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Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm an

ser-zykov [4K]

Answer:

volumetric flow rate = $0.0251 m^3/s$

Velocity in pipe section 1 = $6.513m/s$

velocity in pipe section 2 = 12.79 m/s

Explanation:

We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.

The density of water is = 997 kg/m³

density = mass/ volume

since we are given the mass, therefore, the volume will be mass/density

25/997 = $0.0251 m^3/s$

volumetric flow rate = $0.0251 m^3/s$

Average velocity calculations:

<em>Pipe section A:</em>

cross-sectional area =

$\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2$

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

$velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s$

<em>Pipe section B:</em>

cross-sectional area =

$\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2$

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

$velocity = 25/(997 \times 1.96\times10^{-3}) = 12.79m/s$

7 0

2 years ago

To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105

GenaCL600 [577]

Answer:

$62.14\ \text{miles}$

$6213727.37\ \text{miles}$

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = $10^5\ \text{mm}^{-2}$

Volume of the metal = $1000\ \text{mm}^3$

$10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}$

$1\ \text{mile}=1609.34\ \text{m}$

$\dfrac{10^5}{1609.34}=62.14\ \text{miles}$

The chain would extend $62.14\ \text{miles}$

Dislocation density = $10^{10}\ \text{mm}^{-2}$

Volume of the metal = $1000\ \text{mm}^3$

$10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}$

$\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}$

The chain would extend $6213727.37\ \text{miles}$

3 0

2 years ago

Storm sewer backup causes your basement to flood at the steady rate of 1 in. of depth per hour. The basement floor area is 2600

Wittaler [7]

Answer:

attached below

Explanation:

4 0

2 years ago

Can a 1½ " conduit, with a total area of 2.04 square inches, be filled with wires that total 0.93 square inches if the maximum f

Papessa [141]

Answer:

it is not possible to place the wires in the condui

Explanation:

given data

total area = 2.04 square inches

wires total area = 0.93 square inches

maximum fill conduit = 40%

to find out

Can it is possible place wire in conduit conduit

solution

we know maximum fill is 40%

so here first we get total area of conduit that will be

total area of conduit = 40% × 2.04

total area of conduit = 0.816 square inches

but this area is less than required area of wire that is 0.93 square inches

so we can say it is not possible to place the wires in the conduit

4 0

3 years ago

Conditions of special concern: i. Suggest two reasons each why distillation columns are run a.) above or b.) below ambient press

lutik1710 [3]

Solution :

Methods for selling pressure of a distillation column :

a). Set, $\text{based on the pressure required to condensed}$ the overhead stream using cooling water.

(minimum of approximate 45°C condenser temperature)

b). Set, $\text{based on highest temperature}$ of bottom product that avoids decomposition or reaction.

c). Set, $\text{based on available highest }$ not utility for reboiler.

Running the distillation column above the ambient pressure because :

The components to be distilled have very high vapor pressures and the temperature at which they can be condensed at or below the ambient pressure.

Run the reactor at an evaluated temperature because :

a). The rate of reaction is taster. This results in a small reactor or high phase conversion.

b). The reaction is endothermic and equilibrium limited increasing the temperature shifts the equilibrium to the right.

Run the reaction at an evaluated pressure because :

The reaction is gas phase and the concentration and hence the rate is increased as the pressure is increased. This results in a smaller reactor and /or higher reactor conversion.

The reaction is equilibrium limited and there are few products moles than react moles. As increase in pressure shifts the equilibrium to the right.

7 0

2 years ago