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Gelneren [198K]
3 years ago
6

Consider a hot working operation in which there is no work hardening, but the flow stress is sensitive to the strain rate. Calcu

late the maximum strain rate that can be used in the deformation if the flow stress cannot exceed 210 MPa. The strength constant is 175 MPa, and the strain rate sensitivity exponent is 0.1. Assume that there is no effect of friction with the tooling. (2 points)
Engineering
1 answer:
alexandr402 [8]3 years ago
6 0

Explanation:

You can use Hollomon's equation to estimate this.

Y_f is the flow stress.

K is the strenght coefficient or constant.

e is the strain

n is the strain hardening exponent.

You should be able to solve for e given the data in the problem statement.

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Entropy change is evaluated using Eq. 6.2a based on an internally reversible process. Can the entropy change between two states
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3 years ago
A steam power plant operating on a simple ideal Rankine cycle maintains the boiler at 6000 kPa, the turbine inlet at 600 °C, and
ohaa [14]

Answer:

ηa=0.349

ηb=0.345

Explanation:

The enthalpy and entropy at state 3 are determined from the given pressure and temperature with data from table:

h_{3}=3658.8kJ/kg\\ s_{3}=7.1693kJ/kg

The quality at state 4 is determined from the condition  s_{4} =s_{3} and the entropies of the components at the condenser pressure taken from table:

 q_{4} =\frac{s_{4}-s_{liq50}  }{s_{evap,50} } \\=\frac{7.1693-1.0912}{6.5019}=0.935

The enthalpy at state 4 then is:  

h_{4} =h_{liq50} +q_{4} h_{evap,50}\\ (340.54+0.935*2304.7)kJ/kg\\=2495.43kJ/kg\\

Part A

In the case when the water is in a saturated liquid state at the entrance of the pump the enthalpy and specific volume are determined from A-5 for the given pressure:  

h_{1}=340.54kJ/kg\\ \alpha _{1}=0.00103m^3/kg

The enthalpy at state 2 is determined from an energy balance on the pump:

h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1}  )

    =346.67 kJ/kg

The thermal efficiency is then determined from the heat input and output in the cycle:  

=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.349

Part B  

In the case when the water is at a lower temperature than the saturation temperature at the condenser pressure we look into table and see the water is in a compressed liquid state. Then we take the enthalpy and specific volume for that temperature with data from  and the saturated liquid values:  

h_{1}=293.7kJ/kg\\ \alpha _{1}=0.001023m^3/kg

The enthalpy at state 2 is then determined from an energy balance on the pump:  

h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1}  )

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The thermal efficiency in this case then is:  

=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.345

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Answer:

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Explanation:

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