PLEASE HELP! (8.01 MC)

S_A_V [24]

The answer is the number of protons.

explanation: The number of protons in one atom of an element determines the atoms identity, and the number of electrons determines its electrical charge.

5 0

3 years ago

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A pump with an 80% efficiency drives water up between two reservoirs through a piping system of total length L = 15 and circular

Sunny_sXe [5.5K]

The losses in the pipe increases the power requirement of the pipe to

maintain a given flowrate.

Responses (approximate value);

(a) 2.598 m/s

(b) 181,058.58

(c) 0.025

(d) 227:1000

(e) 1,216.67 W

<h3>Which methods can be used to calculate the pressure head in the pipe?</h3>

The given parameters are;

Pump efficiency, η = 80%

Length of the pipe, L = 15 m

Cross-sectional diameter, d = 7 cm

Reservoir temperature, T = 20°C = 293.15 K

$\mathbf{K_{entrance}}$ ≈ $\mathbf{K_{exit}}$ ≈ 1.0, $\mathbf{K_{elbow}}$ ≈ 0.4

Volumetric flow rate, Q = 10 Liters/s = 0.01 m³/s

Surface roughness, ∈ = 0.15 mm

(a) The cross sectional area of the pipe, A = π·r²

Where;

$r = \mathbf{\dfrac{d}{2}}$

Which gives;

$r = \dfrac{0.07 \, cm}{2} = \mathbf{0.035 \, cm}$

$Average \ water \ velocity, \ v =\mathbf{ \dfrac{Q}{A}}$

Therefore;

$v = \dfrac{0.01}{ \pi \times 0.035^2} \approx 2.598$

The average velocity of the water, v ≈ 2.598 m/s

(b) The viscosity of water at 20°C is 0.001003 kg/(m·s) given as follows;

Density of water at 20°C, ρ = 998.23 kg/m³

Reynolds' number, Re, is found as follows;

$Re = \mathbf{\dfrac{\rho \cdot V \cdot D}{\mu}}$

Which gives;

$Re = \dfrac{998.58 \times 2.598 \times 0.07 }{0.001003} \approx \underline{181,058.58}$

(c) The friction factor is given by the following formula;

$\dfrac{1}{\sqrt{f} } = \mathbf{-2.0 \cdot log \left(\dfrac{\epsilon/D}{3.7} + \dfrac{5.74}{Re^{0.9}} } \right)}$

Which gives;

f ≈ 0.025

(d) Friction head loss is given as follows;

$h_f = \mathbf{f \times \dfrac{L}{D} \times \dfrac{V^2}{2 \cdot g}}$

Which gives;

$h_f = 0.025 \times \dfrac{15}{0.07} \times \dfrac{2.598^2}{2 \times 9.81} \approx \mathbf{1.84}$

$Other \ head \ losses, \ h_l= \sum K \cdot \dfrac{V^2}{2}$

Which gives;

$h_l=(1 + 1+0.4) \times \dfrac{ 2.598^2}{2} \approx \mathbf{8.0995}$

Ratio between friction head loss and other head loss is therefore;

$\dfrac{h_f}{h_l} \approx \dfrac{1.84}{8.0995} \approx \underline{0.227}$

The ratio between friction head loss and other head loss is approximately 227:1000

(e) The power required P is found as follows;

$P= \mathbf{ \dfrac{\rho \cdot g \cdot Q \cdot H}{\eta}}$

Which gives;

$P= \dfrac{998.23 \times 9.81 \times 0.01 \times (1.84 + 8.0995)}{0.8} \approx \mathbf{ 1216.67}$

The power required to drive the pump, P ≈ 1,216.67 W

Learn more about flow in pipes here:

brainly.com/question/7246532

6 0

2 years ago

What weather would create the smaller waves?

jeyben [28]

Calm weather would be the answer

8 0

3 years ago

If you start with 163 g of water at 29◦C, how much heat must you add to convert all the liquid into vapor at 100◦C? Assume no he

Snowcat [4.5K]

Answer:

48.37514 kj

Explanation:

Given data:

Mass of water = 163 g

Initial temperature = 29°C

Final temperature = 100°C

Heat added = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

ΔT = 100°C - 29°C

ΔT = 71°C

Q = 163 g × 4.18 j/g.°C × 71°C

Q = 48375.14 j

Joule to Kj conversion:

48375.14 /1000 = 48.37514 kj

3 0

3 years ago

Zebra mussels have no???

olchik [2.2K]

Zebras do ave muscles

4 0

3 years ago