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Andreyy89
2 months ago
13

A rectangular concrete (n=0.013) channel 20 feet wide, on a 2.5% slope, is discharging 400 ft3 /sec into a stilling basin. The s

tilling basin is also 20 feet wide and has a water depth of 8 ft determined from the downstream channel condition. What is the length of the stilling basin? What is the height of the endsill?
Engineering
1 answer:
grin007 [323]2 months ago
8 0

Response:

Length of stilling basin = 32.9 feet

Height of end sill = 6.58 feet

Clarification:

Discharge = Q = 400 ft^3 /sec

Slope = 2.5 ft

Width = 20 feet

n = 0.013

Assuming depth of flow as "d"

Q = 1/n (R)^2/3 (slope)^1/2 A   ( R represents the hydraulic Radius)

By inserting the provided data into the above equation:

400 = 1/0.013 * (R)^2/3 * sqrt (2.5/100) * 20d

R = A/P

In this case, A stands for the flow area and P indicates the wetted perimeter

400*0.013 = (20d/(20+20d))^2/3 * sqrt(2.5/100) * 20d

d = 1.42 feet

Depth of stilling channel before the jump = d1 = 8 feet

Depth of stilling channel after the jump = d2 = 1.42 feet

Length of stilling basin = 5(d2 - d1)

= 5( 8 - 1.42)

Length of stilling basin = 32.9 feet

Now calculating the height of end sill:

Jump height = (8 - 1.42)

Height of end sill = 6.58 feet

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Consider a rectangular fin that is used to cool a motorcycle engine. The fin is 0.15m long and at a temperature of 250C, while t
Kisachek [356]

Answer:

q' = 5826 W/m

Explanation:

Given:-

- The length of the fin in question, L = 0.15 m

- The fin's surface temperature, Ts = 250°C

- The velocity of free stream air, U = 80 km/h

- The air temperature, Ta = 27°C

- The flow is parallel over both sides of the fin, assuming turbulent flow conditions throughout.

Find:-

What is the heat removal rate per unit width of the fin?

Solution:-

- Steady state conditions are assumed, along with negligible radiation and turbulent flow conditions.

- From Table A-4, we gather air properties (T = 412 K, P = 1 atm ):

    Dynamic viscosity, v = 27.85 * 10^-6 m²/s  

    Thermal conductivity, k = 0.0346 W / m.K

    Prandtl number Pr = 0.69

- Compute the Nusselt Number (Nu) corresponding to - turbulent conditions - using the relevant relationship as follows:

                          Nu = 0.037*Re_L^\frac{4}{5} * Pr^\frac{1}{3}

Where,    Re_L: Average Reynolds number for the entire length of fin:

                          Re_L = \frac{U*L}{v} \\\\Re_L = \frac{80*\frac{1000}{3600} * 0.15}{27.85*10^-^6} \\\\Re_L = 119688.80909

Consequently,

                         

Nu = 0.037*(119688.80909)^\frac{4}{5} * 0.69^\frac{1}{3}\\\\Nu = 378

- The convective heat transfer coefficient (h) can now be derived from:

                          h = \frac{k*Nu}{L} \\\\h = \frac{0.0346*378}{0.15} \\\\h = 87 \frac{W}{m^2K}

- The heat loss rate q' per unit width can be established using the convection heat transfer formula and should be multiplied by (x2) since the airflow is present on both sides of the fin:

                          q' = 2*[h*L*(T_s - T_a)]\\\\q' = 2*[87*0.15*(250 - 27)]\\\\q' = 5826\frac{W}{m}

- Ultimately, the heat loss per unit width from the rectangular fin is q' = 5826 W/m

- The thermal loss per unit width (q') attributed to radiation:

                  q' = 2*a*T_s^4*L

Where, a signifies the Stefan-Boltzmann constant = 5.67*10^-8

                  q' = 2*5.67*10^-^8*(523)^4*0.15\\\\q' = 1273 \frac{W}{m}

- It is observed that radiation losses are not insignificant, accounting for 20% of thermal loss by convection. As the emissivity (e) of the fin is unspecified, this value is dismissed from the calculations as it pertains to the provided information.

7 0
2 months ago
PDAs with two stacks are strictly more powerful than PDAs with one stack. Prove that 2-stack PDAs are not a valid model for CFLs
iogann1982 [368]

Answer:

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Explanation:

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6 0
2 months ago
Design a decimal arithmetic unit with two selection variables, V1, and Vo, and two BCD digits, A and B. The unit should have fou
choli [298]
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4 0
2 months ago
You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an
pantera1 [306]

Answer:

t'_{1\2} = 6.6 sec

Explanation:

The half-life for the specified RC circuit can be expressed as

t_{1\2} =\tau ln2

where [/tex]\tau = RC[/tex]

t_{1\2} = RCln2

Given t_{1\2} = 3 sec

The circuit has a resistance of 40 ohms, and by adding a new resistor of 48 ohms, the total resistance becomes 40 + 48 = 88 ohms.

Thus, the new half-life is

t'_{1\2} =R'Cln2

Now, divide equation 2 by 1

\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}

t'_{1\2} = t'_{1\2}\frac{R'}{R}

After substituting all values, we can calculate the revised half-life

t'_{1\2} = 3 * \frac{88}{40} = 6.6 sec

t'_{1\2} = 6.6 sec

7 0
2 months ago
An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,
Mrrafil [318]

Answer:

Volume change percentage is 2.60%

Water level increase is 4.138 mm

Explanation:

Provided data

Water volume V = 500 L

Initial temperature T1 = 20°C

Final temperature T2 = 80°C

Diameter of the vat = 2 m

Objective

We aim to determine percentage change in volume and the rise in water level.

Solution

We will apply the bulk modulus equation, which relates the change in pressure to the change in volume.

It can similarly relate to density changes.

Thus,

E = -\frac{dp}{dV/V}................1

And -\frac{dV}{V} = \frac{d\rho}{\rho}............2

Here, ρ denotes density. The density at 20°C = 998 kg/m³.

The density at 80°C = 972 kg/m³.

Plugging in these values into equation 2 gives

-\frac{dV}{V} = \frac{d\rho}{\rho}

-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}

dV = 0.0130 m³

Therefore, the percentage change in volume will be

dV % = -\frac{dV}{V}  × 100

dV % = -\frac{0.0130}{500*10^{-3} }  × 100

dV % = 2.60 %

Hence, the percentage change in volume is 2.60%

Initial volume v1 = \frac{\pi }{4} *d^2*l(i)................3

Final volume v2 = \frac{\pi }{4} *d^2*l(f)................4

From equations 3 and 4, subtract v1 from v2.

v2 - v1 =  \frac{\pi }{4} *d^2*(l(f)-l(i))

dV = \frac{\pi }{4} *d^2*dl

Substituting all values yields

0.0130 = \frac{\pi }{4} *2^2*dl

Thus, dl = 0.004138 m.

Consequently, the water level rises by 4.138 mm.

8 0
2 months ago
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