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3 years ago

13 Which pump is the most appropriate for low head high discharge capacity installations?

Engineering

1 answer:

gavmur [86]3 years ago

5 0

Options are missing and they are;

A. Propeller pump

B. Jet (mixed flow) pump

C. Centrifugal pump

D. Positive displacement pump

Answer:

C. Centrifugal pump

Explanation:

Dynamic pumps which are also known as centrifugal pumps are the ones widely used when we require high discharge. This is because they deal with transfer of fluids with low viscosity in flow rates that are high. Another reason is that they are also very good with low pressure installation.

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Chemical materials that are transported are called..

svetlana [45]

Answer:

Transport Molecules

Explanation:

I believe the answer is transport molecules because development of a cell membrane that could allow some materials to pass.

3 0

3 years ago

Air is to be heated steadily by an 8-kW electric resistance heater as it flows through an insulated duct. If the air enters at 5

Furkat [3]

To solve this problem it is necessary to apply the concepts related to the heat exchange of a body.

By definition heat exchange in terms of mass flow can be expressed as

$W = \dot{m}c_p \Delta T$

Where

$C_p =$ Specific heat

$\dot{m}$ = Mass flow rate

$\Delta T$ = Change in Temperature

Our values are given as

$C_p = 1.005kJ/kgK \rightarrow$ Specific heat of air

$T_1 = 50\°C$

$\dot{m} = 2kg/s$

$W = 8kW$

From our equation we have that

$W = \dot{m}c_p \Delta T$

$W = \dot{m}c_p (T_2-T_1)$

Rearrange to find $T_2$

$T_2 = \frac{W}{\dot{m}c_p}+T_1$

Replacing

$T_2 = \frac{8}{2*1.005}+(50+273)$

$T_2 = 326.98K \approx 53.98\°C$

Therefore the exit temperature of air is 53.98°C

6 0

3 years ago

The stagnation chamber of a wind tunnel is connected to a high-pressure airbottle farm which is outside the laboratory building.

Natasha2012 [34]

This question is not complete, the complete question is;

The stagnation chamber of a wind tunnel is connected to a high-pressure air bottle farm which is outside the laboratory building. The two are connected by a long pipe of 4-in inside diameter. If the static pressure ratio between the bottle farm and the stagnation chamber is 10, and the bottle-farm static pressure is 100 atm, how long can the pipe be without choking? Assume adiabatic, subsonic, one-dimensional flow with a friction coefficient of 0.005

Answer:

the length of the pipe is 11583 in or 965.25 ft

Explanation:

Given the data in the question;

Static pressure ratio; p1/p2 = 10

friction coefficient f = 0.005

diameter of pipe, D =4 inch

first we obtain the value from FANN0 FLOW TABLE for pressure ratio of ( p1/p2 = 10 )so

4fL $_{max}$ / D = 57.915

we substitute

(4×0.005×L $_{max}$ ) / 4 = 57.915

0.005L $_{max}$ = 57.915

L $_{max}$ = 57.915 / 0.005

L $_{max}$ = 11583 in

Therefore, the length of the pipe is 11583 in or 965.25 ft

6 0

3 years ago

The lattice constant of a simple cubic primitive cell is 5.28 Å. Determine the distancebetween the nearest parallel ( a ) (100),

skelet666 [1.2K]

Answer:

a)5.28 Å , b)3.73 Å , c)3.048 Å

Explanation:

the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.

Therefore, a particular unit cell consist only 1/8th part of an atom.

The lattice constant of a simple cubic primitive cell is 5.28 Å

We know formula of distance,

d = $\frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}$

a)(100)

a=5.28 Å

Distance = $\frac{5.28 Å}{\sqrt{1^{2}+0^{2}+0^{2}}}$ =5.28 Å

b)(110)

Distance = $\frac{5.28}{\sqrt{1^{2}+1^{2}+0^{2}}}$ = 3.73 Å

c)(111)

Distance= $\frac{5.28}{\sqrt{1^{2}+1^{2}+1^{2}}}$ = 3.048 Å

6 0

3 years ago

4. An aluminum alloy fin of 12 mm thick, 10 mm width and 50 mm long protrudes from a wall, which is maintained at 120C. The amb

Minchanka [31]

Answer:

a) 84.034°C

b) 92.56°C

c) ≈ 88 watts

Explanation:

Thickness of aluminum alloy fin = 12 mm

width = 10 mm

length = 50 mm

Ambient air temperature = 22°C

Temperature of aluminum alloy is maintained at 120°C

<u>a) Determine temperature at end of fin</u>

m = √ hp/Ka

= √( 140*2 ) / ( 12 * 10^-3 * 55 )

= √ 280 / 0.66 = 20.60

Attached below is the remaining answers

7 0

3 years ago