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torisob [31]
3 years ago
15

A solid steel shaft has to transmit 100 kW at 160 RPM. Taking allowable shear stress at 70 Mpa, find the suitable diameter of th

e shaft. The maximum torque transmitted in each revolution exceeds the mean by 20%.
Engineering
1 answer:
MA_775_DIABLO [31]3 years ago
7 0

Answer:

The diameter of the shaft is 80.5 mm.

Explanation:

Torsion equation is applied for the diameter of the solid shaft.

Step1

Given:

Power of the shaft is 100 kw.

Revolution per minute is 160 RPM.

Allowable shear stress is 70 Mpa.

Maximum torque is 20% more than the mean torque.  

Step2

Mean torque is calculated as follows:

P=T\omega

P=T(\frac{2\pi N}{60})

100\times 1000=T(\frac{2\pi 160}{60})

T=5968.31 N-m

Step3

Maximum torque is calculated as follows:

T_{max}=(1+\frac{20}{100})T

T_{max}=1.2T

T_{max}=1.2\times 5968.31

T_{max}=7161.97 N-m

Step4

Apply torsional equation for diameter of shaft as follows:

\tau _{max}=\frac{T_{max}}{\frac{\pi d^{3}}{16}}

70\times 10^{6}=\frac{7161.97}{\frac{\pi d^{3}}{16}}

d^{3}=5.211\times 10^{-4}

d=0.0805 m

or,

d=80.5 mm

Thus, the diameter of the shaft is 80.5 mm.

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viktelen [127]

Answer:

a) 69,630KW

b) 203 KW

Explanation:

The data obtained from Tables A-4, A-5 and A-6 is as follows:

h_{1} = h_{f,@20KPa} = 251.42 KJ/kg\\v_{1} = v_{f,@20KPa} = 0.001017 KJ/kgK\\\\w_{p,in} = v_{1} * (P_{2} - P_{1})\\w_{p,in} = (0.001017)*(4000-20)\\\\w_{p,in} = 4.05 KJ/kg\\\\h_{2} = h_{1} - w_{p,in} \\h_{2} = 251.42 + 4.05\\\\h_{2}  =  255.47KJ/kg\\\\P_{3} = 4000KPa\\T_{3} = 700 C\\s_{3} = 7.6214 KJ/kgK\\\\h_{3} = 3906.3 KJ/kg\\\\P_{4} = 20 KPa\\s_{3} = s_{4} = 7.6214KJ/kgK\\s_{f} = 0.8320 KJ/kgK\\s_{fg} = 7.0752 KJ/kgK\\\\

x_{4} = \frac{s_{4} - s_{f} }{s_{fg} }  \\\\x_{4} = \frac{7.6214-0.8320}{7.0752} = 0.9596\\\\h_{f} = 251.42KJ/kg \\h_{fg} = 2357.5KJ/kg \\\\h_{4} = h_{f} + x_{4}*h_{fg} = 251.42 + 0.9596*2357.5 = 2513.7KJ/kg\\\\

The power produced and consumed by turbine and pump respectively are:

W_{T,out} = flow(m) *(h_{3} - h_{4}) \\W_{T,out} = 50 *(3906.3-2513.7)\\\\W_{T,out} = 69,630 KW\\\\W_{p,in} = flow(m) *w_{p,in} = 50*4.05 = 203 KW

7 0
3 years ago
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Answer:

Springs store energy when compressed and release energy when they rebound

Explanation:

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Answer:

0.064 mg/kg/day

6.25% from water, 93.75% from fish

Explanation:

Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.

The BCF = 10³, so the concentration of the chemical in the fish is:

10³ = x / (0.1 mg/kg)

x = 100 mg/kg

For 2 L of water and 30 g of fish:

2 kg × 0.1 mg/kg = 0.2 mg

0.030 kg × 100 mg/kg = 3 mg

The total daily intake is 3.2 mg.  Divided by the woman's mass of 50 kg, the dosage is:

(3.2 mg/day) / (50 kg) = 0.064 mg/kg/day

b) The percent from the water is:

0.2 mg / 3.2 mg = 6.25%

And the percent from the fish is:

3 mg / 3.2 mg = 93.75%

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The Hoover Dam is 221 m tall and 379 m wide. Approximating it as a flat plate, determine the effective resultant force on the da
IrinaVladis [17]

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2165800 Pa

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See it in the pic.

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