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2 years ago

The device whose operation closely matches the way the clamp-on ammeter works is

Engineering

1 answer:

Dmitrij [34]2 years ago

6 0

Answer:

The answer is

C. Split phase motor

Explanation:

Clamp meters rely on the principle of magnetic induction to make non contact AC current measurements. Electric current flowing through a wire produces a magnetic field.

Which is similar to basic mode of operation of electric motor and split phase motor is a type of electric motor.

What is a a clamp on meter?

Clamp meters are electrical testers which have wide jaws that are able to clamp around an electrical conductor. Originally designed as a single purpose tool for measuring AC current, clamp meters now include inputs for accepting test leads and other probes that support a wide range of electrical measurements, the jaws of a clamp meter permit work in tight spaces and permits current measurements on live conductors without circuit interruption.

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The solid aluminum shaft has a diameter of 50 mm. Determine the absolute maximum shear stress in the shaft and sketch the shear-

jarptica [38.1K]

Answer:

Max shear = 8.15 x 10^7 N/m2

Explanation:

In order to find the maximum stress for a solid shaft having radius r, we will be applying the Torsion formula which can be written as;

Allowable Shear Stress = Torque x Radius / pi/2 x radius^4

Putting the values we have;

T = 2000 N/m

Radius = Diameter/2 = 0.05 / 2 = 0.025 m

Putting values in formula;

Max shear = 2000 x 0.025 / 3.14/2 x (0.025)^4

Max shear = 8.15 x 10^7 N/m2

5 0

2 years ago

A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calcula

makkiz [27]

Answer:

F_y = 151319.01N = 15.132 KN

Explanation:

From the linear momentum equation theory, since flow is steady, the y components would be;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

We are given;

Length; L = 5ft = 1.52.

Initial diameter;d1 = 12in = 0.3m

Exit diameter; d2 = 24 in = 0.6m

Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s

Initial pressure;p1 = 30 psi = 206843 pa

Thus,

initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²

Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²

Now, we know that volume flow rate of water is given by; Q = A•V

Thus,

At exit, Q2 = A2•V2

So, 0.28 = 0.28•V2

So,V2 = 1 m/s

When flow is incompressible, we often say that ;

Initial mass flow rate = exit mass flow rate.

Thus,

ρ1 = ρ2 = 1000 kg/m³

Density of water is 1000 kg/m³

And A1•V1 = A2•V2

So, V1 = A2•V2/A1

So, V1 = 0.28 x 1/0.07

V1 = 4 m/s

So, from initial equation of y components;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

Where F_y is vertical force of enlargement pressure and P2 = 0

Thus, making F_y the subject;

F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2

Plugging in the relevant values to get;

F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)

F_y = 151319.01N = 15.132 KN

6 0

2 years ago

Which situation is an enabler for the rise of Artificial Intelligence (AI) in recent years?

Vladimir [108]

Explanation:

Which situation is an enabler for the rise of Artificial Intelligence (A in recent years? availability of cloud-based, hosted machine learning platforms.

8 0

2 years ago

Read 2 more answers

A plane wall of thickness 2L = 60 mm and thermal conductivity k= 5W/m.K experiences uniform volumetric heat generation at a rate

aniked [119]

Answer:

Explanation:

A plane wall of thickness 2L=40 mm and thermal conductivity k=5W/m⋅Kk=5W/m⋅K experiences uniform volumetric heat generation at a rateq

, while convection heat transfer occurs at both of its surfaces (x=-L, +L), each of which is exposed to a fluid of temperature T∞=20∘CT

∞

=20

∘

C. Under steady-state conditions, the temperature distribution in the wall is of the form T(x)=a+bx+cx2T(x)=a+bx+cx

where a=82.0∘C,b=−210∘C/m,c=−2×104C/m2a=82.0

∘

C,b=−210

∘

C/m,c=−2×10

C/m

, and x is in meters. The origin of the x-coordinate is at the midplane of the wall. (a) Sketch the temperature distribution and identify significant physical features. (b) What is the volumetric rate of heat generation q in the wall? (c) Determine the surface heat fluxes, q

′′

(−L)q

′′

(−L) and q

′′

(+L)q

′′

(+L). How are these fluxes related to the heat generation rate? (d) What are the convection coefficients for the surfaces at x=-L and x=+L? (e) Obtain an expression for the heat flux distribution q

′′

(x)q

′′

(x). Is the heat flux zero at any location? Explain any significant features of the distribution. (f) If the source of the heat generation is suddenly deactivated (q=0), what is the rate of change of energy stored in the wall at this instant? (g) What temperature will the wall eventually reach with q=0? How much energy must be removed by the fluid per unit area of the wall (J/m2)(J/m

) to reach this state? The density and specific heat of the wall material are 2600kg/m32600kg/m

and 800J/kg⋅K800J/kg⋅K, respectively.

6 0

2 years ago

The uniform beam is supported by two rods AB and CD that have crosssectional areas of 10 mm2 and 15 mm2 , respectively. Determin

ddd [48]

Answer:

w=2.25

Explanation:

It is necessary to determine the maximum w so that the normal stress in the AB and CD rods does not exceed the permitted normal stress.

The surface of the cross-section of the stapes was determined:

A_ab= 10 mm^2

A-cd= 15 mm^2

The maximum load is determined from the condition that the normal stresses is not higher than the permitted normal stress σ_allow.

σ_ab = F_ab/A_ab $\leq$ σ_allow

σ_cd = F_cd/A_cd $\leq$ σ_allow

In the next step we will determine the static size: Picture b).

We apply the conditions of equilibrium:

∑F_x=0

∑F_y=0

∑M=0

∑M_a=0 ==> -w*6*0.5*6*0.75*F_cd*6 =0

==> F_cd = 2*w*k*N

∑F_y=0 ==> F_cd+F_ab - 6*w*0.5 ==>2*w+F_ab -6*w*0.5 =0

==> F_ab = w*k*N

Now we determine the load w

Sector AB:

σ_ab = F_ab/A_ab $\leq$ σ_allow=300 KPa

= w/10*10^-6 $\leq$ σ_allow=300 KPa

w_ab = 3*10^-3 kN/m

Sector CD:

σ_cd = F_cd/A_cd $\leq$ σ_allow=300 KPa

= 2*w/15*10^-6 $\leq$ σ_allow=300 KPa

w_cd = 2.25*10^-3 kN/m

w=min{w_ab;w_cd} ==> w=min{3*10^-3;2.25*10^-3}

==> w=2.25 * 10^-3 kN/m

The solution is:

w=2.25 N/m

note:

find the attached graph

6 0

2 years ago