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Aleonysh [2.5K]
3 years ago
6

Consider a fully developed laminar flow in a circular pipe. The velocity at R/2 (midway between the wall surface and the centerl

ine) is given by _____ provided that Vmax is the maximum velocity in the tube.
Engineering
1 answer:
ryzh [129]3 years ago
4 0

Answer:

The velocity at R/2 (midway between the wall surface and the centerline) is given by (3/4)(Vmax) provided that Vmax is the maximum velocity in the tube.

Explanation:

Starting from the shell momentum balance equation, it can be proved that the velocity profile for fully developedblaminar low in a circular pipe of internal radius R and a radial axis starting from the centre of the pipe at r=0 to r=R is given as

v = (ΔPR²/4μL) [1 - (r²/R²)]

where v = fluid velocity at any point in the radial direction

ΔP = Pressure drop across the pipe

μ = fluid viscosity

L = pipe length

But the maximum velocity of the fluid occurs at the middle of the pipe when r=0

Hence, maximum veloxity is

v(max) = (ΔPR²/4μL)

So, velocity at any point in the radial direction is

v = v(max) [1 - (r²/R²)]

At the point r = (R/2)

r² = (R²/4)

(r²/R²) = r² ÷ R² = (R²/4) ÷ (R²) = (1/4)

So,

1 - (r²/R²) = 1 - (1/4) = (3/4)

Hence, v at r = (R/2) is given as

v = v(max) × (3/4)

Hope this Helps!!!

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solmaris [256]

Answer:

0.064 mg/kg/day

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Explanation:

Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.

The BCF = 10³, so the concentration of the chemical in the fish is:

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x = 100 mg/kg

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0.030 kg × 100 mg/kg = 3 mg

The total daily intake is 3.2 mg.  Divided by the woman's mass of 50 kg, the dosage is:

(3.2 mg/day) / (50 kg) = 0.064 mg/kg/day

b) The percent from the water is:

0.2 mg / 3.2 mg = 6.25%

And the percent from the fish is:

3 mg / 3.2 mg = 93.75%

3 0
3 years ago
A fuel cell vehicle draws 50 kW of power at 70 mph and is 40% efficient at rated power. You are asked to size the fuel cell syst
svetoff [14.1K]

Answer:

  a) fuel cell weight and volume: 50 kg, 33.3 L

  b) fuel tank weight and volume: 321 kg, 643 L

Explanation:

<u>Fuel Cell</u>

Delivery of 50 kW from a source with a power density of 1.5 kW/L requires a source that has a volume of ...

  (50 kW)/(1.5 kW/L) = 33.3 L

The weight of the power source is 1 kW/kg, so will be ...

  (50 kW)/(1 kW/kg) = 50 kg

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<u>Fuel Tank</u>

400 miles at 70 mph will take (400/70) h ≈ 5.71429 h. In that time, the energy used by the vehicle power plant is ...

  (50 kW)(5.71429 h)(3600 s/h) = 1028.57 MJ

Since the power plant is 40% efficient, it must be supplied with 2.5 times that amount of energy, or 2571.43 MJ.

The tank volume and mass will then be ...

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  mass = 2571.43 MJ/(8 MJ/L) = 321.4 kg

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It appears the fuel tank would need to be equivalent in size to a sphere about 1.1 m in diameter.

4 0
3 years ago
A belt drive was designed to transmit the power of P=7.5 kW with the velocity v=10m/s. The tensile load of the tight side is twi
Leviafan [203]

Answer:

F₁ = 1500 N

F₂ = 750 N

F_{e} = 500 N

Explanation:

Given :

Power transmission, P = 7.5 kW

                                      = 7.5 x 1000 W

                                      = 7500 W

Belt velocity, V = 10 m/s

F₁ = 2 F₂

Now we know from power transmission equation

P = ( F₁ - F₂ ) x V

7500 = ( F₁ - F₂ ) x 10

750 =  F₁ - F₂

750 = 2 F₂ - F₂      ( ∵F₁ = 2 F₂ )

∴F₂  = 750 N

Now F₁ = 2 F₂

        F₁ = 2 x F₂

        F₁ = 2 x 750

        F₁ = 1500 N   ,   this is the maximum force.

Therefore we know,

F_{max} = 3 x F_{e}

where F_{e} is centrifugal force

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                          = 1500 / 3

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8 0
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Answer:

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V_o = 10 liters

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c1 = 1.03

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n =1.4

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V_o  = \frac{5}{[\frac{72}{80}]^{1/1.4} -[\frac{72}{180}]^{1/1.4}}

V_o = 12.26 liters

calculate pressure ratio to determine correction factor

\frac{p_2}{p_1} =\frac{180}{80} = 2.25

correction factor for calculate dpressure ration  for isothermal process is

c1 = 1.15

actual \volume = c1\times 10 = 11.5 liters

8 0
3 years ago
The AGC control voltage: ___________
lyudmila [28]

Answer:

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Explanation:

3 0
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