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Art [367]
3 years ago
6

he Weather Channel reports that it is a hot, muggy day with an air temperature of 90????F, a 10 mph breeze out of the southwest,

and bright sunshine with a solar insolation of 400 W/m2. Consider the wall of a metal building over which the prevailing wind blows. The length of the wall in the wind direction is 10 m, and the emissivity is 0.93. Assume that all the solar irradiation is absorbed, that irradiation from the sky is negligible, and that flow is fully turbulent over the wall. Estimate the average wall temperature.
Engineering
1 answer:
Lilit [14]3 years ago
7 0

Answer:

29\°c

Explanation:

We need to define a couple of properties

T_f = 305KP= 1atm\\v=16.27*10^{-6}m^2/s\\k=0.02658W/mK\\Pr=0.707

For this problem I took Steady-state conditions with isothermal temperature in T_s

As well as a flow turbulent over the wall and negligible heat transfer into the building.

We start making a energy balance on the wall,

E'_{in}-E'_{out}=0

-q'_{cv}+(\alpha_S G_S-E_S)L=0

-\bar{h}_L L(T_s-T_{\infty})+(\alpha_S G_S - \epsilon \sigma T_s^4)L = 0

Again, assuming fully turbulent flow over the leng of the wall,

\bar{Nu}_L = \frac{\bar{h}_L}{k} = 0.037 Re_L^{4/4} Pr^{1/3}

Re_L = u_{\infty}\frac{L}{v} = 4.47*\frac{10}{16.27*10^{-6}}= 2.748*10^6

\bar{h}_L = (0.02658/10)0.037(2.748*10^6)^{4/5}(0.707)^{1/3}=12-4W/m^K

Substituting to fin T_s,

-12.4W/m^2*10m[T_s-(32.2+273)]K+[1*400W/m^2-0.93*5.67*10^{-8}W/m^2.K^4T_s^4]*10m=0

T_s=302.2K=29\°c

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Explanation:

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2 years ago
A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate
kherson [118]

Explanation:

thermal expansion ∝L = (δL/δT)÷L ----(1)

δL = L∝L + δT ----(2)

we have δL = 12.5x10⁻⁶

length l = 200mm

δT = 115°c - 15°c = 100°c

putting these values into equation 1, we have

δL = 200*12.5X10⁻⁶x100

= 0.25 MM

L₂ = L + δ L

= 200 + 0.25

L₂ = 200.25mm

12.5X10⁻⁶ *115-15 * 20

= 0.025

20 +0.025

D₂ = 20.025

as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0

3 0
3 years ago
Assume the following LTI system where the input signal is an impulse train (i.e.,x(t)=∑????(t−nT0)[infinity]n=−[infinity].a)Find
Igoryamba

Answer:

See explaination

Explanation:

The Fourier transform of y(t) = x(t - to) is Y(w) = e- jwto X(w) . Therefore the magnitude spectrum of y(t) is given by

|Y(w)| = |X(w)|

The phase spectrum of y(t) is given by

<Y(w) = -wto + <X(w)

please kindly see attachment for the step by step solution of the given problem.

4 0
3 years ago
If a vacuum gau ge reads 9.62 psi, it means that: a. the very highest column of mercury it could support would be 19.58 inches.
scZoUnD [109]

Answer:All of the above

Explanation:

9.62 psi means 497.49 mm of Hg pressure

for (a)19.58 inches is equals to 497.49 mm of Hg

(b)atmospheric pressure is 14.69 psi

vaccum gauge is 9.62psi

absolute pressure is=14.69-9.62=5.07

(c)vaccum means air is sucked and there is negative pressure so it tells about below atmospheric pressure.

thus all are correct

8 0
3 years ago
A 1000-turn coil of wire 1.0 cm in diameter is in a magnetic field that increases from 0.10 T to 0.30 T in 10 ms. The axis of th
ddd [48]

emf generated by the coil is 1.57 V

Explanation:

Given details-

Number of turns of wire- 1000 turns

The diameter of the wire coil- 1 cm

Magnetic field (Initial)= 0.10 T

Magnetic Field (Final)=0.30 T

Time=10 ms

The orientation of the axis of the coil= parallel to the field.

We know that EMF of the coil is mathematically represented as –

E=N(ΔФ/Δt)

Where E= emf generated

ΔФ= change inmagnetic flux

Δt= change in time

N= no of turns*area of the coil

Substituting the values of the above variables

=1000*3.14*0.5*10-4

=.0785

E=0.0785(.2/10*10-3)

=1.57 V

Thus, the emf generated is 1.57 V

4 0
3 years ago
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