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aleksklad [387]

3 years ago

12

Calculate the torque produced by a motor that has 500 windings across a 30cm diameter rotating core. The core is 0.6 M long insi

de a 50 tesla magnetic field. The resistance of the wire is 30 ohms when 12 volts are applied.

1 answer:

zalisa [80]3 years ago

7 0

Answer:

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Are the wooden pillars shown in the image below, a dead load?

Brrunno [24]

Answer:

no

Explanation:

it's not a dead load because when load is put on the pillars it's not fully straining it's been slowly getting to be heavier in that period of time before it falls

6 0

3 years ago

A long corridor has a single light bulb and two doors with light switch at each door. design logic circuit for the light; assume

sattari [20]

Answer and Explanation:

Let A denote its switch first after that we will assume B which denotes the next switch and then we will assume C stand for both the bulb. we assume 0 mean turn off while 1 mean turn on, too. The light is off, as both switches are in the same place. This may be illustrated with the below table of truth:

A B C (output)

0 0 0

0 1 1

1 0 1

1 1 0

The logic circuit is shown below

C = A'B + AB'

If the switches are in multiple places the bulb outcome will be on on the other hand if another switches are all in the same place, the result of the bulb will be off. This gate is XOR. The gate is shown in the diagram adjoining below.

3 0

3 years ago

A cylindrical bar of metal having a diameter of 20.9 mm and a length of 205 mm is deformed elastically in tension with a force o

natita [175]

Answer:

a. 1.91 b. -8.13 mm

Explanation:

Modulus =stress/strain; calculating stress =F/A, hence determine the strain

Poisson's ratio =(change in diameter/diameter)/strain

8 0

3 years ago

For an Otto cycle, plot the cycle efficiency as a function of compression ratio from 4 to 16.

Elza [17]

Assumptions:

Steady state.
Air as working fluid.
Ideal gas.
Reversible process.
Ideal Otto Cycle.

Explanation:

Otto cycle is a thermodynamic cycle widely used in automobile engines, in which an amount of gas (air) experiences changes of pressure, temperature, volume, addition of heat, and removal of heat. The cycle is composed by (following the P-V diagram):

Intake <em>0-1</em>: the mass of working fluid is drawn into the piston at a constant pressure.

Adiabatic compression <em>1-2</em>: the mass of working fluid is compressed isentropically from State 1 to State 2 through compression ratio (r).

$r =\frac{V_1}{V_2}$

Ignition 2-3: the volume remains constant while heat is added to the mass of gas.
Expansion 3-4: the working fluid does work on the piston due to the high pressure within it, thus the working fluid reaches the maximum volume through the compression ratio.

$r = \frac{V_4}{V_3} = \frac{V_1}{V_2}$

Heat Rejection 4-1: heat is removed from the working fluid as the pressure drops instantaneously.
Exhaust 1-0: the working fluid is vented to the atmosphere.

If the system produces enough work, the automobile and its occupants will propel. On the other hand, the efficiency of the Otto Cycle is defined as follows:

$\eta = 1-(\frac{1}{r^{\gamma - 1} } )$

where:

$\gamma = \frac{C_{p} }{C_{v}} : specific heat ratio$

Ideal air is the working fluid, as stated before, for which its specific heat ratio can be considered constant.

$\gamma = 1.4$

Answer:

See image attached.

5 0

3 years ago

Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans

aksik [14]

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, $T_{i} = 20^{\circ}C$ = 273 + 20 = 293 K

Temperature at the outlet, $T_{o} = 200{\circ}C$ = 273 + 200 = 473 K

Pressure at inlet, $P_{i} = 80 kPa = 80\times 10^{3} Pa$

Pressure at outlet, $P_{o} = 800 kPa = 800\times 10^{3} Pa$

Speed at the outlet, $v_{o} = 20 m/s$

Diameter of the tube, $D = 10 cm = 10\times 10^{- 2} m = 0.1 m$

Input power, $P_{i} = 400 kW = 400\times 10^{3} W$

Now,

To calculate the heat transfer, $Q$ , we make use of the steady flow eqn:

$h_{i} + \frac{v_{i}^{2}}{2} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}$

where

$h_{i}$ = specific enthalpy at inlet

$h_{o}$ = specific enthalpy at outlet

$v_{i}$ = air speed at inlet

$p_{s}$ = specific power input

H and H' = Elevation of inlet and outlet

Now, if

$v_{i} = 0$ and H = H'

Then the above eqn reduces to:

$h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}$

$Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}$ (1)

Also,

$p_{s} = \frac{P_{i}}{ mass, m}$

Area of cross-section, A = $\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}$

Specific Volume at outlet, $V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s$

From the eqn:

$P_{o}V_{o} = mRT_{o}$

$m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s$

Now,

$p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg$

Also,

$\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg$

Now, using these values in eqn (1):

$Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW$

Now, rate of heat transfer, q:

q = mQ = $0.925\times 813.33 = 752.33 kW$

4 0

3 years ago