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Ira Lisetskai [31]
3 years ago
15

Saturated steam coming off the turbine of a steam power plant at 30 °C condenses on the outside of a 3-cm-outer-diameter, 35-m-l

ong pipe at a rate of 45 kg/h. Determine the rate of heat transfer from the steam to the cooling water flowing through the pipe.
Engineering
1 answer:
saw5 [17]3 years ago
5 0

Answer:

The answer 30.07J/sec

Explanation:

GIVEN DATA:

temperature of steam = 40 degree celcius

mass flow rate = 63 kg/h

from saturated water tables,

from temperature 40 °, enthalpy of evaporation value is 2406 kj/kg

rate of heat transfer (Q) can be determine by using following relation

putting all value to get Q value

Q = 45 *2406

Q = 108270 kJ/h

Q = 30.07 kJ/sec

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This program will store roster and rating information for a soccer team. Coaches rate players during tryouts to ensure a balance
Flauer [41]

Answer:

#include <iostream>

#include <vector>

using namespace std;

int main() {

   vector<int> jerseyNumber;

   vector<int> rating;

   int temp;

   for (int i = 1; i <= 5; i++) {

       cout << "Enter player " << i

            << "'s jersey number: ";

       cin >> temp;

       jerseyNumber.push_back(temp);

       cout << "Enter player " << i

            << "'s rating: ";

       cin >> temp;

       rating.push_back(temp);

       cout << endl;

   }

   cout << "ROSTER" << endl;

   for (int i = 0; i < 5; i++)

       cout << "Player " << i + 1 << " -- "

            << "Jersey number: " << jerseyNumber.at(i)

            << ", Rating: " << rating.at(i) << endl;

   char option;

   '

   while (true) {

       cout << "MENU" << endl;

       cout << "a - Add player" << endl;

       cout << "d - Remove player" << endl;

       cout << "u - Update player rating" << endl;

       cout << "r - Output players above a rating"

            << endl;

       cout << "o - Output roster" << endl;

       cout << "q - Quit" << endl << endl;

       cout << "Choose an option: ";

       cin >> option;

       switch (option) {

           case 'a':

           case 'A':

               cout << "Enter a new player's"

                    << "jersey number: ";

               cin >> temp;

               jerseyNumber.push_back(temp);

               cout << "Enter the player's rating: ";

               cin >> temp;

               rating.push_back(temp);

               break;

           case 'd':

           case 'D':

               cout << "Enter a jersey number: ";

               cin >> temp;

               int i;

               for (i = 0; i < jerseyNumber.size();

                    i++) {

                   if (jerseyNumber.at(i) == temp) {

                       jerseyNumber.erase(

                               jerseyNumber.begin() + i);

                       rating.erase(rating.begin() + i);

                       break;

                   }

               }

               break;

           case 'u':

           case 'U':

               cout << "Enter a jersey number: ";

               cin >> temp;

               for (int i = 0; i < jerseyNumber.size();

                    i++) {

                   if (jerseyNumber.at(i) == temp) {

                       cout << "Enter a new rating "

                            << "for player: ";

                       cin >> temp;

                       rating.at(i) = temp;

                       break;

                   }

               }

               break;

           case 'r':

           case 'R':

               cout << "Enter a rating: ";

               cin >> temp;

               cout << "\nABOVE " << temp << endl;

               for (int i = 0; i < jerseyNumber.size();

                    i++)

                   if (rating.at(i) > temp)

                       cout << "Player " << i + 1

                            << " -- "

                            << "Jersey number: "

                            << jerseyNumber.at(i)

                            << ", Rating: "

                            << rating.at(i) << endl;

               break;

           case 'o':

           case 'O':

               cout << "ROSTER" << endl;

               for (int i = 0; i < jerseyNumber.size();

                    i++)

                   cout << "Player " << i + 1 << " -- "

                        << "Jersey number: "

                        << jerseyNumber.at(i) << ", Rating: "

                        << rating.at(i) << endl;

               break;

           case 'q':

               return 0;

           default:

               cout << "Invalid menu option."

                    << " Try again." << endl;

       }

   }

}

Explanation:

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.

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3 years ago
Name the main classes of polymer and define their characteristic properties
Svetlanka [38]

Answer:

Polymers are the naturally occurring or synthetic macromolecules that are composed of repeating subunits, called monomers.

The three main classes of polymers are: thermoplastic, thermosetting, and the elastomers.

Thermoplastic polymers have linear bonding. These polymers can be melted again and thus can recycled.

Thermosetting polymers have cross-linked bonding. These polymers decompose when heated and thus can not be remelted and recycled.

Elastomers have linear bonding with some cross-linking. These polymers extreme elastic extensibility and thus can revert back to its original shape after deformation, without causing any permanent damage.

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Nataly [62]

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Metal

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Metal

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