Question

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

Answer 1

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

Answer:

(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

(c) the thermal efficiency = 55.23%

Step-by-step 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)` =
`((V_(1) )/(V_(2) ) * (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

(c) the thermal efficiency η

η =
`W_(cycle)`/m/
`Q_(in)`/m = 828.41/1500 = 0.552 (55.23%)

Answer 2

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%

Step-by-step 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:

`(T_(2) )/(T_(1) ) =((P_(2) )/(P_(1) ) )^{(r-1)/(r) } =((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=(2)/(3) *1500=1000kJ/kg`

q = m * Cv * (T₃ - T₂)

Where Cv = 0.718 kJ/kg K

Replacing and clearing T₃:

`T_(3) =(1000)/(0.718) +T_(2) =2146.3K`

`P_(3) =(T_(3)*P_(2) )/(T_(2) ) =(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) =(500)/(1.005) +2146.3=2643K`

`V_(3) =(RT_(3) )/(P_(3) ) =(0.287*2146.3)/(7154) =0.086m^(3) /kg`

`V_(4) =(T_(4)V_(3) )/(T_(3) ) =(2643*0.086)/(2146.3) =0.1059m^(3) /kg`

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

`V_(1) =(RT_(1) )/(P_(1) ) =(0.287*300)/(100) =0.861m^(3) /kg`

if V₁ = V₅

`T_(5) =T_(4) ((V_(4) )/(V_(5) ) )^(r-1) =2643((0.1059)/(0.861) )^(1.4-1) =1143.5K`

d) The efficiency of the cycle is:

`n=1-((1)/(r) )^(r-1) (\alpha L-1)/(r(L-1)\alpha +(\alpha -1))`

Where L = V₄/V₃

α = P₃/P₂

Replacing values:

`n=1-((1)/(10) )^(1.4-1) ((2.848(1.2317)^(1.4)-1 )/(1.41(1.2317-1)*2.848+(2.848-1)) )=0.596` = 59.6%