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Temka [501]
3 years ago
13

Gshyevdnmdcicm gngjc

Engineering
1 answer:
Len [333]3 years ago
4 0

This is not a valid question. Please try again.

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Consider a fully developed laminar flow in a circular pipe. The velocity at R/2 (midway between the wall surface and the centerl
ryzh [129]

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!!!

4 0
3 years ago
Water of dynamic viscosity 1.12E-3 N*s/m2 flows in a pipe of 30 mm diameter. Calculate the largest flowrate for which laminar fl
Naya [18.7K]

Answer:

For water

Flow rate= 0.79128*10^-3 Ns

For Air

Flow rate =1.2717*10^-3 Ns

Explanation:

For the flow rate of water in pipe.

Area of the pipe= πd²/4

Diameter = 30/1000

Diameter= 0.03 m

Area= 3.14*(0.03)²/4

Area= 7.065*10^-4

Flow rate = 7.065*10^-4*1.12E-3

Flow rate= 0.79128*10^-3 Ns

For the flow rate of air in pipe.

Flow rate = 7.065*10^-4*1.8E-5

Flow rate =1.2717*10^-3 Ns

7 0
4 years ago
A proposed piping and pumping system has 20-psig static pressure, and the piping discharges to atmosphere 160 ft above the pump.
larisa86 [58]

Answer: (B) 100

Explanation:

Given that;

Pstatic = 20 psig , hz = 160ft, hf = 20ft

Now total head will be;

T.h = hz + hf

T.h= 160 + 20

T.h = 180ft

Minimum pressure = Psatic + egh

we know that specific weight of water is 62.4 (lb/ft3)

so

P.min = (20 bf/in² ) + (62.4 b/ft³ × 180 fr

P.min = (20 bf/in² ) + ( 62.4 × 180 × 1 ft²/144 in²)

P.min = 20 + 78

P.min = 98 lbf/in²

Therefore the minimum pressure rating (psi) of the piping system is most nearly B) 100

7 0
3 years ago
The formation of faults in Earth's crust is an effect. What causes faults to form in the crust? Global Positioning System sensor
allsm [11]

Answer: The movement of tectonic plates

Explanation:

Tectonic plates are the part of the earth's crust that both the ocean and land rest on. These plates are constantly moving as a result of currents in the mantle.

These movements cause stress on the surface which has the effect of fracturing rocks and thereby creating/ forming faults in the earth's crust. Sometimes faults form when these plates move away from each other and sometimes they are formed when they push into each other.

7 0
3 years ago
A rectangular open box, 25 ft by 10 ft in plan and 12 ft deep weighs 40 tons. Sufficient amount of stones is placed in the box a
dimulka [17.4K]

Answer:

44.95 tonnes

Explanation:

According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces

It is given that empty weight of box = 40 tons

Let the mass of the stones to be placed be = M tonnes

Thus the combined mass of box and stones = (40+M) tonnes..........(i)

Since the box will displace water equal to it's volume V we have volume of box = 25ft*10ft*12ft= 3000ft^{3}

Volume= 84.95m^{3}

Since 1ft^{3} =0.028m^{3}

Now the weight of water displaced = Weight =\rho \times Volumewhererho is density of water = 1000kg/m^{3}

Thus weight of liquid displaced = \frac{84.95X1000}{1000}tonnes=84.95 tonnes..................(ii)

Equating i and ii we get

40 + M = 84.95

thus Mass of stones = 44.95 tonnes

3 0
3 years ago
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