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Tamiku [17]
3 years ago
11

Calculate the wire pressure for a round copper bar with an original cross-sectional area of 12.56 mm2 to a 30% reduction of area

, given that the included die angle is 300 with a coefficient of friction (μ) of 0.08 and the yield stress for copper is 350 MPa.
Engineering
1 answer:
dybincka [34]3 years ago
8 0

Answer:153.76 MPa

Explanation:

Initial Area\left ( A_0\right )=12.56 mm^2

Final Area\left ( A_f\right )=0.7\times 12.56 mm^2=8.792 mm^2

Die angle=30^{\circ}

\alpha =\frac{30}{2}=15^{\circ}

\mu =0.08

Yield stress\left ( \sigma _y \right )=350 MPa

B=\mu cot\left ( \aplha\right )=0.2985

\sigma _{pressure}=\sigma _y\left [\frac{1+B}{B}\right ]\left [ 1-\frac{A_f}{A_0}\right ]^B

\sigma _{pressure}=350\left [\frac{1+0.2985}{0.2985}\right ]\left [ 1-\frac{8.792}{12.56}\right ]^{0.2985}

\sigma _{pressure}=153.76 MPa

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Answer:

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Explanation:

a) of the attached figure we have:

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Q_{l}=14days(\frac{24 h}{1 day})(\frac{120000 kJ}{1 h})-2000 kWh(\frac{3600 s}{1 h})=33120000 kJ

b) The coefficient of performance is:

COP=\frac{Q_{h} }{W}=\frac{120000 kJ/h*14(\frac{24 h}{1 day}) }{2000 kWh(\frac{3600 s}{1 h}) } = 5.6

c) The coefficient of performance of a reversible heat pump is:

COP_{reversible}=\frac{T_{h} }{T_{h}-T_{l}  }

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COP_{reversible}=\frac{293}{293-283}=29.3

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How is air pressure affected by the shape of an aircraft wing
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Answer:

\eta_{turbine} = 0.603 = 60.3\%

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

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At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in  vapor state:

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Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than s_g and greater than s_f at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88

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h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg

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