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Masja [62]
4 years ago
12

Determine size of a standard square key made of 1045 hot rolled steel for a 2 inch DIA shaft transmitting 100 HP at 500 rpm with

factor of safety 2.5 for yield in direct shear. W____inch L_____inch
Engineering
1 answer:
user100 [1]4 years ago
3 0

Answer:

Width = Length = 1.148 inches

Explanation:

We have been given the following data:

D = diameter = 2 inch = 0.0508m

P = Power = 100 HP = 74570 W

N = 500 rpm

Safety Factor = 2.5

<h3>Step 1:</h3>

We need to find yield strength which is represented by σ(y).

σ(y) for 1045 hot rolled steel = 330MPa

<h3 /><h3>Step 2: </h3>

Find Shear Strength. Formula is given:

τ(y) = σ(y) / 2

τ(y) = 330 / 2

τ(y) = 165 MPa

τ(y) = 165 × 10⁶ Pa

τ(y) = 165 × 10⁶ kg.m⁻¹.s⁻²

<h3>Step 3:</h3>

Find Torque. Formula is given:

T = 60P / 2πN

T= (60)(74570) / 2π(500)

T = 1424.9 Nm

<h3 /><h3>Step 4:</h3>

Find Shear Force. Formula is given:

F = 2T/d

F = 2(1424.9)/0.0508

F = 56098.43 N

F = 56098.43 kg.m.s⁻²

<h3 /><h3>Step 5:</h3>

Find length by the given formula:

F/L² = τ(y)/Safety factor

Rearrange for L:

L = √(F· Safety factor / τ(y))

Substitute the values found in previous steps to calculate L.

L = 0.02915 meters

Convert it into inches:

L = 1.148 in

As it si a square key:

L = W

Width = 1.148 in

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Read 2 more answers
Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe
NemiM [27]

Answer:

(a) attached below

(b) pf_{C}=0.85 lagging

(c) I_{C} =32.37 A

(d) X_{C} =49.37 Ω

(e) I_{cap} =9.72 A and I_{line} =27.66 A

Explanation:

Given data:

P_{1}=15 kW

S_{2} =10 kVA

pf_{1} =0.6 lagging

pf_{2}=0.8 leading

V=480 Volts

(a) Draw the power triangle for each load and for the combined load.

\alpha_{1}=cos^{-1} (0.6)=53.13°

\alpha_{2}=cos^{-1} (0.8)=36.86°

S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA

Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99 ≅ 20kVAR

P_{2} =S_{2}*pf_{2} =10*0.8=8 kW

Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99 ≅ -6 kVAR

The negative sign means that the load 2 is providing reactive power rather than consuming  

Then the combined load will be

P_{c} =P_{1} +P_{2} =15+8=23 kW

Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR

(b) Determine the power factor of the combined load and state whether lagging or leading.

S_{c} =P_{c} +jQ_{c} =23+14j

or in the polar form

S_{c} =26.92°

pf_{C}=cos(31.32) =0.85 lagging

The relationship between Apparent power S and Current I is

S=VI^{*}

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

I_{C} =S_{C}/\sqrt{3}*V

I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

Q_{C} =3*V^2/X_{C}

X_{C} =3*V^2/Q_{C}

X_{C} =3*(480)^2/14*10^3 Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is  

I_{cap} =V/X_{C} =480/49.37=9.72 A

Line current flowing from the source is

I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A

8 0
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