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2 years ago

Which property would cesium most likely have?

Chemistry

2 answers:

Romashka-Z-Leto [24]2 years ago

8 0

The answer is a because its not low ductile or gasst

ivolga24 [154]2 years ago

8 0

Answer: Option (B) is the correct answer.

Explanation:

Cesium is a group 1A element and therefore, it is a metal. Alkali metals are soft, malleable and ductile.

And, group 1A elements being metals are good conductors of heat and electricity.

Group 1A elements are highly reactive in nature. Hence, cesium is also reactive in nature. Also, cesium is not a gas.

Therefore, we can conclude that a property cesium would most likely have is that it is ductile.

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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

(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;

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

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

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:

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