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anastassius [24]
3 years 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:
Sonbull [250]3 years ago
8 0

Answer:

Length of stilling basin = 32.9 feet

Height of end sill = 6.58 feet

Explanation:

Discharge = Q = 400 ft^3 /sec

Slope = 2.5 ft

Width = 20 feet

n = 0.013

we will assume the depth of flow as "d"

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

by substituting the given data in above formula we get:

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

R = A/P

here, A is the flow area and P is the wetted perimeter

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

<u>d = 1.42 feet</u>

<u></u>

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

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

Length of stilling basin = 5(d2 - d1)

                                      = 5( 8 - 1.42)

<u>Length of stilling basin = 32.9 feet </u>

Now calculating the height of end sill:

Jump height = (8 - 1.42)

<u>Height of end sill = 6.58 feet</u>

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Answer:

All will be Explained below.

Explanation:

1) Under which Condition can a soil be chemically Stabilize.

Answer

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3 years ago
A well insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30 C. A valve is opened, allowing the argon to escape unti
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Answer:

Final mass of Argon=  2.46 kg

Explanation:

Initial mass of Argon gas ( M1 ) = 4 kg

P1 = 450 kPa

T1 = 30°C = 303 K

P2 = 200 kPa

k ( specific heat ratio of Argon ) = 1.667

assuming a reversible adiabatic process

<u>Calculate the value of the M2 </u>

Applying ideal gas equation ( PV = mRT )

P₁V / P₂V = m₁ RT₁ / m₂ RT₂

hence : m2 = P₂T₁ / P₁T₂ * m₁

                   = (200 * 303 ) / (450 * 219 ) * 4

                   = 2.46 kg

<em>Note: Calculation for T2 is attached below</em>

5 0
3 years ago
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7 0
2 years ago
A person walks into a refrigerated warehouse with head uncovered. Model the head as a 25- cm diameter sphere at 35°C with a surf
galina1969 [7]

Answer:

Hello some parts of your question is missing below is the missing part

Convection coefficient = 11 w/m^2. °c

answer : 44.83 watts

Explanation:

Given data :

surface emissivity ( ε )= 0.95

head ( sphere) diameter( D )  = 0.25 m

Temperature of sphere( T )  = 35° C

Temperature of surrounding ( T∞ )  = 25°C

Temperature of surrounding surface ( Ts ) = 15°C

б  = ( 5.67 * 10^-8 )

Determine the total rate of heat loss

First we calculate the surface area of the sphere

As = \pi D^{2}  

= \pi * 0.25^2 =  0.2 m^2

next we calculate heat loss due to radiation

Qrad = ε * б * As( T^{4} - T^{4} _{s} )  ---- ( 1 )

where ;

ε = 0.95

б = ( 5.67 * 10^-8 )

As = 0.2 m^2

T = 35 + 273 = 308 k

Ts = 15 + 273 = 288 k

input values into equation 1

Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )

         = 22.83  watts

Qrad ( heat loss due to radiation ) = 22.83 watts

calculate the heat loss due to convection

Qconv = h* As ( ΔT )

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Hence total rate of heat loss

=  22 + 22.83

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5 0
3 years ago
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almond37 [142]

Answer:

a) The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump is 245.175 kilopascals.

Explanation:

a) Let suppose that pump works at steady state. The mass flow rate of the water (\dot m), in kilograms per second, is determined by following formula:

\dot m = \frac{\eta \cdot \dot W}{g\cdot H} (1)

Where:

\dot W - Pump power, in watts.

\eta - Efficiency, no unit.

g - Gravitational acceleration, in meters per square second.

H - Hydrostatic column, in meters.

If we know that \eta = 0.72, \dot W = 5000\,W, g = 9.807\,\frac{m}{s^{2}} and H = 25\,m, then the mass flow rate of water is:

\dot m = 14.683\,\frac{kg}{s}

The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump (\Delta P), in pascals, is determined by this equation:

\Delta P = \rho\cdot g\cdot H (2)

Where \rho is the density of water, in kilograms per cubic meter.

If we know that \rho = 1000\,\frac{kg}{m^{3}}, g = 9.807\,\frac{m}{s^{2}} and H = 25\,m, then the pressure difference is:

\Delta P = 245175\,Pa

The pressure difference across the pump is 245.175 kilopascals.

4 0
3 years ago
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