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xxMikexx [17]
3 years ago
10

Determine F12 and F21 for the following configurations: (a) A long semicircular duct with diameter of 0.1 meters: (b) A hemisphe

re-disk arrangement, with hemisphere diameter of 0.1 meters:
Engineering
1 answer:
uysha [10]3 years ago
6 0

Answer:

long duct: 1.0 and 0.424

Hemisphere 1.0 ; 0.125; 0.5

Explanation:

For a long duct:

By inspection, F_{12} = 1.0

Calculating by reciprocity, F_{21} = \frac{A_{1} }{A_{2}F_{12}  }  = \frac{2RL}{\frac{3}{4}*2\pi RL  }* 1.0\\                                                         = 0.424

Hemisphere:

By reciprocity gives = 0.125

using the summation rule: F_{21} + F_{22} + F_{23} = 1

However, because this is a hemisphere, the value will be= 0.5 * 1

                                                                                               = 0.5

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What is the value of the work interaction in this process?
Cloud [144]

Answer:

The answer is "-121\  \frac{KJ}{Kg}".

Explanation:

Please find the correct question in the attachment file.

using formula:

\to W=-P_1V_1+P_2V_2 \\\\When \\\\\to W= \frac{P_1V_1-P_2V_2}{n-1}\ \   or \ \  \frac{RT_1 -RT_2}{n-1}\\\\

W =\frac{R(T_1 -T_2)}{n-1}\\\\

    =\frac{0.287(25 -237)}{1.5-1}\\\\=\frac{0.287(-212)}{0.5}\\\\=\frac{-60.844}{0.5}\\\\=-121.688 \frac{KJ}{Kg}\\\\=-121 \frac{KJ}{Kg}\\\\

7 0
3 years ago
Air (cp = 1.005 kJ/kg·°C) is to be preheated by hot exhaust gases in a cross-flow heat exchanger before it enters the furnace. A
uysha [10]

Answer:

Q=67.95 W

T=119.83°C

Explanation:

Given that

For air

Cp = 1.005 kJ/kg·°C

T= 20°C

V=0.6 m³/s

P= 95 KPa

We know that for air

P V = m' R T

95 x 0.6 = m x 0.287 x 293

m=0.677 kg/s

For gas

Cp = 1.10 kJ/kg·°C

m'=0.95 kg/s

Ti=160°C   ,To= 95°C

Heat loose by gas = Heat gain by air

[m Cp ΔT] for air =[m Cp ΔT] for gas

by putting the values

0.677 x 1.005 ( T - 20)= 0.95 x 1.1 x ( 160 -95 )

T=119.83°C

T is the exit temperature of the air.

Heat transfer

Q=[m Cp ΔT] for gas

Q=0.95 x 1.1 x ( 160 -95 )

Q=67.95 W

7 0
3 years ago
There are four distinct events that occur in the Carnot cycle. Name the events and describe each one.
OLEGan [10]

Answer and Explanation:

There four events or stages of reversible nature in a Carnot cycle are:

1. Isothermal expansion of gas

2. Adiabatic expansion

3. Isothermal compression of gas

4.Adiabatic compression of gas

All these 4 processes are reversible processes.

1. A reversible Isothermal expansion of gas takes place in Carnot cycle where an ideal gas  absorbs or intake certain quantity of heat from a heat reservoir or source at an elevated temperature which results in the expansion of gas and work is done on the  surroundings.

2. A reversible expansion of gas adiabatically takes place in Carnot cycle in an environment with thermal insulation where the gas expand and work is done on the surrounding which results in the lowering of the temperature of the system.

3. A reversible compression of gas  isothermally takes place such that the work is done on the system by the surrounding resulting in heat loss.

4. A reversible compression of gas adiabatically, takes place in an environment with thermal insulation and the work is done continuously on the system by the surroundings resulting in the rise in temperature.

8 0
3 years ago
If a system of pulleys results in a force of 25% of the load, how far will the rope need to move to pull the load a distance of
GaryK [48]

Answer:

  40 ft

Explanation:

Assuming no loss of energy in the system of pulleys, the work done is the same whether you move the load directly or through the pulleys.

  W = Fd . . . . . . . . work is the product of force and distance

  F(10 ft) = (0.25F)(d) . . . . . where d is the distance we want to find

  d = 10F/(0.25F) = 40

The rope will need to move 40 feet.

8 0
2 years ago
Write a simple phonebook program that reads in a series of name-number pairs from the user (that is, name and number on one line
Vlad1618 [11]

Answer:

import java.util.HashMap;

import java.util.Map;

import java.util.Scanner;

public class PhoneBook {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       Map<String, String> map = new HashMap<>();

       String name, number, choice;

       do {

           System.out.print("Enter name: ");

           name = in.next();

           System.out.print("Enter number: ");

           number = in.next();

           map.put(name, number);

           System.out.print("Do you want to try again(y or n): ");

           choice = in.next();

       } while (!choice.equalsIgnoreCase("n"));

       System.out.print("Enter name to search for: ");

       name = in.next();

       if (map.containsKey(name)) {

           System.out.println(map.get(name));

       } else {

           System.out.println(name + " is not in the phone book");

       }

   }

}

6 0
3 years ago
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