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Alexeev081 [22]
4 years ago
14

Smoke rises from a chimney on a day with variable winds and produces a visible pattern. Does this pattern represent a pathline o

r a streakline? Under what conditions might the pathline and streakline be the same?
Engineering
1 answer:
jeyben [28]4 years ago
5 0

Answer:

Pathline

Explanation:

- The smoke coming out of the chimney on a day with variable winds would most closely relate to a path-line. Consider a smoke particle " grey " color particle that is injected into air at every instant in time and follows a continuous "path" along the flow of the of the fluid.

- A streakline concentrates on fluid particles that have gone through a fixed station or point. At some instant of time the position of all these particles are marked and a line is drawn through them.

- The streakline and pathline can coincide on the conditions of flow to be steady that is the rate of change of flow of the air is continuous and does not change with time in a day i.e the variation in the winds or any discontinuity is removed.

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The water behind Hoover Dam in Nevada is 221 m higher than the Colorado River below it. At what rate must water pass through the
dexar [7]

Answer:

23.06262m^3/s

Explanation:

The volume flow rate of the water is determined from the needed power output and the elevation difference:

Where, height (h) =221m, power(w)=50MW=50*10^6w

Density of water (ρ)=1000kg/m^3

Efficiency of turbine(η)=100%=1

V=W/ρηgh

=50*10^6m^3/(1)*(1000)*(9.81)*(221)s=23.06262m^3/s

3 0
3 years ago
What is 12 cm theodolite​
Klio2033 [76]

Answer:

Answer below :)

Explanation:

<u>SIZE OF THEODOLITE:</u> A theodolite is designated by diameter of the graduated circle on the lower plate. The common sizes are 8 cm to 12 cm while<em> 14 cm</em> to <em>25 cm</em> instrument are used for triangulation work.

4 0
3 years ago
Read 2 more answers
Refrigerant 22 flows in a theoretical single-stage compression refrigeration cycle with a mass flow rate of 0.05 kg/s. The conde
Debora [2.8K]

Answer:

a.  The work done by the compressor is 447.81 Kj/kg

b. The heat rejected from the condenser in kJ/kg is 187.3 kJ/kg

c. The heat absorbed by the evaporator in kJ/kg is 397.81 Kj/Kg

d. The coefficient of performance is 2.746

e. The refrigerating efficiency is 71.14%

Explanation:

According to the given data we would need first the conversion of temperaturte from C to K as follows:

Temperature at evaporator inlet= Te=-16+273=257 K

Temperatue at condenser exit=Te=48+273=321 K

Enthalpy at evaporator inlet of Te -16=i3=397.81 Kj/Kg

Enthalpy at evaporator exit of Te 48=i1=260.51 Kj/Kg

b. To calculate the the heat rejected from the condenser in kJ/kg we would need to calculate the Enthalpy at the compressor exit by using the compressor equation as follows:

w=i4-i3

W/M=i4-i3

i4=W/M + i3

i4=2.5/0.05 + 397.81

i4=447.81 Kj/kg

a. Enthalpy at the compressor exit=447.81 Kj/kg

Therefore, the heat rejected from the condenser in kJ/kg=i4-i1

the heat rejected from the condenser in kJ/kg=447.81-260.51

the heat rejected from the condenser in kJ/kg=187.3 kJ/kg

c. Temperature at evaporator inlet= Te=-16+273=257 K

The heat absorbed by the evaporator in kJ/kg is Enthalpy at evaporator inlet of Te -16=i3=397.81 Kj/Kg

d. To calculate the coefficient of performance we use the following formula:

coefficient of performance=Refrigerating effect/Energy input

coefficient of performance=137.3/50

coefficient of performance=2.746

the coefficient of performance is 2.746

e. The refrigerating efficiency = COP/COPc

COPc=Te/(Tc-Te)

COPc=255/(321-255)

COPc=3.86

refrigerating efficiency=2.746/3.86

refrigerating efficiency=0.7114=71.14%

8 0
4 years ago
Germanium forms a substitutional solid solution with silicon. Compute the number of germanium atoms per cubic centimeter for a g
brilliants [131]

Answer:

The number of germanium atoms per cubic centimeter for this germanium-silicon alloy is 3.16 x 10²¹ atoms/cm³.

Explanation:

Concentration of Ge (C_{Ge}) = 15%

Concentration of Si (C_{Si}) = 85%

Density of Germanium (ρ_{Ge}) = 5.32 g/cm³

Density of Silicon (ρ_{Si}) = 2.33 g/cm³

Atomic mass of Ge (A_{Ge})= 72.64 g/mol

To calculate the number of Ge atoms per cubic centimeter for the alloy, we will use the formula:

No of Ge atoms/cm³=[Avogadro's Number*C_{Ge}]/([C_{Ge}*A_{Ge}/ρ_{Ge})+(C_{Si}*A_{Ge}/ρ_{Si})]

                              = (6.02x10²³ * 15%) / [(15% * 72.64/5.32)+(72.64*85%/2.33)]

                              = (9.03x10²²)/(2.048+26.499)

                              = (9.03x10²²)/(28.547)

No of Ge atoms/cm³ = 3.16 x 10²¹ atoms/cm³

3 0
3 years ago
Which of the following statements about pitot-static systems is FALSE? a). A pitot probe measures the Total Pressure of the free
Pavlova-9 [17]

Answer:

C

Explanation:

Pitot tube:

  Pitot tube is a device which is used to measure the velocity of flow by measuring pressure difference between the points.

As we know that stagnation pressure is the summation of dynamic and static pressure.

Stagnation pressure = Static pressure + Dynamic pressure

So

Dynamic pressure  = Stagnation pressure -  Static pressure

We know that dynamic pressure

P_{dynamic}=\dfrac{\rho V^2}{2}

On the other hand Pitot tube measure the dynamic pressure.

So option C is correct.

5 0
3 years ago
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