness is 136906.78 N/m

damping constant is 718.96 N.s/m

Explanation:

given data

mass = 120 kg

amplitude = 120 N

frequency = 320 r/min

displacement = 5 mm

to find out

equivalent stiffness and damping

solution

we will apply here frequency formula that is

frequency ω = ω(n) √(1-∈ ²) ......................1

here ω(n) is natural frequency i.e = √(k/m)

so from equation 1

320×2π/60 = √(k/120) × √(1-2∈²)

k × ( 1 - 2∈²) = 33.51² ×120

k × ( 1 - 2∈²) = 134752.99 .....................2

and here amplitude ( max ) of displacement is express as

displacement = force / k × ( $\frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}}$ )

put here value

0.005 = 120/k × ( $\frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}}$ )

k ×∈ × √(1-2∈²) = 1200 ......................3

so by equation 3 and 2

$\frac{k\varepsilon \sqrt{1-\varepsilon^2})}{k(1-2\varepsilon^2)} = \frac{12000}{134752.99}$

$\varepsilon^{2} - \varepsilon^{4} = 7.929 * 10^{-3} - 0.01585 * \varepsilon^{2}$

solve it and we get

∈ = 1.00396

and

∈ = 0.08869

here small value we will consider so

by equation 2 we get

k × ( 1 - 2(0.08869)²) = 134752.99

k = 136906.78 N/m

so equivalent stiffness is 136906.78 N/m

and

damping is express as

damping = 2∈ √(mk)

put here all value

damping = 2(0.08869) √(120×136906.78)

so damping constant is 718.96 N.s/m

7 0

3 years ago

A gas stream contains 4.0 mol % NH3 and its ammonia content is reduced to 0.5 mol % in a packed absorption tower at 293 K and 10

bagirrra123 [75]

Answer:

Explanation:

Step by step solved solution is given in the attached document.

8 0

3 years ago

A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. Suppose the actual heat pump has a COP o

AysviL [449]

Answer:

W=2 MW

Explanation:

Given that

COP= 2.5

Heat extracted from 85°C

Qa= 5 MW

Lets heat supplied at 150°C = Qr

The power input to heat pump = W

From first law of thermodynamics

Qr= Qa+ W

We know that COP of heat pump given as

$COP=\dfrac{Qr}{W}$

$2.5=\dfrac{5}{W}$

W=2 MW

For Carnot heat pump

$COP=\dfrac{T_2}{T_2-T_1}$

$2.5=\dfrac{T_2}{T_2-(273+85)}$

2.5 T₂ - 895= T₂

T₂=596.66 K

T₂=323.6 °C

7 0

3 years ago

DRIVERS ED

forsale [732]

Answer:

Explanation:

only if there signal is turned on

8 0

3 years ago

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