or a bronze alloy, the stress at which plastic deformation begins is 267 MPa and the modulus of elasticity is 115 GPa. (a) What is the maximum load (in N) that may be applied to a specimen having a cross-sectional area of 164 mm2 without plastic deformation

Answers

Answer 1

Answer:

Answer:

The maximum load (in N) that may be applied to a specimen with this cross sectional area is $F=43788 N$

Explanation:

We know that the stress at which plastic deformation begins is 267 MPa.

We are going to assume that the stress is homogeneously distributed. As a definition, we know that stress (P) is force (F) over area(A), as follows:

$P=(F)/(A)$ (equation 1)

We need to find the maximum load (or force) that may be applied before plastic deformation. And the problem says that the maximum stress before plastic deformation is 267 MPa. So, we need to find the value of load when P= 267 MPa.

Before apply the equation, we need to convert the units of area in $m^(2)$ . So,

$A[m^(2)]=A[mm^(2) ]((1 m)/(1000 mm)) ^(2)\nA[m^(2)]=164 mm^(2)((1 m)/(1000 mm))^(2) \nA[m^(2)]=0.000164 m^(2)$

And then, from the equation 1,

$F=PA\nF=(267 MPa)(0.000164 m^(2))\nF=(267x10^(6) Pa)(0.000164 m^(2))\nF=(267x10^(6) N/m^(2))(0.000164 m^(2))\nF=43788 N$

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Answers

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Answers

Answer:

☐ NE-SW

Explanation:

Based on the description, the rock direction is North East - South West (NE-SW). Rocks generally can expand or compress depending on the type and magnitude of stress applied on the rocks. However, if the applied stress is sufficiently high, cracks and fractures will be created on the rock and it can ultimately lead to the formation of particles.

Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 39 m3/min and exits at 12 bar, 400 K. Heat transfer occurs at a rate of 6.5 kW from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in kW.

Answers

Answer:

The power input, in kW is -86.396 kW

Explanation:

Given;

initial pressure, P₁ = 1.05 bar

final pressure, P₂ = 12 bar

initial temperature, T₁ = 300 K

final temperature, T₂ = 400 K

Heat transfer, Q = 6.5 kW

volumetric flow rate, V = 39 m³/min = 0.65 m³/s

mass of air, m = 28.97 kg/mol

gas constant, R = 8.314 kJ/mol.k

R' = R/m

R' = 8.314 /28.97 = 0.28699 kJ/kg.K

Step 1:

Determine the specific volume:

p₁v₁ = RT₁

$v_1 = (R'T_1)/(p_1) = ((0.28699.(kJ)/(kg.K) )(300 k))/((1.05 bar *\ (10^5 N/m^2)/(1 bar) *(1kJ)/(1000N.m) )) \n\nv_1 = 0.81997 \ m^3/kg$

Step 2:

determine the mass flow rate; m' = V / v₁

mass flow rate, m' = 0.65 / 0.81997

mass flow rate, m' = 0.7927 kg/s

Step 3:

using steam table, we determine enthalpy change;

h₁ at T₁ = 300.19 kJ /kg

h₂ at T₂ = 400.98 kJ/kg

Δh = h₂ - h₁

Δh = 400.98 - 300.19

Δh = 100.79 kJ/kg

step 4:

determine work input;

W = Q - mΔh

Where;

Q is heat transfer = - 6.5 kW, because heat is lost to surrounding

W = (-6.5) - (0.7927 x 100.79)

W = -6.5 -79.896

W = -86.396 kW

Therefore, the power input, in kW is -86.396 kW

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Answers

Answer:

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Explanation:

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