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3 years ago

Π/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 ...

Engineering

1 answer:

leva [86]3 years ago

6 0

Answer: Pi= 4 - 4/3 + 4/5 - 4/7 + 4/9 ...

Explanation:

Is the same as the example,

If Π/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 ...

Then

(Π/4 )*4= 4*(1 - 1/3 + 1/5 - 1/7 + 1/9 ...)

Π =4 - 4/3 + 4/5 - 4/7 + 4/9 ...

The way to write this is

Sum(from n=0 to n=inf) of (-1)^n 4/(2n+1)

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A 10.2 mm diameter steel circular rod is subjected to a tensile load that reduces its cross- sectional area to 52.7 mm^2. Determ

VMariaS [17]

Answer:

The percentage ductility is 35.5%.

Explanation:

Ductility is the ability of being deform under applied load. Ductility can measure by percentage elongation and percentage reduction in area. Here, percentage reduction in area method is taken to measure the ductility.

Step1

Given:

Diameter of shaft is 10.2 mm.

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Step2

Initial area is calculated as follows:

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2 years ago

Steam at a pressure of 100 bar and temperature of 600 °C enters an adiabatic nozzle with a velocity of 35 m/s. It leaves the noz

butalik [34]

Answer:

Exit velocity $V_2=1472.2$ m/s.

Explanation:

Given:

At inlet:

$P_1=100 bar,T_=600°C,V_1=35m/s$

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$h_1=3624.7\frac{KJ}{Kg}$

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Given that dryness or quality of steam at the exit of nozzle is 0.85 and pressure P=80 bar.So from steam table we can find the other properties.

Properties of saturated steam at 80 bar

$h_f= 1316.61\frac{KJ}{Kg} ,h_g= 2757.8\frac{KJ}{Kg}$

So the enthalpy of steam at the exit of turbine

$h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}$

$h_2=1316.61+0.85(2757.8-1316.61)\frac{KJ}{Kg}$

$h_2=2541.62\frac{KJ}{Kg}$

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$h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w$

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