Ask question

Ask question

All categories

3 years ago

13

A multipurpose transformer has a secondary coil with several points at which a voltage can be extracted, giving outputs of 5.60,

12.0, and 480 V. (a) The input voltage is 140 V to a primary coil of 220 turns. What are the numbers of turns in the parts of the secondary used to produce the output voltages

1 answer:

Rudik [331]3 years ago

4 0

Answer:

for 5.6V 9 turns, for 12.0V 19 turns, for 480V 755 turns

Explanation:

Vp/Vs= Np/Ns

Vp: Primary voltage

Vs: Secondary Voltage

Np: number of turns on primary side

Ns: number of turns on secondary side

for output 5.6V

140/5.6= 220/Ns

Ns= 8.8 or 9 Turns

for output 12.0V

140/12= 220/Ns

Ns= 18.9 or 19 turns

for output 480V

140/480= 220/Ns

Ns= 754.3 or 755 turns

You might be interested in

Which of the following refers to software designed to alter system files and utilities on a victim’s system with the intention o

Oksi-84 [34.3K]

The answer is Rootkits.

6 0

3 years ago

Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320

lesya [120]

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

$n = \frac{S_{uts}}{\tau_{max}}$

Where:

$n$ - Safety factor, dimensionless.

$S_{uts}$ - Ultimate shear strength, measured in pascals.

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: ( $n = 4.2$ , $S_{uts} = 320\times 10^{6}\,Pa$ )

$\tau_{max} = \frac{S_{uts}}{n}$

$\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}$

$\tau_{max} = 76.190\times 10^{6}\,Pa$

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

$\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}$

Where:

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

$V$ - Shear force, measured in kilonewtons.

$A$ - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

$V = \frac{P}{5}$

$V = \frac{450\,kN}{5}$

$V = 90\,kN$

The minimum allowable cross section area is cleared in the shearing stress equation:

$A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}$

If $V = 90\,kN$ and $\tau_{max} = 76.190\times 10^{3}\,kPa$ , the minimum allowable cross section area is:

$A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}$

$A = 1.640\times 10^{-3}\,m^{2}$

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

$A = \frac{\pi}{4}\cdot D^{2}$

The diameter is now cleared and computed:

$D = \sqrt{\frac{4}{\pi}\cdot A}$

$D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})$

$D = 0.0457\,m$

$D = 45.7\,mm$

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

5 0

3 years ago

Quinn’s relatives relayed a story about putting on a headset and seeing a digital world that they could walk around in and explo

Kryger [21]

Answer:

I know it is C)Virtual reality

Explanation:

Look at the clues

story about putting on a headset ( virtual reality head set!)

seeing a digital world (A virtual reality world)

they could walk around in (Fake walking you are basically jogging in place)

explore in order to see what ancient Benin looked like (Looking at a real place only digitally)

as if they were really there ( they think they are actually there)

The only reason I know all of this is because I have done virtual reality multiple times and I LOVED it SUPER fun ( I was doing archery) :) Hope this helps!

6 0

3 years ago

Read 2 more answers

4. Water vapor enters a turbine operating at steady state at 1000oF, 220 lbf/in2 , with a volumetric flow rate of 25 ft3/s, and

hodyreva [135]

Yes i is the time of the day you get to frost the moon and back and then you can come over and then go to hang out with me me and then go to hang out

6 0

3 years ago

A rigid, well-insulated tank of volume 0.9 m is initially evacuated. At time t = 0, air from the surroundings at 1 bar, 27°C beg

Eva8 [605]

Answer:

$\dot{w}$ = -0.303 KW

Explanation:

This is the case of unsteady flow process because properties are changing with time.

From first law of thermodynamics for unsteady flow process

$\dfrac{dU}{dt}=\dot{m_i}h_i+\dot{Q}-\dot{m_e}h_i+\dot{w}$

Given that tank is insulated so $\dot{Q}=0$ and no mass is leaving so

$\dot{m_e}=0$

$\int dU=\int \dot{m_i}h_i\ dt-\int \dot{w}\ dt$

$m_2u_2-m_1u_1=(m_2-m_1)h_i- \dot{w}\Delta t$

Mass conservation $m_2-m_1=m_e-m_i$

$m_1,m_2$ is the initial and final mass in the system respectively.

Initially tank is evacuated so $m_1=0$

We know that for air $u=C_vT ,h=C_p T$ , $P_2v_2=m_2RT_2$

$m_2=0.42 kg$

So now putting values

$0.42 \times 0.71 \times 730=0.42\times 1.005\times 300- \dot{w} \times 300$

$\dot{w}$ = -0.303 KW

3 0

3 years ago