Engineering, 31.03.2020 01:03 Poohpooh1510

An air standard dual cycle has a compression ratio of 10. At the beginning of compression, 1 p kPa =100 and 1T = 300 K . The heat addition per unit mass during the combustion process is 1500 / kJ kg with two third added under constant volume and one third added under constant pressure. You may treat air as an ideal gas. Specific heats of air are v c and p v air c c R = + , respectively

Answers: 2

Show answers

Answers

Answer from: Pen350

a) Process 1-2: adiabatic compression

Process 2-3: heat addition (constant volume)

Process 3-4: heat addition (constant pressure)

Process 4-5: adiabatic expansion

Process 5-1: heat rejection (constant volume)

b) T₃ = 2146.3 K

T₄ = 2643 K

c) T₅ = 1143.5 K

d) n = 59.6%

Explanation:

the data given by the exercise are:

Dual cycle

P₁ = 100 kPa

T₁ = 300 K

r = compression ratio = 10

Heat addition = 1500 kJ/kg

a) according to the attached diagram we have to:

Process 1-2: adiabatic compression

Process 2-3: heat addition (constant volume)

Process 3-4: heat addition (constant pressure)

Process 4-5: adiabatic expansion

Process 5-1: heat rejection (constant volume)

b) If the process 1-2 is adiabatic compression, we have:

$\frac{T_{2} }{T_{1} } =(\frac{P_{2} }{P_{1} } )^{\frac{r-1}{r} } =(\frac{V_{1} }{V_{2} } )^{r-1}$

Where r = Cp/Cv = 1.4

r = V₁/V₂ = 10

$T_{2} =T_{1} (10)^{1.4-1} =300(10)^{1.4-1} =753.6K$

$P_{2} =P_{1} (10)^{1.4} =100(10)^{1.4} =2511.9kPa$

The process 2-3 is heat addition (constant volume). The total heat is:

$q=\frac{2}{3} *1500=1000kJ/kg$

q = m * Cv * (T₃ - T₂)

Where Cv = 0.718 kJ/kg K

Replacing and clearing T₃:

$T_{3} =\frac{1000}{0.718} +T_{2} =2146.3K$

$P_{3} =\frac{T_{3}*P_{2} }{T_{2} } =\frac{2146.3*2511.9}{753.6} =7154kPa$

The process 3-4 is heat addition (constant pressure), the heat addition is:

q = 1500/3 = 500 kJ/kg

q = m * Cp * (T₄ - T₃)

Cp = 1.005 kJ/kg*K

Replacing values and clearing T₄

$T_{4} =\frac{500}{1.005} +2146.3=2643K$

$V_{3} =\frac{RT_{3} }{P_{3} } =\frac{0.287*2146.3}{7154} =0.086m^{3} /kg$

$V_{4} =\frac{T_{4}V_{3} }{T_{3} } =\frac{2643*0.086}{2146.3} =0.1059m^{3} /kg$

c) The process 4-5 is adiabatic expansion, thus:

$V_{1} =\frac{RT_{1} }{P_{1} } =\frac{0.287*300}{100} =0.861m^{3} /kg$

if V₁ = V₅

$T_{5} =T_{4} (\frac{V_{4} }{V_{5} } )^{r-1} =2643(\frac{0.1059}{0.861} )^{1.4-1} =1143.5K$

d) The efficiency of the cycle is:

$n=1-(\frac{1}{r} )^{r-1} \frac{\alpha L-1}{r(L-1)\alpha +(\alpha -1)}$

Where L = V₄/V₃

α = P₃/P₂

Replacing values:

$n=1-(\frac{1}{10} )^{1.4-1} (\frac{2.848(1.2317)^{1.4}-1 }{1.41(1.2317-1)*2.848+(2.848-1)} )=0.596$ = 59.6%

An air standard dual cycle has a compression ratio of 10. At the beginning of compression, 1 p kPa =

Answer from: brooke2337

concluding part of the question

Determine;

a.) the temperature at the end of each heat addition process in K

b.) the net work of the cycle per unit mass of air in kJ/kg

c.) the thermal efficiency

(a) $T_{2}$ = 731.3 K, $T_{3}$ = 1865.8 K, $T_{4}$ = 2266.67 K and $T_{5}$ = 1145.75 K

(b) the net work of the cycle per unit mass of air = 828.41 kJ/kg

Explanation:

Given;

compression ratio $V_{1$ / $V_{2}$ =10

Heat per unit mass

$Q_{in}$ /m = 1500 kJ/kg

$Q_{23}$ /m = 1000 kJ/kg

$Q_{34}$ /m = 500 kJ/kg

$P_{1}$ = 100 kPa

$T_{1$ = 300 K

Assumptions:

1. the air in piston cylinder assembly is the closed system

2. all processes are internally reversible

3. the air is modeled as an ideal gas.

4. the compression and expansion processes are adiabatic.

5. kinetic and potential energy effect are negligible.

Note: analysis of the cycle is done by fixing each principle state of the cycle.

State 1:

$T_{1$ = 300 K ⇒ $u_{1}$ = 214.07 kJ/kg, $V_{r1}$ = 621.2

State 2: for isentropic compression

$V_{r2}$ = $V_{r1}$ * $V_{1$ / $V_{2}$ = 621.2/10 = 62.12

thus, $T_{2}$ = 731.3 K and $u_{2}$ = 535.6 kJ/kg

State 3: for the heat addition process from 2 to 3.

$u_{3}$ = $Q_{23}$ /m + $u_{2}$ = 1000 + 535.6 = 1535.6 kJ/kg

(a) thus, $T_{3}$ = 1865.8 K and $h_{3}$ = 2070.52 kJ/kg

State 4: for the heat addition process from 3 to 4

$h_{4}$ = $h_{3}$ + $Q_{34}$ /m = 2070.52 + 500 = 2570.52 kJ/kg

thus, $T_{4}$ = 2266.67 K and $V_{r4}$ = 2.013

State 5: for isentropic expansion

$V_{r5}$ = $(\frac{V_{1} }{V_{2} } * \frac{T_{3} }{T_{4} } )$ $V_{r4}$ = ((10 * (1865.8/2266.67)) = 16.57

thus, $T_{5}$ = 1145.75 K and $u_{5}$ = 885.66 kJ/kg

(b) $W_{cycle}$ = $Q_{cycle}$ , thus

$W_{cycle}$ /m = $Q_{in}$ /m - ( $u_{5}$ - $u_{1}$ ) = 1500 - (885.66 - 214.07) = 828.41 kJ/kg

η = $W_{cycle}$ /m/ $Q_{in}$ /m = 828.41/1500 = 0.552 (55.23%)

Answer from: Quest

answer:

the average distance would be 20ly

Explanation:

100,000ly times 5,000ly=20ly

Answer from: Quest

answer:

Romans

Explanation:

Hope it helps

Other questions on the subject: Engineering

Engineering, 04.07.2019 18:10, DestyHilBan1183

Adouble-strand no. 60 roller chain is used to transmit power between a 13-tooth driving sprocket rotating at 300 rev/min and a 52-tooth driven sprocket. a) what is the allowable horsepower of this drive? b) estimate the center-to-center distance if the chain length is 82 pitches. c) estimate the torque and bending force on the driving shaft by the chain if the actual horsepower transmitted is 30 percent less than the corrected (allowable) power.

Answers: 3

continue

Engineering, 04.07.2019 18:10, salazjlove

Which of the following refers to refers to how well the control system responds to sudden changes in the system. a)-transient regulation b)- distributed regulation c)-constant regulation d)-steady-state regulation

Answers: 1

continue

Engineering, 04.07.2019 18:10, 0436500

Aturning operation is performed with following conditions: rake angle of 12°, a feed of 0.35 mm/rev, and a depth of cut of 1.1 mm. the work piece is aluminum alloy 6061 with t6 heat treatment (a16061-t6). the resultant chip thickness was measured to be 1.0 mm. estimate the cutting force, fc. use shear stress of 207 mpa and coefficient of friction on the tool face of 0.6.

Answers: 1

continue

Engineering, 04.07.2019 18:20, DroctorWellsfan

Inspection for bearing condition will include: (clo4) a)-color b)-smell c)-size d)-none of the above

Answers: 1

continue

You know the right answer?

An air standard dual cycle has a compression ratio of 10. At the beginning of compression, 1 p kPa =...