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frosja888 [35]
3 years ago
7

Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series

and a pump between them. The first pipe is 20 m long and has a 6-cm diameter, while the second pipe is 36 m long and has a 3-cm diameter. The water level in the reservoir is 30 m above the centerline of the pipe. The pipe entrance is sharp-edged, and losses associated with the connection of the pump are negligible. Neglecting the effect of the kinetic energy correction factor, determine the required pumping head and the minimum pumping power to maintain the indicated flow rate. The density and dynamic viscosity of water at 15°C are rho = 999.1 kg/m3 and μ = 1.138 × 10–3 kg/m·s. The loss coefficient is KL = 0.5 for a sharp-edged entrance. The roughness of cast iron pipes is ε = 0.00026 m.
Engineering
1 answer:
love history [14]3 years ago
8 0

Answer:

The required pumping head is 1344.55 m and the pumping power is 236.96 kW

Explanation:

The energy equation is equal to:

\frac{P_{1} }{\gamma } +\frac{V_{1}^{2}  }{2g} +z_{1} =\frac{P_{2} }{\gamma } +\frac{V_{2}^{2}  }{2g} +z_{2}+h_{i} -h_{pump} , if V_{1} =0,z_{2} =0\\h_{pump} =\frac{V_{2}^{2}}{2} +h_{i}-z_{1}

For the pipe 1, the flow velocity is:

V_{1} =\frac{Q}{\frac{\pi D^{2} }{4} }

Q = 18 L/s = 0.018 m³/s

D = 6 cm = 0.06 m

V_{1} =\frac{0.018}{\frac{\pi *0.06^{2} }{4} } =6.366m/s

The Reynold´s number is:

Re=\frac{\rho *V*D}{u} =\frac{999.1*6.366*0.06}{1.138x10^{-3} } =335339.4

\frac{\epsilon }{D} =\frac{0.00026}{0.06} =0.0043

Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941

The head of pipe 1 is:

h_{1} =\frac{V_{1}^{2}  }{2g} (k_{L}+\frac{fL}{D}  )=\frac{6.366^{2} }{2*9.8} *(0.5+\frac{0.0294*20}{0.06} )=21.3m

For the pipe 2, the flow velocity is:

V_{2} =\frac{0.018}{\frac{\pi *0.03^{2} }{4} } =25.46m/s

The Reynold´s number is:

Re=\frac{\rho *V*D}{u} =\frac{999.1*25.46*0.03}{1.138x10^{-3} } =670573.4

\frac{\epsilon }{D} =\frac{0.00026}{0.03} =0.0087

The head of pipe 1 is:

h_{2} =\frac{V_{2}^{2}  }{2g} (k_{L}+\frac{fL}{D}  )=\frac{25.46^{2} }{2*9.8} *(0.5+\frac{0.033*36}{0.03} )=1326.18m

The total head is:

hi = 1326.18 + 21.3 = 1347.48 m

The required pump head is:

h_{pump} =\frac{25.46^{2} }{2*9.8} +1347.48-36=1344.55m

The required pumping power is:

P=Q\rho *g*h_{pump}  =0.018*999.1*9.8*1344.55=236965.16W=236.96kW

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         Weight of the Philae on Mars is = \frac{Weight Of Earth *Gravity Of Mars}{Gravity Of Earth}

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Using r0a=0.1m;

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Va=0+0.1ω0aj

Calculating te velocity of using te equation below

Vb=Va+Vba

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Va=0.1ω0aj

Vb=Va+ωabxrba

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3 years ago
(20pts) Air T[infinity] = 10 °C and u[infinity] = 100 m/s flows over a flat plate. Assume that the density of air is 1.0 kg/m3 a
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3 years 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,
tester [92]

Answer:

percentage change in volume is 2.60%

water level rise is 4.138 mm

Explanation:

given data

volume of water V = 500 L

temperature T1 = 20°C

temperature T2 = 80°C

vat diameter = 2 m

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percentage change in volume and how much water level rise

solution

we will apply here bulk modulus equation that is ratio of change in pressure   to rate of change of volume to change of pressure

and we know that is also in term of change in density also

so

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

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

here ρ is density

and we know ρ  for 20°C = 998 kg/m³

and ρ  for 80°C = 972 kg/m³

so from equation 2 put all value

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

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

dV = 0.0130 m³

so now  % change in volume will be

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so percentage change in volume is 2.60%

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now from equation 3 and 4 , subtract v1 by v2

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

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Answer:

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Explanation:

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m=0.17kg/s

6 0
2 years ago
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