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irga5000 [103]
3 years ago
8

Thick fluids such as asphalt and waxes and the pipes in which they flow are often heated in order to reduce the viscosity of the

fluids and thus to reduce the pumping costs. Consider the flow of such a fluid through a 100-m-long pipe of outer diameter 30 cm in calm ambient air at 0°C. The pipe is heated electrically, and a thermostat keeps the outer surface temperature of the pipe constant at 25°C. The emissivity of the outer surface of the pipe is 0.8, and the effective sky temperature is -30°C.
Required:
Determine the power rating of the electric resistance heater, in kW, that needs to be used.
Engineering
1 answer:
Gala2k [10]3 years ago
7 0

Answer:

what id the answer

Explanation:

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Instructions: For each problem, identify the appropriate test statistic to be use (t test or z-test). Then compute z or t value.
Natalka [10]

The sample of 81 students was selected with a mean score of 90, this illustrates an example of a right tailed one sample z test.

<h3>How to illustrate the sample?</h3>

From the information given, the teacher claims that the mean score of students in his class is greater than 82 with a standard deviation of 20. In this case, the right tailed one sample z test is used.

The z value will be:

= (90 - 82)/20/✓81

= 3.6

Since 3.6 > 1.645, the null hypothesis will be rejected as there's enough evidence to support the teacher's claim.

When an online medicine shop claims that the mean delivery time for medicines is less than 120 minutes with a standard deviation of 30 minutes, the left tailed one sample test is used.

The z value will be:

= (100 - 120)/(30/✓49)

= -4.66

The null hypothesis is rejected as there is enough evidence to support the claim of the medicine shop.

Learn more about sampling on:

brainly.com/question/17831271

#SPJ1

6 0
2 years ago
The cost to become a member of a fitness center is as follows: the membership fee per month is $50.00 the personal training sess
NikAS [45]

Answer:

#include <iostream>

using namespace std;

void information(){

cout<<"Welcome to the portal"<<endl;

cout<<"The membership fee per month is $50.00"<<endl;

cout<<"The personal training session fee is $30.00"<<endl;

cout<<"The senior citizens discount is 30%"<<endl;

cout<<"If the membership is bought and paid for 12 or more months, the discount is 15%;"<<endl;

cout<<"If more than five personal training sessions are bought and paid for, the discount on each session is 20%."<<endl;

}

void getInfo(bool &senior, int &months, int &personal){

cout<<"Are you senior citizen y/n ? : ";

char choice;

cin>>choice;

if(choice=='y' || choice=='Y'){

senior = true;

}else{

senior = false;

}

cout<<"Enter number of months for membership : ";

cin>>months;

cout<<"Enter number of personal training session : ";

cin>>personal;

}

double calcCost(bool senior, int months, int personal){

double cost=0;

double memberShipCost=months*50;

double trainingCost=personal*30;

if(personal>5){

trainingCost*=.80; //discount on personal training

}

if(months>=12){

memberShipCost*=.85; //discount on membership

}

cost = memberShipCost+trainingCost;

//discount for seniors

if(senior){

cost = cost*.70;

}

return cost;

}

int main() {

information();

cout<<"Select an option"<<endl;

char choice;

while(true){

cout<<endl<<endl;

cout<<"a. Calculate membership costs."<<endl;

cout<<"b. Quit program."<<endl;

cout<<"Enter your choice : ";

cin>>choice;

bool senior=true;

int months,personal;

if(choice=='a'){

getInfo(senior,months,personal);

double cost=calcCost(senior, months, personal);

cout<<"The calculated cost is $"<<cost<<endl;

}else if(choice=='b'){

break;

}

}

return 0;

}

Explanation:

8 0
4 years ago
Base course aggregate has a target dry density of 119.7 lb/cu ft in place. It will be laid down and compacted in a rectangular s
natita [175]

Answer:

total weight of aggregate =  5627528 lbs = 2814 tons  

Explanation:

given data

dry density = 119.7 lb/cu ft

area = 2000 ft × 48 ft × 6 in

aggregate = 3.1%

required compaction = 95%

solution

we get  here volume of space to be filled with aggregate that is

volume = 2000 × 48 × 0.5 = 48000 ft³

when here space fill with aggregate of density is

density = 0.95 × 119.7    = 113.72 lb/ft³

and

dry weight of this aggregate will be  is

dry weight = 48000 × 113.72 = 5458320 lbs

and

we consider take percent moisture by weigh so that there weight of moisture in aggregate is express as

weight of moisture = 0.031 × 5458320 = 169208 lbs

and

total weight of aggregate will be

total weight of aggregate = 5458320 + 169208

total weight of aggregate =  5627528 lbs = 2814 tons  

5 0
3 years ago
Under the right conditions, it is possible, due to surface tension, to have metal objects float on water. Consider placing a sho
stiv31 [10]

Answer:

D = 0.060732 in

Explanation:

given data

sp. wt. = 500 lb/ft³

diameter = 0.036 in

solution

we get here maximum diameter of rod that is express as

D = \sqrt{\frac{8 \sigma }{\pi y}}   ......................1

here \sigma surface tension of water at 60⁰f  = 5.03 × 10^{-3}  lb/ft and y = 500 lb/ft³

so put here value and we will get

D = \sqrt{\frac{8 \times 5.03 \times 10^{-3} }{\pi \times 500}}

D = 0.005061 ft

D = 0.060732 in

4 0
3 years ago
A) Describe the operation of a heat pump operating on the theoretical reversed Carnot cycle, with a neat sketch of the layout.
kicyunya [14]

Answer:

a) The operation of a heat pump involves the extraction of energy in the form of heat Q₁ from a cold source

b) The modifications required to convert a plant operating on an ideal Carnot cycle to a plant operating on a Rankine cycle involves

i) Complete condensation of the vapor at the condenser to saturated liquid for pumping to the boiler

ii) Heating of the pumped, pressurized water to the boiler pressure

Explanation:

a) 1 - 2. Wet vapor enters compressor where it undergoes isentropic compression to state 2 by work W₁₂  

2 - 3. The vapor enters the condenser at state 2 where it undergoes isobaric and isothermal condensation to a liquid with the evolution of heat  Q₂

3 - 4. The condensed liquid is expanded isentropically with the work done equal to W₃₋₄

4 - 1. At the state 4, with reduced pressure from the previous expansion, the liquid makes its way to the evaporator where it absorbs heat, Q₁, from the body to be cooled.

b. i) Complete condensation of the vapor at the condenser to saturated liquid for pumping to the boiler

Here the condensation process is modified from partial condensation to complete condensation at the same temperature which reduces the size of the pump required to pump the liquid water as opposed to pumping steam plus liquid

ii) Heating of the pumped, pressurized water to the boiler pressure

The pumped water at state 4 will be required to be heated to saturated water temperature equivalent to the boiler pressure, hence heat will need to be added at state.

Sketches of the schematic of a Basic Rankine cycle is attached  

3 0
3 years ago
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