Answer to Question #163246 in Mechanical Engineering for Abi

Question

An air-standard Otto cycle operates with 300 𝐾 and 95 kPa at the start of the 

compression stroke. The compression ratio is 9.5 and the maximum temperature of 

the cycle is 1100 K. On the basis of a cold air-standard analysis, using 𝑐𝑣 =

0.718

𝑘𝐽

𝑘𝑔−𝐾

and 𝑐𝑝 = 1.005

𝑘𝐽

𝑘𝑔

−𝐾

, determine the 

(a) temperature at the end of the compression stroke, in K

(b)peak pressure, in kPa

(c) Heat Addition, kJ/kg

(d) net work of the cycle, kJ/kg

(e)thermal Efficiency 

Answer 1

Q163246

 Deadline: 16.02.21, 04:18

An air-standard Otto cycle operates with 300 𝐾 and 95 kPa at the start of the compression stroke. The compression ratio is 9.5 and the maximum temperature of the cycle is 1100 K. On the basis of cold air-standard analysis, using 𝑐𝑣 =

0.718𝑘𝐽/ 𝑘𝑔−𝐾 and 𝑐𝑝 = 1.005𝑘𝐽/ 𝑘𝑔−𝐾

, determine the 

(a) the temperature at the end of the compression stroke, in K

(b)peak pressure, in kPa

(c) Heat Addition, kJ/kg

(d) net work of the cycle, kJ/kg

(e)thermal Efficiency.

Solution:

The PV diagram for the Otto cycle is given as

Process 1 -2 is the adiabatic compression.

Process 2 - 3 is the Heat addition at constant volume

Process 3 - 4 is the adiabatic expansion.

Process 4 - 1 is the Heat rejection at constant volume.

At the start of the compression, we are given the pressure and temperature

which is equal to P1 = 95 kPa and T₁ = 300K.

We are already given the compression ratio = 9.5

$So, \frac{V_1}{V_2} = \frac{V_3}{V_4} = 9.5$

V₁ = V₃ and V₂ = V₄ because process 2-3 and 4-1 are at constant volume.

Step 1: To find the temperature at the end of the compression stroke

We can use the formula

$\frac{T_2}{T_1} = (\frac{V_1}{V_2} )^{(k-1)}$

where 'k' is the specific heat ratio of the gas. $k=\frac{C_p}{C_v} = \frac{1.005kJ/kg.K}{0.718kJ/kg.K} = 1.4$

substitute T₁ = 300K, V₁/V₂ = 9.5 and k = 1.4 in the formula we have

$\frac{T_2}{300K} = (9.5 )^{(1.4-1)}$

$\frac{T_2}{300K} = (9.5 )^{(0.4)} = 2.4609;$

T₂ = 2.4609 * 300K = 738.3 K.

Hence the temperature at the end of the compression stroke is 738 K.

Step 2: To find P₂.

P₁ = 95kPa, T₁ = 300K , T₂ = 738.3K V1/V2 = 9.5

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$\frac{P_1T_2V_1}{T_1V_2}=P_2$

$\frac{95kPa*738.3K*9.5}{300K}=P_2$

P₂ = 2221kPa

Step 3: To find peak pressure in 'kPa'

In the diagram, we can see that the peak pressure is P₃. We have to find out this peak pressure.

Process 2-3 is the Heat addition at constant volume.

In this process addition of heat is taking place at constant volume. This increases the temperature, so the temperature will reach maximum in this step.

Hence T₃ = maximum temperature = 1100K.

We can use the formula, $\frac{P_3V_3}{T_3}=\frac{P_2V_2}{T_2}$ and find the Peak pressure.

volume is constant in this step, so V3 = V2, we can cancel out V3 and V2 from both the side.

$\frac{P_3}{T_3}=\frac{P_2}{T_2}$

substitute P₂ = 2221kPa, T₃ = 1100K, T₂= 738.3K and find maximum pressure P3.

$\frac{P_3}{1100K}=\frac{2221kPa}{738.3K}$

$P_3=\frac{2221kPa*1100K}{738.3K}$

P3 = 3309 kPa.

which in 2 significant figures is 3300 kPa.

Hence the peak pressure will be 3300 kPa.

Step 4: To find the heat added in kJ/kg.

Heat is added in the process 2 - 3. This process is taking place at constant volume.

We can use the formula

Q _added = C_v * ( T₃- T₂) = 0.718kJ/kg.K * (1100K-738.3K)

= 0.718 kJ/kg.K* 361.7K = 259.7kJ/kg.

= 260 kJ/kg.

Hence 260 kJ/kg of heat is added in the cycle.

Step 5: To find net work done of the cycle.

The difference between the Q_added and Q _rejected will given us the amount of work done by the cycle.

I process 4-1 heat is rejected by the cycle. This process is also taking place at constant volume. But before that you will need T4. We can use the same formula which we used in

Step 1 and find the T4.

$\frac{T_4}{1100K} = (9.5)^{-(1.4-1)} = 9.5^{-0.4} = 0.4064$

T₃ = 1100K * 0.4064 = 447 K

Q _rejected = Cv * ( T1 - T4 ) = 0.718 kJ/kg.K * ( 300K - 447K)

= 104.08 kJ/kg

Work done = Q added - Q rejected = 260kJ/kg - 104.08 kJ/kg

= 155.9 kilojoules.

Hence work done is 155.9 kilojoules.

Step 6 : To find Thermal efficiency ,

η = work done/ Q added = 155.9kJ / 260kJ = 0.5997 = 0.6

Hence the thermal efficiency of the given cycle will be 0.6 or 60.0 %