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weeeeeb [17]
2 years ago
4

A capillary-tube viscometer is being selected to measure viscosity of a liquid food. The maximum viscosity to be measured will b

e 230 cP, and the maximum flow rate that can be measured accurately is 0.015 kg/min. If the tube length is 10 cm and a maximum pressure of 25 Pa can be measured, determine the tube diameter to be used. The density of the product is 1000 kg/m3.
Engineering
1 answer:
Pavlova-9 [17]2 years ago
8 0

Answer:

The diameter to be used is 0.98 cm

Explanation:

The Poiseuille´s law is equal to:

Q=\frac{\pi Pr^{4} }{8\eta L}

Where

Q = flow rate = 0.015 kg/min = 2.5x10⁻⁷m³/s

P = pressure difference = 25 Pa

r = radius

η = viscosity = 230 cP = 0.23 Pa s

L = length of the tube = 10 cm = 0.1 m

Clearing r:

r=\sqrt[4]{\frac{8\eta LQ}{\pi P} } =\sqrt[4]{\frac{8*0.23*0.1*2.5x10^{-7} }{\pi *25} } =0.0049m

The diameter is:

d=2r=2*0.0049=0.0098m=0.98cm

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Evan notices a small fire in his workplace. Since the fire is small and the atmosphere is not smoky he decides to fight the fire
Norma-Jean [14]

Answer:

not calling the firemean

Explanation:

7 0
2 years ago
Consider insulation on a circular pipe For the same thickness and type of insulation, the thermal resistance of the insulation i
leonid [27]

Answer:

b). The same for all pipes independent of the diameter

Explanation:

We know,

R_{conduction}=\frac{ln(\frac{r_{2}}{r_{1}})}{2\pi LK}

R_{convection}=\frac{1}{h(2\pi r_{2}L)}

From the above formulas we can conclude that the thermal resistance of a substance mainly depends upon heat transfer coefficient,whereas radius has negligible effects on heat transfer coefficient.

We also know,

Factors on which thermal resistance of insulation depends are :

1. Thickness of the insulation

2. Thermal conductivity of the insulating material.

Therefore from above observation we can conclude that the thermal resistance of the insulation is same for all pipes independent of diameter.

5 0
2 years ago
An 80-L vessel contains 4 kg of refrigerant-134a at a pressure of 160kPa. Determine (a) the temperature, (b) the quality, (c) th
makvit [3.9K]

Answer:

temperature -15.6 C, quality x=0.646, enthalpy h=667.20 KJ, volume of vapor phase Vg= 79.8 L

Explanation:

property table for R-134a

https://www.ohio.edu/mechanical/thermo/property_tables/R134a/R134a_PresSat.html

at 160 KPa , temperature = -15.66 C

quality x=mass of vapour/ total mass of liq-vap mixture

alternaternately: x=(v-vf)/(vg-vf)    

v=total volume i.e. volume of container"80L"   80L=0.08 cubic meter

vf=vol of liquid phase  vg=vol of vapor phase vf, vg values at 160Kpa

x=(0.08-0.0007437)/(0.1235-0.0007437)=0.646

enthalpy

h=hf+xhfg          hf, hfg values at 160Kpa

h=hf+xhfg=31.2+0.646(209.9)=166.80 KJ/Kg

for 4Kg R-134a h=m(166.80 KJ/Kg )=667.20 KJ

volume of vapor phase

vg at 160Kpa=0.1235 cubic meter for quality=1.

in this case quality=0.646 , so it will occupy 64.6% space of the vapor phase at quality=1.

vol. of vapor phase=0.646*0.1235=0.0798 cubic meter=79.8 L

7 0
2 years ago
If water molecules pass through a membrane with a steady state flux of 220 mole/(m2 day), how long will it take, in hours, for 0
goblinko [34]

Answer:

<em>0.0386 hr</em>

<em></em>

Explanation:

Area = 565 cm^2 = 0.0565 m^2  (1 cm^2 = 0.0001 m^2)

flux state rate = 220 mole/m^2-day

<em>There are 24 hrs in a day,</em> therefore rate in hrs will be

220/24 = 9.17 mole/m^2-hr

mass of water = 0.4 kg

molar mass of water = (1 x 2) + 16 = 18 kg/mole

therefore,

<em>mole of water = mass of water/molar mass of water</em>

mole of water = 0.4/18 = 0.02 mole

<em>mole flux = mole/area</em> = 0.02/0.0565 = 0.354 mol/m^2

<em>time that will be taken will be for water to pass = mole flux/mole flux rate</em>

time = 0.354/9.17 = <em>0.0386 hr</em>

7 0
2 years ago
Q3: Summation Write a recursive implementation of summation, which takes a positive integer n and a function term. It applies te
harina [27]

Answer:

Here is the recursive function summation:

def summation(n, term):      

   if n == 1:  

       return term(n)

   else:

       return term(n) + summation(n - 1, term)

Explanation:

The function summation() has two arguments where n is a positive integer and term is a function term. term has the lambda function which is a small function having an argument and an expression e.g lambda b: b+20

So the summation() function is a recursive function which returns sum of the first n terms in the sequence defined by term ( a lambda function).

If you want to check if this function works, you can call this function by passing values to it like given in the question.

summation(5, lambda x: 2**x)

Here the value of n is 5 and the term is a lambda function x: 2**x

If you want to see the results of this function on output screen then use:

print(summation(5, lambda x: 2**x))

The print() function will print the results on screen.

This returns the sum of first 5 terms in sequence defined in the function x: 2**x

In recursive methods there are two cases: base case and recursive case. Base case is the stopping case which means that the recursion will stop when the base case/ base condition evaluates to true. The recursive case is when the function keeps calling itself so the recursive function keepsexecuting until the base case becomes true.

Here the base case is if n == 1:  So the recursive function calling itself until the value of n becomes 1.  

Recursive case is:

       return term(n) + summation(n - 1, term)

For the above example with n= 5 and term = x:2**x the recursions starts from n and adds all the terms of the series one by one and the value of n keeps decrementing by 1 at every recursive call.

When the value of n is equal to 1 the base case gets true and the recursion ends and the result of the sum is displayed in output.

This is how the summation() function works for the above function call:

2^1 + 2^2 + 2^3 + 2^4 + 2^5

n is 5 So this term function is called recursively 5 times and at every recursive call its value decreases by 1. Here the term function is used to compute 2 raise to power n. So in first recursive call the 2 raise to the power 5 is computed, then 5 is decremented and then in second recursive call to summation(), 2 raise to the power 4 is calculated, in third recursive call  to summation(), 2 raise to the power 3 is calculated, in fourth recursive call  to summation(), 2 raise to the power 2 is calculated, in fifth recursive call  to summation(), 2 raise to the power 1 is calculated, then the base condition is reached as n==1. So the recursion stops and the sum of the above computed power function results is returned which is 62.

2^1 + 2^2 + 2^3 + 2^4 + 2^5 = 62

The screen shot of recursive function along with the output of explained examples is attached.

6 0
2 years ago
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