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LuckyWell [14K]
3 years ago
15

A 200-mm-long strip of metal is stretched in two steps, first to 300 mm and then to 400 mm. Show that the total true strain is t

he sum of the true strains in each step; in other words, the true strains are additive.
Engineering
1 answer:
Neko [114]3 years ago
4 0

Explanation:

For true Strain:

step 1:

E true = Ln(1 + 0.5 ) = 0.40

Step 2:

E true = Ln(1 + 0.33 ) = 0.29

By single step process:

E true = Ln(1 + 1 ) = 0.69

total strain of step process = 0.40 + 0.29 = 0.69 units

SO TRUE STRAIN IS ADDITIVE.

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A water pump delivers 3 hp of shaft power when operating. If the pressure differential between the outlet and the inlet of the p
Natali [406]

Answer:

Mechanical Efficiency =  83.51%

Explanation:

Given Data:

Pressure difference = ΔP=1.2 Psi

Flow rate = V=8ft^3/s\\

Power of Pump = 3 hp

Required:

Mechanical Efficiency

Solution:

We will first bring the change the units of given data into SI units.

P=1.2*6.895 = 8.274KPa\\V=8*0.00283=0.226 m^3/s\\P=3*0.746=2.238KW

Now we will find the change in energy.

Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.

Thus change in energy is

=(Mass * change in P)/density\\= \frac{M*P}{p}\\\\

As we know that Mass = Volume x density

substituting the value

Energy = Volume * density x ΔP / density

Change in energy = Volumetric flow x ΔP

Change in energy = 0.226 x 8.274 = 1.869 KW

Now mechanical efficiency = change in energy / work done by shaft

Efficiency = 1.869 / 2.238

Efficiency = 0.8351 = 83.51%

5 0
3 years ago
The steady-state data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1200K and 400K, re
Anni [7]

Answer:

a) W_cycle = 200 KW , n_th = 33.33 %  , Irreversible

b) W_cycle = 600 KW , n_th = 100 %     , Impossible

c) W_cycle = 400 KW , n_th = 66.67 %  , Reversible

Explanation:

Given:

- The temperatures for hot and cold reservoirs are as follows:

  TL = 400 K

  TH = 1200 K

Find:

For each case W_cycle , n_th ( Thermal Efficiency ) :

(a) QH = 600 kW, QC = 400 kW

(b) QH = 600 kW, QC = 0 kW

(c) QH = 600 kW, QC = 200kW

- Determine whether the cycle operates reversibly, operates irreversibly, or is impossible.

Solution:

- The work done by the cycle is given by first law of thermodynamics:

                                 W_cycle = QH - QC

- For categorization of cycle is given by second law of thermodynamics which states that:

                                 n_th < n_max     ...... irreversible

                                 n_th = n_max     ...... reversible

                                 n_th > n_max     ...... impossible

- Where n_max is the maximum efficiency that could be achieved by a cycle with Hot and cold reservoirs as follows:

                                n_max = 1 - TL / TH = 1 - 400/1200 = 66.67 %

And,                         n_th = W_cycle / QH

a) QH = 600 kW, QC = 400 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 400 = 200 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 200 / 600 = 33.33 %

   - The type of process according to second Law of thermodynamics:

               n_th = 33.333 %                n_max = 66.67 %

                                       n_th < n_max  

      Hence,                Irreversible Process  

b) QH = 600 kW, QC = 0 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 0 = 600 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 600 / 600 = 100 %

   - The type of process according to second Law of thermodynamics:

                 n_th = 100 %                 n_max = 66.67 %

                                     n_th > n_max  

      Hence,               Impossible Process              

c) QH = 600 kW, QC = 200 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 200 = 400 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 400 / 600 = 66.67 %

   - The type of process according to second Law of thermodynamics:

               n_th = 66.67 %                 n_max = 66.67 %

                                     n_th = n_max  

      Hence,                Reversible Process

7 0
3 years ago
Text that is located between &lt;title &gt;and &lt;/title &gt; appears in the browser's_____​
Vinil7 [7]

Answer:

Tab title

Explanation:

The tab title can be defined as the title that shows up in a browser tab known as page title. The title only has texts, other tags within the texts are not considered.

8 0
3 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
An 80-percent-efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at a speed of 10 ft/s thr
Sladkaya [172]

Answer:

Part A

The mass plow rate, is approximately 97.0 lbm/s

Part B

The power used to overcome friction, is approximately 1.9 hp

Explanation:

The efficiency of the pump, η = 80%

The power input to the pump, P = 20 hp

The speed of the water through the pipe, v = 10 ft./s

The diameter of the pipe, d = 5.2 inches = 13/30 ft.

The free surface of the pool above the lake, h = 80 ft.

The density of the water, ρ = 62.4 lbm/ft.³

Part A

The mass plow rate, \dot m = Q × ρ

Where;

ρ = 62.4 lbm/ft³

Q = A × v

A = The cross-sectional area of the pipe

∴ Q = π·d²/4 × v = π × ((13/30 ft.)²)/4 × 10 ft.s ≈ 1.4748 ft.³/s

∴ The mass plow rate, \dot m ≈ 1.4748 ft.³/s × 62.4 lbm/ft.³ = 97.02752 lbm/s

The mass plow rate, \dot m ≈ 97.0 lbm/s

Part B

The power to pump the water at the given rate, P_w =  \dot m·g·h

∴ P_w =  97.02752 lbm/s × 32.1740 ft./s² × 80 ft. ≈ 14.1130725 Hp

P_w ≈ 14.1130725 Hp

The power output of the pump, P_{out} = 0.8 × 20 hp = 16 hp

Therefore, the power used to overcome friction, P_f = P_{out} - P_w

∴ P_f ≈ 16 hp - 14.1130725 Hp ≈ 1.8869275 hp

The power used to overcome friction,  P_f ≈ 1.9 hp

7 0
3 years ago
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