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kompoz [17]
3 years ago
15

A tension test is carried out on an Al alloy specimen which has an original diameter of 0.505 in and an original gauge length of

2 in. After fracture, the diameter at the fractured neck is measured to be 0.37 in and the gauge length after fracture is 2.32 in. The maximum load during the test was 37 kN whereas at fracture the load was 34 kN.
a. What is the tensile strength?
b. What are the engineering and true stresses at fracture?
c. What are the engineering and true strain at fracture?
d. What is the ductility of this metal?

Engineering
1 answer:
Contact [7]3 years ago
5 0

Answer:

Detailed solution is given in attached image

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Irina-Kira [14]

The important difference between wiring panels in series or in parallel is that electrically it affects the voltage and amperage of the resultant circuit.

In a series circuit you sum the voltage of each panel to get the overall voltage of the array. However, the amperage of the overall circuit stays the same stays the same.

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Calculate the reactions at 4 ends (supports) of this bookshelf. Assume that the weight of each book is approximately 1 lb. The w
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Answer:

R = 4.75 lb  (↑)

Explanation:

Number of books = n = 19

Weight of each book = W = 1 lb

Length of the bookshelf = L = 40 inches

We can get the value of the distributed load as follows

q = n*W/L = 19*1 lb/ 40 inches = 0.475 lb/in

then the reactions at 4 ends (supports) of the bookshelf are

R = (q/2)/2 = 4.75 lb

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Hot carbon dioxide exhaust gas at 1 atm is being cooled by flat plates. The gas at 220 °C flows in parallel over the upper and l
sergeinik [125]

The local convection heat transfer coefficient at 1 m from the leading edge is  0.44 \frac{W}{m^{2} \times K} ,  the average convection heat transfer coefficient over the entire plate is  0.293 \frac{W}{m^{2} \times K}and the total heat flux transfer to the plate is 61.6 KJ.

Explanation:

It is case of heat and mass transfer in which due to temperature difference between gas  and surface. Further temperature  boundary layer will developed on flat plate in longitudinal direction.  

Hot carbon dioxide exhaust gas

physical properties

r= 1.05 \frac{kg}{m^{3}}

c_p = 1.02 \frac{kJ}{Kg \times K}

m= 231 \times 10^{7}  \frac{N \times s }{m^2}

υ = 21.8 \times 10^{6}  \frac{m^2}{s}

k = 32.5 \times 10^{3} \frac{W}{m \times K}

\alpha = 30.1 \times 10^{6} \frac{m^{2}}{s}

Pr = 0.725

Apart from these other data arr given below,

v= 3 \frac{m}{s}  \\ p= 1 atm \\ L_c = 1.5m \\T_g= 220 C \\ T_s = 80 C

To find the local convection heat transfer coefficient at 1 m from the leading edge, we use correlation used for laminar flow over flat plate,

Nu = \frac{ h \times L }{k}  = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })

where h= Average heat transfer coefficient

           L= Length of a plate

           k= Thermal Conductivity of carbon dioxide

           Re = Reynold's Number

           Pr  = Prandtle Number

(a) Convection heat transfer coefficient at 1 m from the leading edge

    is referred as local convection heat transfer coefficient.

   

   To find convection heat transfer coefficient at 1 m from leading edge,

  Nu = \frac{ h_local \times L }{k}  = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })

  Here, first we have to find Re and Pr,

   Re = \frac{r \times v \times L}{m}

   Re = \frac{1.0594 \times 3 \times 1}{231 \times 10^{7}}

   Re = 20.63 \times  10^{-10}

   Pr number is take from physical property data and Pr is 0.725.

   Putting value of Re and Pr in main equation,

   we get

   Nu = \frac{ h_local \times 1 }{32.5 \times 10^{3}}  = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })

    h_local   = 32.5 \times 10^{3} \times  0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })

    h_local   =  0.44 \frac{W}{m^{2} \times K}

(b)  To find average convection heat transfer coefficient,

      it can be find out as case (a), only difference is that instead of L=1 m,        L=1.5 m would come,  

   Therefore,

    Nu = \frac{ h \times 1.5 }{32.5 \times 10^{3}}  = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })

    Finally,

      h  = \frac{0.44}{1.5}

      h  = 0.293 \frac{W}{m^{2} \times K}

(C) Total heat flux transfer to the plate is found out by,

     Q = h \times (T_g - T_s)

     Q = 0.293 \times (220-80) \\ Q= 0.293 \times 140  \\ Q= 61.6 KJ

     

     

   

   

     

   

     

   

   

 

   

   

   

   

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