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Dahasolnce [82]
3 years ago
7

A thin-walled tube with a diameter of 6 mm and length of 20 m is used to carry exhaust gas from a smoke stack to the laboratory

in a nearby building for analysis. The gas enters the tube at 200°C and with a mass flow rate of 0.001 kg/s. Autumn winds at a temperature of 15°C blow directly across the tube at a velocity of 5 m/s. Assume the thermophysical properties of the exhaust gas are those of air. (a) Estimate the average heat transfer coefficient for the exhaust gas flowing inside the tube. (b) Estimate the heat transfer coefficient for the air flowing across the outside of the tube. (c) Estimate the overall heat transfer coefficient U and the temperature of the exhaust gas when it reaches the laboratory.
Engineering
1 answer:
Molodets [167]3 years ago
4 0

Answer:

Explanation:

Mean temperature is given by

T_mean = \frac{T_i + T_ \infinity}{2}\\\\T_mean = \frac{200 + 15}{2}

Tmean = (Ti + T∞)/2

T_mean = 107.5^{0}

Tmean = 107.5⁰C

Tmean = 107.5 + 273 = 380.5K

Properties of air at mean temperature

v = 24.2689 × 10⁻⁶m²/s

α = 35.024 × 10⁻⁶m²/s

\mu = 221.6 × 10⁻⁷N.s/m²

\kappa = 0.0323 W/m.K

Cp = 1012 J/kg.K

Pr = v/α  = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶

              = 0.693

Reynold's number, Re

Pv = 4m/πD²

where Re = (Pv * D)/\mu

Substituting for Pv

Re = 4m/(πD\mu)

     = (4 x 0.003)/( π × 6 ×10⁻³ × 221.6 × 10⁻⁷)

     = 28728.3

Since Re > 2000, the flow is turbulent

For turbulent flows, Use

Dittus - Doeltr correlation with n = 0.03

Nu = 0.023Re⁰⁸Pr⁰³ = (h₁D)/k

(h₁ × 0.006)/0.0323 = 0.023(28728.3)⁰⁸(0.693)⁰³

(h₁ × 0.006)/0.0323 = 75.962

h₁ = (75.962 × 0.0323)/0.006

h₁ = 408.93 W/m².K

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Technician B only is correct

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Explanation:

7 0
3 years ago
3. In the text, we described a multithreaded file server, showing why it is better than a single-threaded server and a finite-st
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Answer and Explanation:

• 1 thread awaits the incoming request

• 1 thread responds to the request

• 1 thread reads the hard disk

A multithreaded file server is better than a single-threaded server and a finite-state machine server because it provides better response compared to the rest and can make use of the shared Web data.

Yes, there are circumstances in which a single-threaded server might be better. If it is designed such that:

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4 0
3 years ago
Express (118)10 and (-49)10 in 8-bit binary one’s complement form and then add the numbers. What would be the representation (-0
stealth61 [152]

Answer:

118_{10}= 0110111 2) -49_{10}=110001_{2} 3) 0_{10}=0:16 \Rightarrow 0_{10}=0_{16}

Explanation:

1) Expressing the Division as the summation of the quotient and the remainder

for

118, knowing it is originally a decimal form:

118:2=59  +(0), 59/2 =29 + 1, 29/2=14+1, 14/2=7+0, 7/2=3+1, 3/2=1+1, 1/2=0+1

118_{10}= 0110111

2) -49_{10}

Similarly, we'll start the process with the absolute value of -49 since we want the positive value of it. Then let's start the successive divisions till zero.

|-49|=49

49:2=24+1, 24:2=12+0,12:2=6+0,6:2=3+0,3:2=1+1,1:2=0+1

100011

-49_{10}=110001_{2}

3) (-0)_{10}

The first step on that is dividing by 16, and then dividing their quotient again by 16, so on and adding their remainders. Simply put:

(-0)_{10}=0:16=0 \Rightarrow (0)_{10}=0_{16} \:or\\(0)_{16}=0000000000000000

5 0
3 years ago
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