An air standard cycle with constant specific heats is executed in a closed system with 0.003 kg of air and consists of the follo
wing three processes:
1-2 - V= Constant heat addition from 95 kPa and 17 Deg Celcius to 380 kPa
2-3 - Isentropic expansion to 95 kPa
3-1 - P= Constant heat rejection to initial state
a) Show cycle on P-v and T-s diagrams
b) Calculate net work per cycle in kJ
c) Determine thermal efficiency
1-2 - V= Constant heat addition from 95 kPa and 17 Deg Celcius to 380 kPa
2-3 - Isentropic expansion to 95 kPa
3-1 - P= Constant heat rejection to initial state
a) Show cycle on P-v and T-s diagrams
b) Calculate net work per cycle in kJ
c) Determine thermal efficiency
1 answer:
Answer:
a) Please see attached copy below
b) 0.39KJ
c) 20.9‰
Explanation:
The three process of an air-standard cycle are described.
Assumptions
1. The air-standard assumptions are applicable.
2. Kinetic and potential energy negligible.
3. Air in an ideal gas with a constant specific heats.
Properties:
The properties of air are gotten from the steam table.
b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.
P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K
T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K
Qin=m(u₂₋u₁)=mCv(T₂-T₁)
=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ
Qout=m(h₃₋h₁)=mCp(T₃₋T₁)
=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ
Wnet, out= Qin-Qout = (1.87-1.48)KJ =0.39KJ
c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰
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Answer:
See explanation
Explanation:
Solution:-
- Three students measure the volume of a liquid sample which is 6.321 L.
- Each student measured the liquid sample 4 times. The data is provided for each measurement taken by each student as follows:
Students
Trial A B C
1 6.35 6.31 6.38
2 6.32 6.31 6.32
3 6.33 6.32 6.36
4 6.36 6.35 6.36
- We will define the two terms stated in the question " precision " and "accuracy"
- Precision refers to how close the values are to the sample mean. The dense cluster of data is termed to be more precise. We will use the knowledge of statistics and determine the sample standard deviation for each student.
- The mean measurement taken by each student would be as follows:
- The precision can be quantize in terms of variance or standard deviation of data. Therefore, we will calculate the variance of each data:
- We will rank each student sample data in term sof precision by using the values of variance. The smallest spread or variance corresponds to highest precision. So we have:
Var ( A ) < Var ( B ) < Var ( C )
most precise Least precise
- Accuracy refers to how close the sample mean is to the actual data value. Where the actual volume of the liquid specimen was given to be 6.321 L. We will evaluate the percentage difference of sample values obtained by each student .
- Now we will rank the sample means values obtained by each student relative to the actual value of the volume of liquid specimen with the help of percentage difference calculated above. The least percentage difference corresponds to the highest accuracy as follows:
P ( B ) < P ( A ) < P ( C )
most accurate least accurate