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elixir [45]
3 years ago
5

In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by

the correlation Nux=0.035Rex^(0.8) Pr^(1/3) Determine the ratio of the average convection heat transfer coefficient (h) over the entire plate length to the local convection heat transfer coefficient (hx) (h/hx = L) at x = L.
Engineering
1 answer:
Stells [14]3 years ago
8 0

Answer:

\dfrac{\bar{h}}{h}=\dfrac{5}{4}

Explanation:

Given that

Nu_x=0.035Re_x^{0.8} Pr^{1/3}

We know that

Rex=ρvx/μ

So

Nu_x=0.035Re_x^{0.8} Pr^{1/3}

Nu_x=0.035\times\left(\dfrac{\rho vx}{\mu}\right)^{0.8}Pr^{1/3}

All other quantities are constant only x is a variable in the above equation .so lets take all other quantities as a constant C

Nu_x=C.x^{0.8}=C.x^{4/5}

We also know that

Nux=hx/K

C.x^{4/5}=\dfrac{hx}{k}

m is the constant

h=mx^{-1/5}

This is local heat transfer coefficient

The average value of h given as

\bar{h}=\dfrac{\int_{0}^{L}hdx}{L}

\bar{h}=\dfrac{5m}{4}\times\dfrac{L^{4/5}}{L}

\bar{h}=\dfrac{5m}{4}L^{-1/5}             ---------1

The value of local heat transfer coefficient at x=L

h=mx^{-1/5}

h=mL^{-1/5}            -----------2

From 1 and 2 we can say that

\dfrac{\bar{h}}{h}=\dfrac{5}{4}

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Write a method called letterCount that takes two String arguments, one containing some text and the other containing a single le
jenyasd209 [6]

Answer:

I am writing a Python program.  Here is the function letterCount. This function takes two string arguments text and letter and return count of all occurrences of a letter in the text.

def letterCount(text, letter):  

 count = 0  # to count occurrences of letter in the text string

 for char in text:  # loop moves through each character in the text

   if letter == char: # if given letter matches with the value in char

     count += 1  # keeps counting occurrence of a letter in text

 return count # returns how many times a letter occurred in text

   

Explanation:

In order to see if this function works you can check by calling this function and passing a text and a letter as following:

print(letterCount('apples are tasty','a'))

Output:

3

Now lets see how this function works using the above text and letter values.

text = apples are tasty

letter = a

So the function has to compute the occurrences of 'a' in the given text 'apples are tasty'.

The loop has a variable char that moves through each character given in the text (from a of apples to y of tasty) so it is used as an index variable.

char checks each character of the text string for the occurrence of letter a.

The if condition checks if the char is positioned at a character which matches the given letter i.e. a. If it is true e.g if char is at character a of apple so the if condition evaluates to true.

When the if condition evaluates to true this means one occurrence is found and this count variable counts this occurrence. So count increments every time the occurrence of letter a is found in apples are tasty text.

The loop breaks when every character in text is traversed and finally the count variable returns all of the occurrences of that letter (a) in the given text (apples are tasty). As a occurs 3 times in text so 3 is returned in output.

The screen shot of program along with output is attached.

4 0
3 years ago
Wattage is:
Ksju [112]

Answer:

c.Both A and B.

Explanation:

the wattage is c and d

7 0
3 years ago
9. A box contains (4) red balls, and (7) white balls ,we draw( two) balls with return , find 1. Show the sample space & n(s)
zzz [600]

Answer:

The answers to your questions are given below.

Explanation:

The following data were obtained from the question:

Red (R) = 4

White (W) = 7

1. Determination of the sample space, S.

The box contains 4 red balls and 7 white balls. Therefore, the sample space (S) can be written as follow:

S = {R, R, R, R, W, W, W, W, W, W, W}

nS = 11

2. Determination of the probability of all results that appeared in the sample space.

From the question, we were told that the two balls was drawn with return. There, the probability of all results that appeared in the sample space can be given as follow:

i. Probability that the first draw is red and the second is also red.

P(R1) = nR/nS

Red (R) = 4

Space space (S) = 11

P(R1) = nR/nS

P(R1) = 4/11

P(R2) = nR/nS

P(R2) = 4/11

P(R1R2) = P(R1) x P(R2)

P(R1R2) = 4/11 x 4/11

P(R1R2) = 16/121

Therefore, the Probability that the first draw is red and the second is also red is 16/121.

ii. Probability that the first draw is red and the second is white.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(R) = nR/nS

P(R) = 4/11

P(W) = nW/nS

P(W) = 7/11

P(RW) = P(R) x P(W)

P(RW) = 4/11 x 7/11

P(RW) = 28/121

Therefore, the probability that the first draw is red and the second is white is 28/121.

iii. Probability that the first draw is white and the second is also white.

White (W) = 7

Space space (S) = 11

P(W1) = nW/nS

P(W1) = 7/11

P(W2) = nW/n/S

P(W2) = 7/11

P(W1W2) = P(W1) x P(W2)

P(W1W2) = 7/11 x 7/11

P(W1W2) = 49/121

Therefore, the probability that the first draw is white and the second is also white is 49/121.

iv. Probability that the first draw is white and the second is red.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(W) = nW/nS

P(W) = 7/11

P(R) = nR/nS

P(R) = 4/11

P(WR) = P(W) x P(R)

P(WR) = 7/11 x 4/11

P(WR) = 28/121

Therefore, the probability that the first draw is white and the second is red is 28/121.

7 0
3 years ago
Identify the correct statements in the context of friction factors of laminar and turbulent flows
soldi70 [24.7K]

Answer:

a) In turbulent flow, the tubes with rough surfaces have much higher friction factors than the tubes with smooth surfaces.

Explanation:

Turbulent flow is a type of fluid flow in which fluid will undergo irregular fluctuations. The tubes with rough surfaces have higher friction factors than the tubes with smooth surfaces. In laminar flow the effect of effect of surface roughness is negligible on friction factors.

5 0
3 years ago
As car production became more efficient, cars became more affordable for american families. One result of this increase in autom
tankabanditka [31]

Answer:

IT IS SIMPLE LIKE HENRY FORD MADE THIS OPTION POSSIBLE

Explanation:

5 0
2 years ago
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