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stira [4]
3 years ago
13

A pipe 100 mm dia and 1 km long is to carry water from a reservoir to a village of 1000 people consuming 200 l/person/day. The p

ipeline works 8 hrs daily. The losses in the pipe including friction has been calculated as 10 m. The pressure to be maintained in the outlet is 25 kN/m2. What should be the height of the top of the reservoir assuming both the village and reservoir are situated in a plain?​
Engineering
2 answers:
natka813 [3]3 years ago
8 0

Answer:

i899999999999999999ijhhh

Explanation:

MrMuchimi3 years ago
8 0

Answer:

do the wam wam

Explanation:

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30 points and brainiest if correct please help A, B, C, D
tatuchka [14]

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B. to lock the tape into place

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3 years ago
Given: A graphite-moderated nuclear reactor. Heat is generated uniformly in uranium rods of 0.05m diameter at the rate of 7.5 x
sineoko [7]

Answer:

The maximum temperature at the center of the rod is found to be 517.24 °C

Explanation:

Assumptions:

1- Heat transfer is steady.

2- Heat transfer is in one dimension, due to axial symmetry.

3- Heat generation is uniform.

Now, we consider an inner imaginary cylinder of radius R inside the actual uranium rod of radius Ro. So, from steady state conditions, we know that, heat generated within the rod will be equal to the heat conducted at any point of the rod. So, from Fourier's Law, we write:

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where,

k = thermal conductivity = 29.5 W/m.K

q = heat generation per unit volume = 7.5 x 10^7 W/m³

V = volume of rod = π r² l

A = area of rod = 2π r l

using these values, we get:

dT = - (q/2k)(r dr)

integrating from r = 0, where T(0) = To = Maximum center temperature, to r = Ro, where, T(Ro) = Ts = surface temperature = 120°C.

To -Ts = qr²/4k

To = Ts + qr²/4k

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To = 120° C + 397.24° C

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5 0
4 years ago
A pump with a power of 5 kW (pump power, and not useful pump power) and an efficiency of 72 percent is used to pump water from a
almond37 [142]

Answer:

a) The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump is 245.175 kilopascals.

Explanation:

a) Let suppose that pump works at steady state. The mass flow rate of the water (\dot m), in kilograms per second, is determined by following formula:

\dot m = \frac{\eta \cdot \dot W}{g\cdot H} (1)

Where:

\dot W - Pump power, in watts.

\eta - Efficiency, no unit.

g - Gravitational acceleration, in meters per square second.

H - Hydrostatic column, in meters.

If we know that \eta = 0.72, \dot W = 5000\,W, g = 9.807\,\frac{m}{s^{2}} and H = 25\,m, then the mass flow rate of water is:

\dot m = 14.683\,\frac{kg}{s}

The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump (\Delta P), in pascals, is determined by this equation:

\Delta P = \rho\cdot g\cdot H (2)

Where \rho is the density of water, in kilograms per cubic meter.

If we know that \rho = 1000\,\frac{kg}{m^{3}}, g = 9.807\,\frac{m}{s^{2}} and H = 25\,m, then the pressure difference is:

\Delta P = 245175\,Pa

The pressure difference across the pump is 245.175 kilopascals.

4 0
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slega [8]

Answer:

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The rest is straightforward...

6 0
3 years ago
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