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Liono4ka [1.6K]
3 years ago
6

A house has a black tar, flat, horizontal roof. The lower surface of the roof is well insulated, while the upper surface is expo

sed to ambient air at 300K through a convective coefficient of 10 W/m2-K. Calculate the roof equilibrium temperature for a) a clear sunny day with an incident solar radiaton flux of 500 W/m2 and the ambient sky at an effective temperature of 50K and b) a clear night with an ambient sky temperature of 50K.
Engineering
1 answer:
myrzilka [38]3 years ago
6 0

Answer a) roof equilibrium temperature is 400K

Explanation:

The roof surface is sorrounded by ambient air whose temperature is 300K and which has a convective coefficient h of 10W/m2-K.

This means that heat will be conducted to the roof from the air around at an heat Flux of 300K x 10W/m2-K = 3000w/m2

For a clear solar day with solar heat Flux of 500W/m2, total heat Flux on roof will be

Q = 500 + 3500 = 3500W/m2

Q = h(Tr-Ta)

Ts is temperature of roof,

Ta is temperature of air

3500 = 10(Tr - 50)

350 = Tr - 50

Tr = 400k

Answer b): roof equilibrium temperature will be 350K

Explanation:

At night total heat Flux is only due to hot ambient air sorrounding the surface of the roof.

Q = 3000W/m2

Q = h(Tr-Ta)

3000 = 10(Tr-50)

300 = Tr - 50

Tr = 350K

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How many volts of electricity would it take to power up an entire city? Take Tokyo for example. Please explain!
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A body is moving with simple harmonic motion. It's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-positio
natima [27]

Answer:

1) A=282.6 mm

2)a_{max}=60.35\ m/s^2

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

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V=3.5 m/s at x=150 mm     ------------1

V=2.5 m/s at x=225 mm   ------------2

Where x measured  from mid position.

We know that velocity in simple harmonic given as

V=\omega \sqrt{A^2-x^2}

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

3.5=\omega \sqrt{A^2-0.15^2}    ------3

2.5=\omega \sqrt{A^2-0.225^2}   --------4

Now by dividing equation 3 by 4

\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}

1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}

So    A=0.2826 m

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3.5=\omega \sqrt{A^2-0.15^2}

3.5=\omega \sqrt{0.2826^2-0.15^2}

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

a_{max}=\omega ^2A

a_{max}=14.61 ^2\times 0.2826\ m/s^2

a_{max}=60.35\ m/s^2

Time period T

T=\dfrac{2\pi}{\omega}

T=\dfrac{2\pi}{14.609}

T=0.42 sec

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