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MrRissso [65]
3 years ago
13

There are 22 gloves in a drawer: 5 pairs of red gloves, 4 pairs of yellow, and 2 pairs of green. You select the gloves in the da

rk and can check them only after a selection has been made. What is the smallest number of gloves you need to select to have at least one matching pair in the best case?
Engineering
1 answer:
antiseptic1488 [7]3 years ago
3 0

Answer:

Best case = 2 gloves

Given Information:

Red gloves = 5 pairs

Yellow gloves = 4 pairs

Green gloves = 2 pairs

smallest number of gloves you need to select to have at least one matching pair in the best case = ?

Explanation:

How many gloves do you need to make one matching pair?

2

Yes, you are right. 2 gloves makes a matching pair and it is the smallest number of gloves you need to select to have at least one matching pair.

But what about worst case?

lets say

you tried all 5 red gloves either all of them were left or right

then you tried 4 yellow gloves either all of them were left or right

then you tried 2 green gloves either all of them were left or right

Now all the left or right gloves are tried (5 + 4 + 2 = 11) and the 12th one will definitely be either matching red, yellow or green.

Therefore, in the worst case scenario, the smallest number of gloves you need to select to have at least one matching pair is 12.

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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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A two-bus power system is interconnected by one transmission line. Bus 1 is a generator bus with specified terminal voltage magn
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Answer:

a) | v2 | ,  β2   ( load, bus voltage at bus 2 )

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b) q2 , β2  

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Explanation:

Attached below is a schematic representation of the solution

<u>a) Identify the variables in the solution vector assume Bus 2 is a load bus</u>

The specified parameters are ; P2 and q2

while | v2 | and β2 are not specified

given that bus 2 is a load bus, bus 1 is a slack bus with ; | v1 |  and β1 been specified while p1 and q1 are not specified

<em>Hence the variables in the solution </em>

<em>= | v2 | ,  β2   ( load, bus voltage at bus 2 )</em>

<em>   p1 ,  q1 ( slack, bus power at bus 1 ) </em>

<u>b) Identify the variables in the solution vector ( assume Bus 2 is a PV bus )</u>

specified at Bus 2 are ; | p2 | , | v2 |

unspecified : q2 , β2

Bus 1 ( still a slack bus )

specified parameter : | v1 |  and β1

unspecified : p1 and q1

<em>Hence the variables in the solution </em>

<em>= q2 , β2  </em>

<em>   p1 and q1 ( slack, bus power at bus 1 ) </em>

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You can safely place a jack on a floor pan to keep a vehicle steady.
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Answer: Yes

Explanation:

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