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Sati [7]
3 years ago
13

Which physical quantity measures fraction of the input energy a machine actually concerts into output energy ?

Engineering
1 answer:
Dvinal [7]3 years ago
4 0
The answer is Kinetic Power.
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A

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Which of the following is true regarding screw gauges and shank?
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correct me if i’m wrong i’m pretty sure it’s B i’ve had the same question
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What is the weight of a steel plate in the shape of a circle with a diameter of 10'? The steel weighs 14 Ib per Ft2
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In the 5 Code of Federal Regulations (C.F.R.), it is recommended that an individual has security awareness training before s/he
GarryVolchara [31]

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Number 3 and 4

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4 years ago
Read 2 more answers
The elementary liquid-phase series reaction
liraira [26]

Answer:

Concentration of A: \frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }

Concentration of B: \frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1}  } (e^{-k_{1}t } -e^{-k_{2}t } )

Concentration of C: \frac{C_{C} }{C_{Ao} } =1+\frac{k_{1} }{k_{2}-k_{1}  } e^{-k_{2}t } -\frac{k_{2} }{k_{2}-k_{1}  } e^{-k_{1} t}

the image shows the graphs of the three concentrations

Explanation:

We have the reaction:

A ------->k1--------->B------------->k2--------->C

Each reaction:

r_{A} =-k_{1} C_{A} \\r_{B} =k_{1} C_{A} -k_{2} C_{B} \\r_{C} =k_{2} C_{C}

Where Cn is the concentration of each specie (A,B,C)

The mass balance for A:

-\frac{dC_{A} }{dt} =-r_{A} \\-\frac{dC_{A} }{dt}=k_{1} C_{A} \\-\int\limits^y_x {\frac{dC_{A} }{dt} } \,=k_{1} t\\\frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }

Where x=CAo and y=CA

The mass balance for B:

-\frac{dC_{B} }{dt} =-r_{B} \\-\frac{dC_{B} }{dt}=k_{2} C_{B} -k_{1} C_{A} \\\frac{dC_{B} }{dt}+k_{2} C_{B}=k_{1} C_{A}\\\frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1}  } (e^{-k_{1}t }-ex^{-k_{2}t }  )

The mass balance for C:

\frac{C_{C} }{C_{Ao} } =1-\frac{C_{A} }{C_{Ao} } -\frac{C_{B} }{C_{Ao} } \\\frac{C_{C} }{C_{Ao} }=1+\frac{k_{1} }{k_{2}-k_{1}  } e^{-k_{2} t}-\frac{k_{2} }{k_{2}-k_{1}  }  e^{-k_{1}t }

The maximum concentration of C is:

C_{Cmax} =C_{Ao} (\frac{k_{2} }{k_{1} } )^{\frac{k_{2} }{k_{2}-k_{1}  }}  =1.6(\frac{0.01}{0.4} )^{\frac{0.01}{0.01-0.4} } =1.76mol/dm^{3}

and the maximum time is:

t_{max} =\frac{ln\frac{k_{2} }{k_{1} } }{k_{2}-k_{1}  } =\frac{ln\frac{0.01}{0.4} }{0.01-0.4} =9.4 h

6 0
3 years ago
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