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Olin [163]
3 years ago
8

Water flows at a uniform velocity of 3 m/s into a nozzle that reduces the diameter from 10 cm to 2 cm. Calculate the water’s vel

ocity leaving the nozzle and the flow rate.

Engineering
1 answer:
katrin2010 [14]3 years ago
4 0

Answer:

See the pictures attached

Explanation:

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Consider a fan located in a 3 ft by 3 ft square duct. Velocities at various points at the outlet are measured, and the average f
natulia [17]

Answer:

minimum electric power consumption of the fan motor is 0.1437 Btu/s

Explanation:

given data

area = 3 ft by 3 ft

air density = 0.075 lbm/ft³

to find out

minimum electric power consumption of the fan motor

solution

we know that energy balance equation that is express as

E in - E out  = \frac{dE \ system}{dt}    ......................1

and at steady state  \frac{dE \ system}{dt} = 0

so we can say from equation 1

E in = E out

so

minimum power required is

E in = W = m \frac{V^2}{2} = \rho A V \frac{V^2}{2}  

put here value

E in =  \rho A V \frac{V^2}{2}  

E in =  0.075 *3*3* 22 \frac{22^2}{2}  

E in = 0.1437 Btu/s

minimum electric power consumption of the fan motor is 0.1437 Btu/s

5 0
3 years ago
The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm,
babunello [35]

Answer:

Qx = 9.109.10^5 \times 10^{-6} m³/s  

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

N = 60 rev/min

pressure p = 11 × 10^6 Pa

viscosity n = 100 Pas

angle = 18°

so  Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA   ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 \times 10^{-3} )²× 9  \times 10^{-3}  × sin18 × cos18

Qd = 94.305 × 10^{-6} m³/s

and

Qb = p × π × D × dc³ × sin²A ÷  12  × n × L    ............2

Qb = 11 × 10^{6} × π × 85 \times 10^{-3}  × ( 9  \times 10^{-3} )³ × sin²18 ÷  12  × 100 × 2

Qb = 85.2 × 10^{-6} m³/s

so here

volume flow rate Qx = Qd - Qb   ..............3

Qx =  94.305 × 10^{-6}  - 85.2 × 10^{-6}  

Qx = 9.109.10^5 \times 10^{-6} m³/s  

8 0
3 years ago
A 132mm diameter solid circular section​
Ganezh [65]

Answer:

not sure if this helps but

5 0
3 years ago
What is the thermal efficiency of this regeneration cycle in terms of enthalpies and fractions of total flow?
irga5000 [103]

Answer:

\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}

Explanation:

generally regeneration of cycle is used in the case of gas turbine. due to regeneration efficiency of turbine is increased but there is no effect on the on the net work out put of turbine.Actually in regeneration net heta input is decreases that is why total efficiency  increase.

 Now from T-S diagram

    W_{net}=W_{out}-W_{in}

   W_{net}=(h_3-h_4)-(h_2-h_1)

  Q_{in}=h_3-h_5

  Due to generation (h_5-h_2) amount of energy has been saved.

  Q_{generation}=Q_{saved}

So efficiency of cycle \eta =\frac{W_{net}}{Q_{in}}

  \eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}

Effectiveness of re-generator

  \varepsilon =\dfrac{(h_5-h_2)}{(h_4-h_2)}

So the efficiency of regenerative cycle

\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}

7 0
3 years ago
U 4. Find 2 bridges in the US and answer the following:
Zarrin [17]

Answer:

Im guessing this is for CEA for PLTW, if so look up the exact assignment number and look at online examples of the exact same assignment.

Explanation:

6 0
3 years ago
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