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A cold air-standard Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the maximum temperature of t

Debora [2.8K]

Answer:

$\dot Q_{in} = 228.659\,kW$ , $T_{3} = 1573.662\,K\,(1300.512\,^{\textdegree}C)$

Explanation:

The ideal efficiency of the Diesel cycle is given by this expression:

$\eta_{th} = \left\{1 - \frac{1}{r^{k-1}} \cdot \left[\frac{r_{c}^{k}-1}{k\cdot (r_{c}-1)} \right]\right\}\times 100\%$

Where $r$ and $r_{c}$ are the compression and cutoff ratios, respectively.

$\eta_{th} = \left\{1-\frac{1}{18^{0.4}}\cdot \left[\frac{1.5^{1.4}-1}{1.4\cdot (1.5-1)} \right] \right\}\times 100\%$

$\eta_{th} = 65.648\,\%$

The heat addition to the cycle is:

$\dot Q_{in} = \frac{\dot W}{\eta_{th}}$

$\dot Q_{in} = \frac{150\,kW}{0.656}$

$\dot Q_{in} = 228.659\,kW$

The temperature at state 2 is:

$T_{2} = T_{1} \cdot r^{k-1}$

$T_{2} = (330.15\,K)\cdot 18^{0.4}$

$T_{2} = 1049.108\,K\,(775.958\,^{\textdegree}C)$

And the temperature at state 3 is:

$T_{3} = T_{2}\cdot r_{c}$

$T_{3} = (1.5)\cdot (1049.108\,K)$

$T_{3} = 1573.662\,K\,(1300.512\,^{\textdegree}C)$

4 0

3 years ago

Among Us Map-Skeld Imposter 3 code- DBZPTQ

Yuki888 [10]

Answer:

0: <em>NOICE </em>:0

Explanation:

7 0

3 years ago

Read 2 more answers

Chemical manufacturers must present which information on the product's label?

Wewaii [24]

Answer:

hazardous chemicals leaving the workplace is labeled, tagged or marked with the following information: product identifier; signal word; hazard statement

Explanation:

this is so you know what chemicals are in it

6 0

4 years ago

A 304 stainless steel (yield strength 30 ksi) cylinder has an inner diameter of 4 in and a wall thickness of 0.1 in. If it is su

Natasha_Volkova [10]

Answer:

The value we got is 1.423 ksi which is less than 30 and so the material hasn't failed and yielding hasn't occured.

Explanation:

This is a cylindrical thin walled vessel. Thus;

Hoop stress; σ1 = pr/t

Where ;

p is internal pressure

r_o is radius = 4/2 = 2 in

t is thickness of wall = 0.1 in

Thus;

σ1 = 70 x 2/0.1 = 1400 ksi

Longitudinal stress; σ2 = pr/2t

σ2 = 70 x 2/(0.1 x 2) = 700 ksi

Now, we want to find the normal stress. The inner radius of the circle will be; r_i = r_o - t = 2 - 0.1 = 1.9 in

So, normal stress by axial force is given by;

σ_fx = F/A = F/(π(r_o² - r_i²))

We are given that F = 500 lb

σ_fx = 500/(π(2² - 1.9²))

σ_fx = 408.1 ksi

We can now find the torsion from the formula;

τ = (T•r_o)/J

We are given that T = 70 lb.ft = 70 x 12 lb.in = 840 lb.in

J is the polar moment of inertia and has a formula; ((π/2)(r_o⁴ - r_i⁴))

So,J = ((π/2)(2⁴ - 1.9⁴)) = 4.662 in⁴

Thus,τ_xy = (840 x 2)/4.662 = 360.4 ksi

σ1 is in the y direction and σ2 is in the x direction. Thus;

σ_x = σ1 + σ_fx = 700 + 408.1 = 1108.1 ksi

Also, σ_y = σ1 = 1400 ksi

Now distortion energy can be expressed as;

σ_Y² = σ_x - σ_x•σ_y + (σ_y)² + 3(τ_xy)²

Plugging in the relevant values, we obtain ;

σ_Y² = 1108.1² - (1108.1*1400) + 1400² + 3(360.4)²

So, σ_Y² = 2.02 x 10^(6) psi²

σ_Y = √2.02 x 10^(6)

σ_Y = 1423 psi = 1.423 ksi

The question says the yield strength of the material is 30 ksi.

The value we got is less than 30 and so the material hasn't failed and yielding hasn't occured.

8 0

3 years ago

Air initially at 120 psia and 500o F is expanded by an adiabatic turbine to 15 psia and 200o F. Assuming air can be treated as a

Goshia [24]

Answer:

a. Wa = 73.14 Btu/lbm

b. Sgen = 0.05042 Btu/lbm °R

c. Isentropic efficiency is 70.76%

d. Minimum specific work for compressor W = -146.2698 Btu/lbm [It is negative because work is being done on the compressor]

Explanation:

Complete question is as follows;

Air initially at 120 psia and 500oF is expanded by an adiabatic turbine to 15 psia and 200oF. Assuming air can be treated as an ideal gas and has variable specific heat.

a) Determine the specific work output of the actual turbine (Btu/lbm).

b) Determine the amount of specific entropy generation during the irreversible process (Btu/lbm R).

c) Determine the isentropic efficiency of this turbine (%).

d) Suppose the turbine now operates as an ideal compressor (reversible and adiabatic) where the initial pressure is 15 psia, the initial temperature is 200 oF, and the ideal exit state is 120 psia. What is the minimum specific work the compressor will be required to operate (Btu/lbm)?

solution;

Please check attachment for complete solution and step by step explanation

8 0

4 years ago