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Refrigeration and Air Standard Cycles - Thermodynamics - Solved Quiz, Exercises of Thermodynamics

Assam Agricultural UniversityThermodynamics

Some of the topic in thermodynamics are: Property tables and ideal gases, First law for closed and open (steady and unsteady) systems, Entropy and maximum work calculations, Isentropic efficiencies, Cycle calculations (Rankine, refrigeration, air standard) with mass flow rate ratios. This quiz is about: Refrigeration and Air Standard Cycles, Dual Cycle, Diesel Cycle, Constant Volume Heat, Isentropic Expansion, Efficiency, Constant Volume Heat, Cycle Efficienc

Typology: Exercises

2013/2014

Uploaded on 02/01/2014

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Solution Refrigeration and Air-Standard Cycles
The dual cycle is an extended
model of internal combustion
piston engines that generalizes the
Otto cycle and the Diesel cycle.
This cycle consists of five steps
with air as the working fluid:
1-2) An initial isentropic
compression from V1 to V2 =
V1/CR, where CR is the
compression ratio.
2-3) A constant volume heat
addition in which a fraction, ,
of the high temperature heat
transfer is added.
3-4) A constant pressure heat
addition in which the
remaining fraction, 1 – , of
the high temperature heat
transfer is added.
4-5) An isentropic expansion to the initial volume, V1 = V5.
5-1) The exhaust process is modeled as a constant volume heat rejection.
Use the attached air tables to compute the efficiency of this cycle for T1 = 540 R, P1 = 10
psia, CR = 8, and qH = (1 – )qh = 400 Btu/lbm.
The total value of qH is 800 Btu/lbm. To get the efficiency we can either find the work from the
three steps that are not constant volume or we can find the heat rejection, which occurs from
point 5 to point 1. In this solution we will do both to show the equivalence of the results.
HH
L
HH q
uu
q
q
q
uuvvPuu
q
w51
54344321 11
Regardless of the approach used, we will have to analyze the entire cycle to obtain properties at
intermediate state points to get us to u5. We start by considering the isentropic expansion from 1
to 2, for which we can use the equation vr(T2) = vr(T1) V2/V1 = vr(T1)/(CR). From the air tables at
T1 = 520 R, we have vr(T1) = 158.59 and u1 = 88.61 Btu/lbm. At point 2, vr(T2) = vr(T1)/(CR) =
158.59 / 8 = 19.82. This value is between temperatures of 1160 R and 1200 R. Interpolation
gives
m
mm
mlb
Btu
lb
Btu
lb
Btu
lb
Btu
u60.203
326.2082.19
326.20546.18
51.20192.208
51.201
2
For the constant volume heat addition step, w = 0 and the first law gives u3 = u2 + qH = 203.60
Btu/lbm + 400 Btu/lbm = 603.60 Btu/lbm. Interpolation at this value of u gives.
R
lb
Btu
lb
Btu
lb
Btu
lb
Btu
RR
RT
mm
mm
3088
78.601608.603
78.60176.610 30803150
3080
3
V
P
1
2
3
4
5
docsity.com
pf3

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Solution Refrigeration and Air-Standard Cycles

The dual cycle is an extended model of internal combustion piston engines that generalizes the Otto cycle and the Diesel cycle. This cycle consists of five steps with air as the working fluid: 1-2) An initial isentropic compression from V 1 to V 2 = V 1 /CR, where CR is the compression ratio. 2-3) A constant volume heat addition in which a fraction,, of the high temperature heat transfer is added. 3-4) A constant pressure heat addition in which the remaining fraction, 1 –, of the high temperature heat transfer is added. 4-5) An isentropic expansion to the initial volume, V 1 = V 5. 5-1) The exhaust process is modeled as a constant volume heat rejection. Use the attached air tables to compute the efficiency of this cycle for T 1 = 540 R, P 1 = 10 psia, CR = 8, andqH = (1 –)qh = 400 Btu/lbm. The total value of qH is 800 Btu/lbm. To get the efficiency we can either find the work from the three steps that are not constant volume or we can find the heat rejection, which occurs from point 5 to point 1. In this solution we will do both to show the equivalence of the results.

H H

L H H q

u u q

q q

u u P v v u u q

 w^ ^1 ^2 ^3 ^44 ^3 ^4 ^5  1   1 ^1 ^5

Regardless of the approach used, we will have to analyze the entire cycle to obtain properties at intermediate state points to get us to u 5. We start by considering the isentropic expansion from 1 to 2, for which we can use the equation vr(T 2 ) = vr(T 1 ) V 2 /V 1 = vr(T 1 )/(CR). From the air tables at T 1 = 520 R, we have vr(T 1 ) = 158.59 and u 1 = 88.61 Btu/lbm. At point 2, vr(T 2 ) = vr(T 1 )/(CR) = 158.59 / 8 = 19.82. This value is between temperatures of 1160 R and 1200 R. Interpolation gives

m

m m m lb

lb Btu

Btu lb

Btu

lb

u Btu 19. 82 20. 326 203.^60

  1. 546 20. 326

For the constant volume heat addition step, w = 0 and the first law gives u 3 = u 2 + qH = 203. Btu/lbm + 400 Btu/lbm = 603.60 Btu/lbm. Interpolation at this value of u gives.

R lb

Btu lb

Btu

lb

Btu lb

Btu

T R R R

m m m m

3 3080610.^3150763080601. 78 603.^608601.^78 ^3088

V

P

1

2

3 4

5

The constant pressure heat addition step gives the value of h 4 = h 3 + qH. We can find h 3 = u 3 + RT 3. Combining these equations gives the result below.

m m m m

H (^) lb

Btu lb

R Btu lb R

Btu lb

h u RT q^603.^60 Btu^0.^0685530884001215.^29

Interpolation at this value of h gives.

  0. 3307312190. 17.^3206412070.^33073. 03 1215.^291207.^0340. 32386

(^44001219).^44001712074400. 03 1215.^291207.^0344427

4

4

^  

 

 (^)  

  

 

 

 (^)  

  

m m m m

r

m m m m

lb

Btu lb

Btu

lb

Btu lb

v T Btu

lb Btu lb Btu R lb

Btu lb

Btu T R R R

We can compute the isentropic expansion from v 4 to v 5 using the equation have vr(T 5 ) = vr(T 4 ) v 5 /v 4 , but we have to know the ratio v 5 /v 4 for this calculation. We can find this ratio from the ideal gas law and the cycle structure which has v 3 = v 2 = v 1 /(CR), v 5 = v 1 , and P 4 = P 3. This gives the value of v 4 /v 5 as follows

3

4 5

4 3

5 4 3

1 4 3

2 4 3

3 4

3

3

4 3

4 4

4 4 ( ) ( ) ( CR ) T

T

v

v CRT

vT CRT

vT T

vT T

vT

v

RT

RT

P

RT

P

vRT        

From the final equation we can find v 4 /v 5 = T 4 /[(CR)T 3 = (4427 R)/[(8)(3088 R)] = 0.17920. We can then find vr(T 5 ) = vr(T 4 ) v 5 /v 4 0.32386 / 0.17920 = 1.8072. Interpolation in the tables for this vr value gives.

m

m m m lb

lb Btu

Btu lb

Btu

lb

u Btu 1. 8072 1. 8899 503.^42

  1. 7999 1. 8899

We can compute the heat rejection as |qL| = u 5 – u 1 = 503.42 Btu/lbm – 88.61 Btu/lbm = 414. Btu/lbm. The total heat added, |qH| = 400 Btu/lbm + 400 Btu/lbm = 800 Btu/lbm. So we can finally compute the cycle efficiency.

m

m H

L

lb

Btu

lb

Btu

q

q 800

 1   1  n = 48.1%

If use the work to find the efficiency, we have to compute the term P 3  4 (v 4 – v 3 ). For an ideal gas, P 3  4 (v 4 – v 3 ) = R(T 4 – T 3 ) and our work-based efficiency equation becomes

H q H

u u RT T u u q

 w^ ^1 ^2 ^4 ^3 ^4 ^5

We have all the required values in this equation except for u 4. We can find this from the values of h 4 and T 4. Once u 4 is known we have all the information required to find the efficiency.

m m lbm

R Btu lb R

Btu lb

u h RT^1215.^29 Btu^0.^068554427911.^81 4  4  4    