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DIA [1.3K]
3 years ago
8

A centrifugal pump is required to pump water to an open water link situated 4 km away from the location of the pump through a pi

pe of diameter 0.2 m having Darcy's friction factor of 0.01. The average speed of water in the pipe is 2 m/s. If it maintains a constant head of 5 m in the tank, what is the corresponding pressure in bar abs.
Engineering
1 answer:
11111nata11111 [884]3 years ago
5 0

Answer:

P= 5.5 bar

Explanation:

Given that

L= 4000 m

d= 0.2 m

Friction factor(F) = 0.01

speed V= 2 m/s

Head = 5 m

Head loss due to friction

h_f=\dfrac{FLV^2}{2gd}

h_f=\dfrac{0.01\times 4000\times 2^2}{2\times 9.81\times 0.2}

h_f=40.77m

So the total head(H) = 5 + 40.77 + 10.3 =56.07

Where 10.3 m is the atmospheric head.

We know that

P=ρ g H

So total Pressure

P= 1000 x 9.81 x 56.07 Pa

P=5.5\times 10^5\ Pa

P= 5.5 bar

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The components of an electronic system dissipating 180 W are located in a 1-m-long horizontal duct whose cross section is 16 cm
oee [108]

Answer:

a) The exit temperature is 39.25°C

b) The highest component surface is 132.22°C

c) The average temperature for air equal to 35°C is a good assumption because the air temperature at the inlet will increase due to the result in the heat gain produced by the duct and whose surface is exposed to a flow of hot.

Explanation:

a) The properties of the air at 35°C:

p = density = 1.145 kg/m³

v = 1.655x10⁻⁵m²/s

k = 0.02625 W/m°C

Pr = 0.7268

cp = 1007 J/kg°C

a) The mass flow rate of air is equal to:

m=\rho *V = 1.145*0.65=0.7443kg/min=0.0124kg/s

The exit temperature is:

T=T_{i} +\frac{Q}{m*c_{p} } =27+\frac{0.85*180}{0.0124*1007} =39.25°C

b) The mean fluid velocity is:

V_{m} =\frac{V}{A} =\frac{0.65}{0.16*0.16} =25.4m/min=0.4232m/s

The hydraulic diameter is:

D_{h} =\frac{4A}{p} =\frac{4*0.16*0.16}{4*0.16} =0.16m

The Reynold´s number is:

Re=\frac{VD_{h} }{v} =\frac{0.4232*0.16}{1.655x10^{-5} } =4091.36

Assuming fully developed turbulent flow, the Nusselt number is:

Nu=0.023Re^{0.8} *Pr^{0.4} =0.023*4091.36^{0.8} *0.7268^{0.4} =15.69

h=\frac{k*Nu}{D_{h} } =\frac{0.02625*15.69}{0.16} =2.57W/m^{2} C

The highest component surface temperature is:

T=T_{e} +\frac{\frac{Q}{A} }{h} =39.2+\frac{0.85*\frac{180}{4*0.16*1} }{2.57} =132.22°C

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Answer:

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2 years ago
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Answer =

dial bore gauge

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Answer:

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2 years ago
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mixer [17]

Answer:

Option D

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