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Bad White [126]
3 years ago
12

1. A piston having a diameter of 5.48 inches and a length of 9.50 in slides downward with a velocity, V, through a vertical pipe

. The downward motion is resisted by an oil film between the piston and the pipe wall. The film thickness is 0.002 inches and the cylinder weighs 0.5 lb. Compute the velocity, V if the oil viscosity is 0.016 lb*s/ft
2. Assume the velocity distribution in the gap is linear. (answer: 0.0046 ft/sec) I need to understand this for my quiz so please work steps clearly. Will Rate!!!"
Engineering
1 answer:
const2013 [10]3 years ago
3 0

Answer:

V = 0.00459 ft/s  

Explanation:

Since the Piston is moving downwards with a constant velocity V, from the first Newton’s law we know that all vertical forces, must have zero resultant (their sum over vertical axis must equal to zero). Therefore, force that pulls the piston down, is equalized by force of viscous friction Fd= Fvf = 0.5lb (lb here is the pound-force unit). We will relate F ѵ f  with τ and from that derive the equation for V.

Fѵf = τ  . A

Where τ  = µ. du/dy = µ . V/b  , and A = π . D . l from this Follows:

Fѵf= (V.  A .µ )/b     V= ( Fѵf .b )/(A.µ)    

Placing all the known values in the equation ( remember  to transform inches to feet, by multiplying inches values with the factor 1/12), we obtain :  

ft2

V = ((0.5lb)   .   (0.002/12 ft))/(π   .   (5.48/12 ft)  .  (9.50/12 ft)  .  (0.016 lb.s/(ft^2 )))

V = 0.00459 ft/s  

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The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of
bixtya [17]

Answer:

h = 287.1 m

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the density of mercury \rho =13570 kg/m3

the atmospheric pressure at the top of the building is

p_t = \rho gh  = 13570*908*0.73 = 97.08 kPa

the atmospheric pressure at bottom

p_b = \rho gh  = 13570*908*0.75 = 100.4 kPa

\frac{w_{air}}{A} =p_b -p_t

we have also

(\rho gh)_{air} = p_b - p_t

1.18*9.81*h = (100.4 -97.08)*10^3

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7 0
3 years ago
Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging wat
zlopas [31]

Answer: 10.631\times 10^3\ N/m^2

Explanation:

Given

Discharge is Q=12.5\ L

Diameter of pipe d=150\ mm

Distance between two ends of pipe L=800\ m

friction factor f=0.008

Average velocity is given by

\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s

Pressure difference is given by

\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow  \Delta P=10.631\ kPa

8 0
2 years ago
A pipeline (NPS = 14 in; schedule = 80) has a length of 200 m. Water (15℃) is flowing at 0.16 m3/s. What is the pipe head loss f
dangina [55]

Answer:

Head loss is 1.64

Explanation:

Given data:

Length (L) = 200 m

Discharge (Q) = 0.16 m3/s

According to table of nominal pipe size , for schedule 80 , NPS 14,  pipe has diameter (D)= 12.5 in or 31.8 cm 0.318 m

We know, head\ loss  = \frac{f L V^2}{( 2 g D)}

where, f = Darcy friction factor

V = flow velocity

g = acceleration due to gravity

We know, flow rate Q = A x V

solving for V

V = \frac{Q}{A}

    = \frac{0.16}{\frac{\pi}{4} (0.318)^2} = 2.015 m/s

obtained Darcy friction factor  

calculate Reynold number (Re) ,

Re = \frac{\rho V D}{\mu}

where,\rho = density of water

\mu = Dynamic viscosity of water at 15 degree  C = 0.001 Ns/m2

so reynold number is

Re = \frac{1000\times 2.015\times 0.318}{0.001}

            = 6.4 x 10^5

For Schedule 80 PVC pipes , roughness (e) is  0.0015 mm

Relative roughness (e/D) = 0.0015 / 318 = 0.00005

from Moody diagram, for Re = 640000 and e/D = 0.00005 , Darcy friction factor , f = 0.0126

Therefore head loss is

HL = \frac{0.0126 (200)(2.015)^2}{( 2 \times 9.81 \times 0.318)}

HL = 1.64 m

7 0
3 years ago
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