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3 years ago

8

Can a 1½ " conduit, with a total area of 2.04 square inches, be filled with wires that total 0.93 square inches if the maximum f

ill for that conduit is 40%?

1 answer:

Papessa [141]3 years ago

4 0

Answer:

it is not possible to place the wires in the condui

Explanation:

given data

total area = 2.04 square inches

wires total area = 0.93 square inches

maximum fill conduit = 40%

to find out

Can it is possible place wire in conduit conduit

solution

we know maximum fill is 40%

so here first we get total area of conduit that will be

total area of conduit = 40% × 2.04

total area of conduit = 0.816 square inches

but this area is less than required area of wire that is 0.93 square inches

so we can say it is not possible to place the wires in the conduit

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A hydraulic cylinder is to be used to move a workpiece in a manufacturing operation through a distance of 50 mm in 10 s. A force

svetlana [45]

Answer:

The answer to this question is 1273885.3 ∅

Explanation:

<em>The first step is to determine the required hydraulic flow rate liquid if working pressure and if a cylinder with a piston diameter of 100 mm is available.</em>

<em>Given that,</em>

<em>The distance = 50mm</em>

<em>The time t =10 seconds</em>

<em>The force F = 10kN</em>

<em>The piston diameter is = 100mm</em>

<em>The pressure = F/A</em>

<em> 10 * 10^3/Δ/Δ </em>

<em> P = 1273885.3503 pa</em>

<em>Then</em>

<em>Power = work/time = Force * distance /time</em>

<em> = 10 * 1000 * 0.050/10</em>

<em>which is =50 watt</em>

<em>Power =∅ΔP</em>

<em>50 = 1273885.3 ∅</em>

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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 = $0.925\times 813.33 = 752.33 kW$

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3 years ago

1: A baseball is hit 4 feet above the ground leaves the bat with an initial speed of 98 ft/sec at an angle of 0 45 is caught by

zvonat [6]

Answer:

299.36 feet

Explanation:

$To \ find \ the \ distance \ of \ the \ ball \ from \ the \ home \ plate. \\ \\ From \ the \ given \ information:$

$Height \ h = 4 \ ft$

$Initial \ speed \ V_o = 98 \ ft/s ec$

$The \ angle \ \theta = 45^0$

$Acceleration \ due \ to \ gravity (g)= 32.2 \ ft/s$

$U_x = V_o \ cos 45 = \dfrac{98}{\sqrt{2}}$

$U_y = V_o \ sin 45 = \dfrac{98}{\sqrt{2}}$

So;

$S_y = u_y t - \dfrac{1}{2}gt^2$

$-1 =\dfrac{98}{\sqrt{2}}t - \dfrac{1}{2}*32*1.85t^2$

By solving:

$t_1 = 4.32 \ sec$

Thus;

$horizontal \ distance = U_x t$

$= \dfrac{98}{\sqrt{2}}\times 4.32$

$\mathbf{=299.36 \ feet}$

$\mathbf{Thus \ , the \ distance \ from \ the \ home \ plate \ = \ 299.36 \ feet}$

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3 years ago

Shear _______ can be directly computed from the angle of twist and the specimen dimensions. (a)- Stress (b)- Modulus (c)- Strain

fgiga [73]

Answer:

c) Strain

Explanation:

For example, the shear strain “γ” on the surface of the rod is determined by measuring the relative angle of twist “φg” over a gage length “Lg”.

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Answer:

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