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olga55 [171]
3 years ago
7

Define the hydraulic diameter for a rectangular duct

Engineering
1 answer:
8090 [49]3 years ago
4 0

Answer with Explanation:

Hydraulic diameter is a term analogous to the diameter of the circular sectional pipe but used for the cases when the cross sectional shape of the pipe is non circular.

It serves as an equivalent diameter that is used to calculate the Reynolds number for the flow.

The hydraulic diameter is 4 times the hydraulic radius of any section.

For a rectangular duct as shown in the attached figure

R_{h}=\frac{Wetted_{Area}}{Wetted_{perimeter}}\\\\R_h=\frac{d\times b}{2(d+b)}\\\\\therefore D_{h}=4\times R_{h}=4\times \frac{db}{2(d+b)}=\frac{2db}{(d+b)}

Where

D_{h} is the hydraulic diameter of the duct with depth 'd' and width 'b'

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Answer:

a. technician a only

Explanation:

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3 years ago
Consider a very long, cylindrical fin. The temperature of the fin at the tip and base are 25 °C and 50 °C, respectively. The dia
nekit [7.7K]

Answer:

98°C

Explanation:

Total surface area of cylindrical fin = πr² + 2πrl , r = 0.015m; l= 0.1m; π =22/7

22/7*(0.015)² + 22/7*0.015*0.1 = 7.07 X 10∧-4 + 47.1 X 10∧-4 = (54.17 X 10∧-4)m²

Temperature change, t = (50 - 25)°C = 25°C = 298K

Hence, Temperature =  150 X (54.17 X 10∧-4) X 298/123 = 242.14/124 = 2.00K =

∴ Temperature change = 2.00K

But temperature, T= (373 - 2)K = 371 K

In °C = (371 - 273)K = 98°C         

7 0
3 years ago
In DC electrode positive, how much power is at the work clamp?
Korolek [52]

Answer:

1/3 power

Explanation:

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7 0
3 years ago
QUESTION 3
lianna [129]
D D D D D D D D D D D D D D D DdDdddddf
6 0
3 years ago
A westbound section of freeway currently has three 12-ft wide lanes, a 6-ft right shoulder, and no ramps within 3 miles upstream
Tresset [83]

Answer:

•Estimated density = 39.685Pc/mi/en

•Level of service, LOS frequency = LOSC

Explanation:

We are given:

•Freeway current lane width,B = 12ft

• freeway current shoulder width,b = 6ft

• percentage of heavy vehicle, Ptb = 10℅

• peak hour factor, PHF = 0.9

Let's consider,

•Number of lanes N = 4

• flow of traffic V = 7500vph

• percentage of Rv = 0, therefore the freeflow speed in freeway FFS = 70mph

• cars equivalent for recreational purpose Er= 2

•cars to be used for trucks and busses Etb= 2.5

Let's first calculate for the heavy adjustment factor.

We have:

F_H_v = \frac{1}{1+P_t_b(E_t_b-1)+Pr(Er-1)}

Substituting figures in the equation we have:

= \frac{1}{1+0.1(2.5-1)+0(2-1)}

= 0.75

Let's now calculate equivalent flow rate of the car using:

Vp = \frac{V}{(P_H_F)*N)*(F_H_v)*(F_p)}

= \frac{7500}{0.9*4*0.75*1}

= 2777.7 pc/h/en

Calculating for traffic density, we have:

D = \frac{Vp}{FFS}

D = \frac{2777.7}{70}

D = 39.685 Pc/mi/en

Using the table for LOS criteria of basic frequency segment, the level of service LOS of frequency is LOSC

4 0
3 years ago
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