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Jlenok [28]
3 years ago
13

Water at 20oC, with a free-stream velocity of 1.5 m/s, flows over a circular pipe with diameter of 2.0 cm and surface temperatur

e of 80oC. Calculate the average heat transfer coefficient and the heat transfer rate per meter length of pipe.\
Engineering
1 answer:
Lisa [10]3 years ago
3 0

Answer:

Average heat transfer coefficient =  31 kw/m^2 k

Heat transfer rate per meter length of pipe =  116.808 KW

Explanation:

water temperature = 20⁰c,  

free-stream velocity = 1.5 m/s

circular pipe diameter = 2.0 cm = 0.02 m

surface temperature = 80⁰c

A) calculate average heat transfer coefficient

we apply the formula below :

m = αAv

A (area) = \pi /4 (d)^2

m = 10^3 * \pi  / 4 ( 0.02)^2 * 1.5

   = 10^3 * 0.7857( 0.0004) * 1.5

   = 0.4714 kg/s

Average heat transfer coefficient  

h = \frac{m(cp)}{A}  ,  A = \pi DL

L = 1 m , m = 0.4714 kgs , cp = 4.18

back to equation

h = \frac{0.4714*4.18}{\pi * 0.02 }   = 1.970 / 0.0628 = 31.369 ≈ 31 kw/m^2 k

B) Heat transfer rate per meter length of pipe

Q = ha( ΔT ),  a = \pi DL

   = 31 * 0.0628 * ( 80 - 20 )

  = 31 * 0.0628 * 60 = 116.808 KW

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nikitadnepr [17]

Answer:

Poissons ratio = -0.3367

Explanation:

Poissons ratio = Lateral Strain / Longitudinal Strain

In this case, the longitudinal strain will be:

Strain (longitudinal) = Change in length / total length

Strain (longitudinal) = (8.40392 - 8.4) / 8.4

Strain (longitudinal) = 4.666 * 10^(-4)

While the lateral strain will be:

Strain (Lateral) = Change in length / total length

Strain (Lateral) = (2.09967 - 2.1) / 2.1

Strain (Lateral) = -1.571 * 10^(-4)

Solving the poisson equation at the top we get:

Poissons ratio = -1.571 / 4.666                                     <u>( 10^(-4) cancels out )</u>

Poissons ratio = -0.3367

6 0
4 years ago
Which of these passages is an example of cohesive writing?
Oksana_A [137]
The answer mostly likely is: “1” due to the fact it is well ordered, and follows a similar thought process, best of luck!
3 0
3 years ago
Read 2 more answers
6.27 LAB: Convert to binary - functions Write a program that takes in a positive integer as input, and outputs a string of 1's a
Olin [163]

Answer:

//Convert any decimal number to binary number

//Program is written in C++ Programming Language

// Comments are used for explanatory purpose

// Program starts here

#include <iostream>

using namespace std;

// Main Method declared here

int main()

{

    int x;

   cout<<"Enter any integer number: ";

    cin>>x;

   DecBin(x);

   return 0;

}

// Here a function named DecBin is declared along with an integer variable, x

void DecBin(int x)

{

   // Declare an array to store the resulting binary digits

   int bindigit[32];

   // counter for binary array

   int kount = 0;

   while (x > 0) {

       // Store the remainder of each division in the declared array

       bindigit[kount] = x % 2;

       x = x / 2;

       kount++;

   }

   // Loop to print the binary digits in reverse order

   for (int j = i - 1; j >= 0; j--)

          {

       cout << bindigit[j];

        }

}

// End of Program

4 0
3 years ago
Poly(cis-1,4-isoprene), or natural rubber (NR), has a tendency crystallize. The Tm of this polymer is slightly below room temper
Nonamiya [84]

Answer:

Explanation:

Crystalline melting temperature (Tm) is termed as the temperature required for a  crystalline polymer to change to a fluid or glasslike crystalline spaces of a semi-crystalline polymer liquefy (expanded sub-atomic movement).  

Crystallization of polymers is an interaction related with incomplete arrangement of their atomic and molecular chains. These chains crease together and structure requested districts called lamellae, which form bigger spheroidal designs named spherulites. Polymers can solidify after cooling from melting, mechanical extending, or dissolvable dissipation. Crystallization influences the optical, mechanical, and synthetic chemical properties of the polymer.  

For a crystalline polymer, a required polymer chain is present in or goes along a few crystalline and amorphous zones. The crystalline zones are comprised of intermolecular & intramolecular arrangements or deliberate and thus firmly stuffed plan of atoms or chain fragments, and an absence of it brings about the development of amorphous zones.  

The mechanical property boundary, for example, shear modulus expansions in the temperature of perception for polymer material framework.  

The temperature reaction of direct linear polymers might be seen as partitioned into three particularly separate fragments:  

1. Above Tm: The polymer stays as fluid whose consistency & viscosity would rely upon atomic molecular weight and temperature.  

2. Between Tm and Tg: This area may go between close to 100% crystalline & 100% amorphous chain atomic bunches relying upon the polymer underlying consistency. The amorphous part carries on similar to supercooled fluid in this section. The generally actual conduct of the polymer in this moderate portion is similar to an elastic rubber.  

3.Below Tg: The polymer material saw as glass is hard and inflexible, showing and emanating a predetermined coefficient of thermal extension. The glass is more like a crystalline strong than the fluid in personal conduct standard regarding mechanical property boundaries. In regard to the molecular atomic request, in any case, the glass all the more intently takes after the fluid. There is little contrast between the direct linear and cross-connected polymer beneath Tg.

8 0
3 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
3 years ago
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