1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
almond37 [142]
3 years ago
10

Consider a modification of the air-standard Otto cycle in which the isentropic compression and expansion processes are each repl

aced with polytropic processes having n = 1.3. The compression ratio is 10 for the modified cycle. At the beginning of compression, p1 = 1 bar and T1 = 310 K. The maximum temperature during the cycle is 2200 K. Determine
(a) The heat transfer and work in kJ per kg of air for each process in the modified cycle.
(b) The thermal efficiency.
(c) The mean effective pressure, in bar.

Engineering
1 answer:
Ulleksa [173]3 years ago
8 0

Answer:

The answers to the question are

(1) Process 1 to 2

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

(2) Process 2 to 3

W = 0

Q = 1135.376 kJ/kg

(3) Process 3 to 4

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

(4) Process 4 to 3

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency = 49.9 %

(c) The mean effective pressure is 9.44 bar

Explanation:

(a) Volume compression ratio \frac{v_1}{v_2}  = 10

Initial pressure p₁ = 1 bar

Initial temperature, T₁ = 310 K

cp = 1.005 kJ/kg⋅K

Temperature T₃ = 2200 K from the isentropic chart of the Otto cycle

For a polytropic process we have

\frac{p_1}{p_2}  = (\frac{v_2}{v_1} )^n Therefore p₂ = p₁ ÷ (\frac{v_2}{v_1} )^n = (1 bar) ÷ (\frac{1}{10} )^{1.3} = 19.953 bar

Similarly for a polytropic process we have

\frac{T_1}{T_2}  = (\frac{v_2}{v_1} )^{n-1} or T₂ = T₁ ÷ (\frac{v_2}{v_1} )^{n-1} = \frac{310}{0.1^{0.3}} = 618.531 K

The molar mass of air is 28.9628 g/mol.

Therefore R = \frac{8.3145}{28.9628} = 0.287 kJ/kg⋅K

cp = 1.005 kJ/kg⋅K Therefore cv = cp - R =  1.005- 0.287 = 0.718 kJ/kg⋅K

1). For process 1 to 2 which is polytropic process we have

W = \frac{R(T_2-T_1)}{n-1} = \frac{0.287(618.531-310)}{1.3 - 1}= 295.16 kJ/kg

Q =(\frac{n-\gamma}{\gamma - 1} )W = (\frac{1.3-1.4}{1.4-1} ) 295.16 kJ/kg = -73.79 kJ/kg

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

2). For process 2 to 3 which is reversible constant volume heating we have

W = 0 and Q = cv×(T₃ - T₂) = 0.718× (2200-618.531) = 1135.376 kJ/kg

W = 0

Q = 1135.376 kJ/kg

3). For process 3 to 4 which is polytropic process we have

W = \frac{R(T_4-T_3)}{n-1} = Where T₄ is given by  \frac{T_4}{T_3}  = (\frac{v_3}{v_4} )^{n-1} or T₄ = T₃ ×0.1^{0.3}

= 2200 ×0.1^{0.3}  T₄ = 1102.611 K

W =  \frac{0.287(1102.611-2200)}{1.3 - 1}= -1049.835 kJ/kg

and Q = 262.459 kJ/kg

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

4). For process 4 to 1 which is reversible constant volume cooling we have

W = 0 and Q = cv×(T₁ - T₄) = 0.718×(310 - 1102.611) = -569.09 kJ/kg

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency is given by

\eta = 1-\frac{T_4-T_1}{T_3-T_2} =1-\frac{1102.611-310}{2200-618.531} = 0.499 or 49.9 % Efficient

(c) The mean effective pressure is given by

p_{m}  = \frac{p_1r[(r^{n-1}-1)(r_p-1)]}{ (n-1)(r-1)}  where r = compression ratio and r_p = \frac{p_3}{p_2}

However p₃ = \frac{p_2T_3}{T_2} =\frac{(19.953)(2200)}{618.531} =70.97 atm

r_p = \frac{p_3}{p_2} = \frac{70.97}{19.953}  = 3.56

Therefore p_m =\frac{1*10*[(10^{0.3}-1)(3.56-1)]}{0.3*9} = 9.44 bar

Please find attached generalized diagrams of the Otto cycle

You might be interested in
Which of the following color schemes is composed of hues next to eachother on the color wheel ?
Softa [21]
Analogous color schemes are composed of hues next to each other
3 0
3 years ago
Someone has suggested that the air-standard Otto cycle is more accurate if the two polytropic processes are replaced with isentr
omeli [17]

Answer:

q_net,in = 585.8 KJ/kg

q_net,out = 304 KJ/kg

n = 0.481

Explanation:

Given:

- The compression ratio r = 8

- The pressure at state 1, P_1 = 95 KPa

- The minimum temperature at state 1, T_L = 15 C

- The maximum temperature T_H = 900 C

- Poly tropic index n = 1.3

Find:

a) Determine the heat transferred to and rejected from this cycle

b) cycle’s thermal efficiency

Solution:

- For process 1-2, heat is rejected to sink throughout. The Amount of heat rejected q_1,2, can be computed by performing a Energy balance as follows:

                                   W_out - Q_out = Δ u_1,2

- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:

                         c_v*(T_2 - T_L) = R*(T_2 - T_L)/n-1 - q_1,2

- Using polytropic relation we will convert T_2 = T_L*r^(n-1):

                  c_v*(T_L*r^(n-1) - T_L) = R*(T_1*r^(n-1) - T_L)/n-1 - q_1,2

- Hence, we have:

                             q_1,2 = T_L *(r^(n-1) - 1)* ( (R/n-1) - c_v)

- Plug in the values:

                             q_1,2 = 288 *(8^(1.3-1) - 1)* ( (0.287/1.3-1) - 0.718)

                            q_1,2= 60 KJ/kg

- For process 2-3, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:

                                          Q_in = Δ u_2,3

                                         q_2,3 = u_3 - u_2

                                         q_2,3 = c_v*(T_H - T_2)  

- Again, using polytropic relation we will convert T_2 = T_L*r^(n-1):

                                         q_2,3 = c_v*(T_H - T_L*r^(n-1) )    

                                         q_2,3 = 0.718*(1173-288*8(1.3-1) )

                                        q_2,3 = 456 KJ/kg

- For process 3-4, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:

                                     q_3,4 - w_in = Δ u_3,4

- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:

                           c_v*(T_4 - T_H) = - R*(T_4 - T_H)/1-n +  q_3,4

- Using polytropic relation we will convert T_4 = T_H*r^(1-n):

                  c_v*(T_H*r^(1-n) - T_H) = -R*(T_H*r^(1-n) - T_H)/n-1 + q_3,4

- Hence, we have:

                             q_3,4 = T_H *(r^(1-n) - 1)* ( (R/1-n) + c_v)

- Plug in the values:

                             q_3,4 = 1173 *(8^(1-1.3) - 1)* ( (0.287/1-1.3) - 0.718)

                            q_3,4= 129.8 KJ/kg

- For process 4-1, heat is lost from the system. The Amount of heat rejected q_4,1, can be computed by performing a Energy balance as follows:

                                          Q_out = Δ u_4,1

                                         q_4,1 = u_4 - u_1

                                         q_4,1 = c_v*(T_4 - T_L)  

- Again, using polytropic relation we will convert T_4 = T_H*r^(1-n):

                                         q_4,1 = c_v*(T_H*r^(1-n) - T_L )    

                                         q_4,1 = 0.718*(1173*8^(1-1.3) - 288 )

                                        q_4,1 = 244 KJ/kg

- The net gain in heat can be determined from process q_3,4 & q_2,3:

                                         q_net,in = q_3,4+q_2,3

                                         q_net,in = 129.8+456

                                         q_net,in = 585.8 KJ/kg

- The net loss of heat can be determined from process q_1,2 & q_4,1:

                                         q_net,out = q_4,1+q_1,2

                                         q_net,out = 244+60

                                         q_net,out = 304 KJ/kg

- The thermal Efficiency of a Otto Cycle can be calculated:

                                         n = 1 - q_net,out / q_net,in

                                         n = 1 - 304/585.8

                                         n = 0.481

6 0
3 years ago
Các đặc điểm chính của đường dây dài siêu cao áp .
rodikova [14]

Answer:

Đường dây siêu cao áp 500kV: Những chuyện giờ mới kể ... ​Ngày 27/5/1994, hệ thống đường dây điện siêu cao áp 500kV Bắc - Nam chính thức đưa ... Tại thời điểm đó, các nước như Pháp, Úc, Mỹ khi xây dựng đường dây dài nhất ... và chế ra các máy kéo dây theo đặc thù công việc của từng đơn vị.

Explanation:

8 0
3 years ago
A three-point flexure test is conducted on a cylindrical specimen of aluminum oxide. The specimen radius is 5.0 mm and the dista
kondaur [170]

Answer:

Detailed solution is given in the attached diagram

7 0
3 years ago
Which term defines the amount of mechanical work an engine can do per unit of heat energy it uses?
skad [1K]

Answer:

d

Explanation:

is the because that's the amount of work in making machine can do producing heat

7 0
3 years ago
Other questions:
  • PLZ HURRY IM ON A TIMER
    6·1 answer
  • Give two methods on how powder is produced in powder metallurgy.
    5·2 answers
  • A device that helps increase field worker productivity by providing reliable location and time
    13·1 answer
  • What is the difference between a Datum and a Datum Feature? a) A Datum and Datum Feature are synonymous. b) A Datum is theoretic
    14·1 answer
  • Reusable refrigerant containers under high-pressure must be hydrostatically tested how often?
    10·1 answer
  • Do you think the industrial revolution helped or hurt workers? Why?
    8·1 answer
  • a poorly tighten terminal is often the cause of a/an ? a) open circuit b) circuit breaker interrupt c)short circuit d) ground fa
    10·1 answer
  • Determine the wattmeter reading when it is connected to resistor load.​
    11·1 answer
  • What is the answer???
    10·1 answer
  • Stress that acts in the plane of a cut section, rather than at right angles to the section is called:_______
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!