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marshall27 [118]

3 years ago

14

An Otto cycle with air as the working fluid has a compression ratio of 8.2. Under cold air standard conditions, what is the ther

mal efficiency of this cycle?

1 answer:

Tresset [83]3 years ago

5 0

Answer:

Under cold air standard conditions, the thermal efficiency of this cycle is 56.9 percent.

Explanation:

From Thermodynamics we remember that thermal efficiency of the ideal Otto cycle ( $\eta_{th}$ ), dimensionless, is defined by the following formula:

$\eta_{th} = 1-\frac{1}{r^{\gamma-1}}$ (Eq. 1)

Where:

$r$ - Compression ratio, dimensionless.

$\gamma$ - Specific heat ratio, dimensionless.

Please notice that specific heat ratio under cold air standard conditions is $\gamma = 1.4$ .

If we know that $r = 8.2$ and $\gamma = 1.4$ , then thermal efficiency of the ideal Otto cycle is:

$\eta_{th} = 1-\frac{1}{8.2^{1.4-1}}$

$\eta_{th} = 0.569$

Under cold air standard conditions, the thermal efficiency of this cycle is 56.9 percent.

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______________ help protect the lower legs and feet from heat hazards like molten metal and welding sparks. A) Safety shoesB) Le

vodka [1.7K]

Answer:

It's leggings. They help protect the lower legs and feet from heat hazards like molten metal and welding sparks.

6 0

3 years ago

Write a program to calculate overtime pay of 10 employees. Overtime is paid at the rate of Rs. 12.00

fgiga [73]

Answer:

Here is the code.

Explanation:

#include<stdio.h>

int main()

{

int i, time_worked, over_time, overtime_pay = 0;

for (i = 1; i <= 10; i++)

{

printf("\nEnter the time employee worked in hr ");

scanf("%d", &time_worked);

if (time_worked>40)

{

over_time = time_worked - 40;

overtime_pay = overtime_pay + (12 * over_time);

}

}

printf("\nTotal Overtime Pay Of 10 Employees Is %d", overtime_pay);

return 0;

}

Output :

Enter the time employee worked in hr 42

Enter the time employee worked in hr 45

Enter the time employee worked in hr 42

Enter the time employee worked in hr 41

Enter the time employee worked in hr 50

Enter the time employee worked in hr 51

Enter the time employee worked in hr 52

Enter the time employee worked in hr 53

Enter the time employee worked in hr 54

Enter the time employee worked in hr 55

Total Overtime Pay Of 10 Employees Is 1020.

6 0

3 years ago

A cast-iron tube is used to support a compressive load. Knowing that E 5 10 3 106 psi and that the maximum allowable change in l

Papessa [141]

Answer:

(a) 2.5 ksi

(b) 0.1075 in

Explanation:

(a)

$E=\frac {\sigma}{\epsilon}$

Making $\sigma$ the subject then

$\sigma=E\epsilon$

where $\sigma$ is the stress and $\epsilon$ is the strain

Since strain is given as 0.025% of the length then strain is $\frac {0.025}{100}=0.00025$

Now substituting E for $10\times 10^{6} psi$ then

$\sigma=(10\times 10^{6} psi)\times 0.00025=2500 si= 2.5 ksi$

(b)

Stress, $\sigma= \frac {F}{A}$ making A the subject then

$A=\frac {F}{\sigma}$

$A=\frac {\pi(d_o^{2}-d_i^{2})}{4}$

where d is the diameter and subscripts o and i denote outer and inner respectively.

We know that $2t=d_o - d_i$ where t is thickness

Now substituting

$\frac {\pi(d_o^{2}-d_i^{2})}{4}=\frac {1600}{2500}$

$\pi(d_o^{2}-d_i^{2})=\frac {1600}{2500}\times 4$

$(d_o^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4$

But the outer diameter is given as 2 in hence

$(2^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4$

$2^{2}-(\frac {1600}{2500\times \pi}\times 4)=d_i^{2}$

$d_i=\sqrt {2^{2}-(\frac {1600}{2500\times \pi}\times 4)}=1.784692324 in\approx 1.785 in$

As already mentioned, $2t=d_o - d_i hence t=0.5(d_o - d_i)$

$t=0.5(2-1.785)=0.1075 in$

3 0

4 years ago

What is an air mass?

kotegsom [21]

Answer:

An air mass is a body of air with horizontally uniform temperature, humidity, and pressure.

Explanation:

Because it is

8 0

3 years ago

Read 2 more answers

A 3.5-m3 rigid tank initially contains air whose density is 2 kg/m3 . The tank is connected to a high-pressure supply line throu

Mumz [18]

Answer:

Explanation:

First, we find the mass of the air originally in the tank.

Density is given as mass divided by volume. It is given as:

$Density = \frac{mass}{volume}$

Therefore, mass is:

$mass = denisty *volume$

Density of air = $2 kg/m^3$ ; Volume of the tank = $3.5 m^3$

$=> Mass = 3.5 * 2 = 7 kg$

The mass of the air initially in the tank is 7 kg.

After air is allowed to enter, the mass changes.

New density = $6.5 kg/m^3$

The new mass will be:

$Mass = 6.5 * 3.5 = 22.75 kg$

We can now find the mass of air that has entered the tank:

Mass of air that entered tank = New mass of air - Original mass of air

M = 22.75 - 7.0 = 15.75 kg

The mass of air that entered the tank is 15.75 kg.

6 0

3 years ago