Ask question

Ask question

All categories

MA_775_DIABLO [31]

3 years ago

10

The theoretical maximum specific gravity of a mix at 5.0% binder content is 2.495. Using a binder specific gravity of 1.0, find

the theoretical maximum specific gravity at 6.5% binder content. (Hint: Both use the same stone, therefore have the same Gmb)

1 answer:

PSYCHO15rus [73]3 years ago

3 0

Answer:

The theoretical maximum specific gravity at 6.5% binder content is 2.44.

Explanation:

Given the specific gravity at 5.0 % binder content 2.495

Therefore

95 % mix + 5 % binder gives S.G. = 2.495

Where the binder is S.G. = 1, Therefore

Per 100 mass unit we have (Mx + 5)/(Vx + 5) = 2.495

(95 +5)/(Vx +5) = 2.495

2.495 × (Vx + 5) = 100

Vx =35.08 to 95

Or density of mix = Mx/Vx = 95/35.08 = 2.7081

Therefore when we have 6.5 % binder content, we get

Per 100 mass unit

93.5 Mass unit of Mx has a volume of

Mass/Density = 93.5/2.7081 = 34.526 volume units

Therefore we have

At 6.5 % binder content.

(100 mass unit)/(34.526 + 6.5) = 2.44

The theoretical maximum specific gravity at 6.5% binder content = 2.44.

You might be interested in

An air-standard dual cycle has a compression ratio of 9.1 and displacement of Vd = 2.2 L. At the beginning of compression, p1 =

jok3333 [9.3K]

Answer:

a) T₂ is 701.479 K

T₃ is 1226.05 K

T₄ is 2350.34 K

T₅ is 1260.56 K

b) The net work of the cycle in kJ is 2.28 kJ

c) The power developed is 114.2 kW

d) The thermal efficiency, $\eta _{dual}$ is 53.78%

e) The mean effective pressure is 1038.25 kPa

Explanation:

a) Here we have;

$\frac{T_{2}}{T_{1}}=\left (\frac{v_{1}}{v_{2}} \right )^{\gamma -1} = \left (r \right )^{\gamma -1} = \left (\frac{p_{2}}{p_{1}} \right )^{\frac{\gamma -1}{\gamma }}$

Where:

p₁ = Initial pressure = 95 kPa

p₂ = Final pressure =

T₁ = Initial temperature = 290 K

T₂ = Final temperature

v₁ = Initial volume

v₂ = Final volume

$v_d$ = Displacement volume =

γ = Ratio of specific heats at constant pressure and constant volume cp/cv = 1.4 for air

r = Compression ratio = 9.1

Total heat added = 4.25 kJ

1/4 × Total heat added = $c_v \times (T_3 - T_2)$

3/4 × Total heat added = $c_p \times (T_4 - T_3)$

$c_v$ = Specific heat at constant volume = 0.718×2.821× 10⁻³

$c_p$ = Specific heat at constant pressure = 1.005×2.821× 10⁻³

v₁ - v₂ = 2.2 L

$\left \frac{v_{1}}{v_{2}} \right =r \right = 9.1$

v₁ = v₂·9.1

∴ 9.1·v₂ - v₂ = 2.2 L = 2.2 × 10⁻³ m³

8.1·v₂ = 2.2 × 10⁻³ m³

v₂ = 2.2 × 10⁻³ m³ ÷ 8.1 = 2.72 × 10⁻⁴ m³

v₁ = v₂×9.1 = 2.72 × 10⁻⁴ m³ × 9.1 = 2.47 × 10⁻³ m³

Plugging in the values, we have;

${T_{2}}= T_{1} \times \left (r \right )^{\gamma -1} = 290 \times 9.1^{1.4 - 1} = 701.479 \, K$

From;

$\left (\frac{p_{2}}{p_{1}} \right )^{\frac{\gamma -1}{\gamma }}= \left (r \right )^{\gamma -1}$ we have;

$p_{2} = p_{1}} \times \left (r \right )^{\gamma } = 95 \times \left (9.1 \right )^{1.4} = 2091.13 \ kPa$

1/4×4.25 = $0.718 \times 2.821 \times 10^{-3}\times (T_3 - 701.479)$

∴ T₃ = 1226.05 K

Also;

3/4 × Total heat added = $c_p \times (T_4 - T_3)$ gives;

3/4 × 4.25 = $1.005 \times 2.821 \times 10^{-3} \times (T_4 - 1226.05)$ gives;

T₄ = 2350.34 K

$\frac{T_{4}}{T_{5}}=\left (\frac{v_{5}}{v_{4}} \right )^{\gamma -1} = \left (\frac{r}{\rho } \right )^{\gamma -1}$

$\rho = \frac{T_4}{T_3} = \frac{2350.34}{1226.04} = 1.92$

$T_{5} = \frac{T_{4}}{\left (\frac{r}{\rho } \right )^{\gamma -1}}= \frac{2350.34 }{\left (\frac{9.1}{1.92 } \right )^{1.4-1}} =1260.56 \ K$

b) Heat rejected = $c_v \times (T_5 - T_1)$

$Therefore \ heat \ rejected = 0.718 \times 2.821 \times 10^{-3}\times (1260.56 - 290) = 1.966 kJ$

The net work done = Heat added - Heat rejected

∴ The net work done = 4.25 - 1.966 = 2.28 kJ

The net work of the cycle in kJ = 2.28 kJ

c) Power = Work done per each cycle × Number of cycles completed each second

Where we have 3000 cycles per minute, we have 3000/60 = 50 cycles per second

Hence, the power developed = 2.28 kJ/cycle × 50 cycle/second = 114.2 kW

d)

$Thermal \ efficiency, \, \eta _{dual} = \frac{Work \ done}{Heat \ supplied} = \frac{2.28}{4.25} \times 100 = 53.74 \%$

The thermal efficiency, $\eta _{dual}$ = 53.78%

e) The mean effective pressure, $p_m$ , is found as follows;

$p_m = \frac{W}{v_1 - v_2} =\frac{2.28}{2.2 \times 10^{-3}} = 1038.25 \ kPa$

The mean effective pressure = 1038.25 kPa.

3 0

4 years ago

PROBLEM IN PICTURE HELP ME DEAR GODDDDDD UGHHH NONONO I HAVE 2 MINUTES TO FINISH THIS ❕❗️❕❗️❗️❕❕❕❕❗️❕❕❗️❕❗️❗️❗️❕‼️‼️‼️‼️❗️‼️❗️

Elan Coil [88]

Thx :) so much :))))))

4 0

3 years ago

Read 2 more answers

You will create three classes, the first two being Student and LineAtOfficeHour. The instances of the first class defines a sing

White raven [17]

Answer:

Complete solution is given below:

Explanation:

//student class

class Student{

private String firstname,lastname;

//constructor

Student(String first,String last){

this.firstname=first;

this.lastname=last;

}

//getters and setters

public String getFirstname() {

return firstname;

}

public void setFirstname(String firstname) {

this.firstname = firstname;

}

public String getLastname() {

return lastname;

}

public void setLastname(String lastname) {

this.lastname = lastname;

}

//function to get the fullname of student

public String fullName() {

return this.firstname+" "+this.lastname;

}

}

//class for line at office hour

class LineAtOfficeHour{

private Student line[];

private int N=0;

private int back=0;

//empty constructor

LineAtOfficeHour() {

line=new Student[5];

}

//parameterized constructor

LineAtOfficeHour(Student st[]) {

int i=0;

line=new Student[5];

while(i<st.length && i<5) {

line[i]=st[i];

i++;

}

this.N=i;

this.back=i-1;

}

//function to check if line is empty or not

public boolean isEmpty() {

if(this.N==0)

return true;

else

return false;

}

//function to check if line is full

public boolean isFull() {

if(this.N==5) {

return true;

}else

return false;

}

///function to get the size of line

public int size() {

return this.N;

}

//function to add a student to the line

public void enterLine(Student s) {

if(isFull())

System.out.println("Line is full!!!!");

else {

line[++back]=s;

this.N++;

}

}

public Student seeTeacher() {

Student result=null;

if(this.N>=0) {

result=line[0];

int i=0;

for(i=1;i<N;i++) {

line[i-1]=line[i];

}

line[i-1]=null;

this.N--;

this.back--;

}

return result;

}

//function to print students in line

public String whosInLine() {

String result ="";

for(int i=0;i<this.N;i++) {

result+=line[i].fullName()+",";

}

return result;

}

}

//driver method

public class TestLine {

public static void main(String[] args) {

LineAtOfficeHour list=new LineAtOfficeHour();

if(list.isEmpty()) {

System.out.println("Line is empty!!!!!!!!!");

}

Student s1[]=new Student[3];

s1[0]=new Student("John","Smith");

s1[1]=new Student("Sam","Zung");

s1[2]=new Student("Peter","Louis");

list=new LineAtOfficeHour(s1);

if(list.isEmpty()) {

System.out.println("Line is empty!!!!!!!!!");

}else {

System.out.println("Line is not empty.........");

}

System.out.println("Students in line: "+list.whosInLine());

System.out.println("Student removed: "+list.seeTeacher().fullName());

System.out.println("Students in line: "+list.whosInLine());

}

}

6 0

4 years ago

A customer asks why rogue has a vc-turbo engine when turbo engines are for sports cars. What are two ways to respond?.

aniked [119]

Best-in-class gas engine 1 and torque 2 fuel economy and All-new engine to be assembled at Nissan's powertrain facility in Decherd, Tenn.

<h3>Rogue</h3>

Nissan has introduced an all-new Rogue for the 2021 model year, delivering high quality and customer satisfaction through its strong combination of design, safety and technology. Now, for 2022, Nissan is making its best-selling model more powerful, more fuel efficient and more fun to drive with a new 1.5-liter Variable Compression (VC) Turbo engine across the Rogue lineup.

With this information, we can conclude that the vc-turbo engine is a great advance for fuel economy.

Learn more about Nissan in brainly.com/question/24814294

7 0

3 years ago

Errors in the output voltage of an actual integrated circuit operational amplifier can be caused by : Select one:

natta225 [31]

Answer:

Option B

Explanation:

An operational amplifier usually has a high open loop gain of around 10^5 which allows a wide range get of feed back levels in order to achieve the desired performance so therefore a low open loop gain reduces the range feed back level thereby reducing the performance which can cause errors in the output voltage.

7 0

3 years ago