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3 years ago

In python, how would I randomize numbers and insert them into a file?

1 answer:

7 0

Random integer values can be generated with the randint() function. This function takes two arguments: the start and the end of the range for the generated integer values. Random integers are generated within and including the start and end of range values, specifically in the interval [start, end].

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A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl

lawyer [7]

Answer:

110 m or 11,000 cm

Explanation:

let mass flow rate for cold and hot fluid = Mc and Mh respectively
let specific heat for cold and hot fluid = Cpc and Cph respectively
let heat capacity rate for cold and hot fluid = Cc and Ch respectively

Mc = 1.2 kg/s and Mh = 2 kg/s

Cpc = 4.18 kj/kg °c and Cph = 4.31 kj/kg °c

Using effectiveness-NUT method

First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate

Cc = Mc × Cpc = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c

Ch = Mh × Cph = 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= Cmin/Cmax

C = 5.016/8.62 = 0.582

.2 Second, we determine the maximum heat transfer rate, Qmax

Qmax = Cmin (Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Qmax = (5.016 kW/°c)(160 - 20) °c

Qmax = (5.016 kW/°c)(140) °c = 702.24 kW

.3 Third, we determine the actual heat transfer rate, Q

Q = Cmin (outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Qmax = (5.016 kW/°c)(60) °c = 303.66 kW

.4 Fourth, we determine Effectiveness of the heat exchanger, ε

ε = Q/Qmax

ε = 303.66 kW/702.24 kW

ε = 0.432

.5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger

NTU = $\\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )$

NTU = $\frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582 -1} )$

NTU = 0.661

.6 sixth, we determine Heat Exchanger surface area, As

From the question, the overall heat transfer coefficient U = 640 W/m²

As = $\frac{NTU C{min} }{U}$

As = $\frac{0.661 x 5016 W. °c }{640 W/m²}$

As = 5.18 m²

.7 Finally, we determine the length of the heat exchanger, L

L = $\frac{As}{\pi D}$

L = $\frac{5.18 m² }{\pi (0.015 m)}$

L= 109.91 m

L ≅ 110 m = 11,000 cm

3 0

3 years ago

1) Find the time in seconds to reach full charge in an RL circuit with L = 5 H and R = 100 ohms

pav-90 [236]

The time constant to reach full charge in an RL circuit is 0.05 ms.

Explanation:

To find the time constant,

The time constant for an RL circuit is defined by τ = L/R.

The given data is

L= 5 H

R= 100 ohms

by using the formula,

τ = L/R

= 5/100

= 0.05 ms

τ = 0.05 ms

Thus, the time constant to reach full charge in an RL circuit is 0.05 ms.

8 0

3 years ago

Consider the following incomplete code segment, which is intended to print the sum of the digits in num. For example, when num i

ValentinkaMS [17]

Answer:

A) while (num >= 0)

Explanation:

To understand why we need to focus on the module and division operation inside the loop. num % 10 divide the number by ten and take its remainder to then add this remainder to sum, the important here is that we are adding up the number in reverse order and wee need to repeat this process until we get the first number (1%10 = 1), therefore, num need to be one to compute the last operation.

A) this is the correct option because num = 1 > 0 and the last operation will be performed, and after the last operation, num = 1 will be divided by 10 resulting in 0 and 0 is not greater than 0, therefore, the cycle end and the result will be printed.

B) This can not be the option because this way the program will never ends -> 0%10 = 0 and num = 0/10 = 0

C) This can not be the option because num = 1 > 1 will produce an early end of the loop printing an incomplete result

D) The same problem than C

E) There is a point, before the operations finish, where sum > num, this will produce an early end of the loop, printing an incomplete result

6 0

3 years ago

What is the last step to the design process?

ryzh [129]

Answer:

B: Present solution

4 0

4 years ago

An element has two naturally occurring isotopes, isotope 1 with an atomic weight of 78.918 amu and isotope 2 with an atomic weig

Sidana [21]

Answer:

The fraction of isotope 1 is 50.7 % and fraction fraction of isotope 1 is 49.2 %.

Explanation:

Given that

Weight of isotope 1 = 78.918 amu

Weight of isotope 2 = 80.916 amu

Average atomic weight= 79.903 amu

Lets take fraction of isotope 1 is x then fraction of isotope 2 will be 1-x.

The total weight will be summation of these two isotopes

79.903 = 78.918 x + 80.916(1-x)

By solving above equation

80.916 - 79.903 = (80.916-78.918) x

x=0.507

So the fraction of isotope 1 is 50.7 % and fraction fraction of isotope 1 is 49.2 %.

4 0

4 years ago