![aalexissm](/avatars/37382.jpg)
aalexissm
20.06.2020 •
Engineering
A piston-cylinder assembly contains 5kg of water that undergoes a series of processes to form a thermodynamic cycle. Process 1à2: Constant pressure cooling from p1=20bar and T1=360°C to saturated vapor Process 2à3: Constant volume cooling to p3=5 bar Process 3à4: Constant pressure heating Process 4à1: Polytropic process following Pv =constant back to the initial state Kinetic and potential energy effects are negligible. Calculate the net work for the cycle in kJ.
Solved
Show answers
More tips
- O Other What happens if you get scared half to death twice?...
- H Horoscopes, Magic, Divination Where Did Tarot Cards Come From?...
- S Style and Beauty How to Make Your Lips Fuller? Ideas and Tips for Beautiful Lips...
- S Style and Beauty How are artificial nails removed?...
- F Family and Home How to Sew Curtain Tapes: Best Tips from Professionals...
- H Horoscopes, Magic, Divination How to Cast a Love Spell on a Guy? Guide for Guys...
- F Family and Home How to Properly Use a Water Level?...
- L Legal consultation What Documents Are Required for a Russian Passport?...
- H Health and Medicine How to Treat Styes: Causes, Symptoms, and Home Remedies...
- F Family and Home Protect Your Home or Apartment from Pesky Ants...
Answers on questions: Engineering
- B Business Exercise 3. Labor Supply (Cobb-Douglas) Each day you are endowed with 24 hours of time (T=24) that you can either spend working (L) or in leisure (l), so that T = L + l. The reason...
- M Mathematics The area of the rectangle is 10. Find the value of x. (Give answer to 3 decimals.) 5x-4 8x+1...
Ответ:
The net work done is 272.38 kJ
Explanation:
The parameters given are;
Mass of water = 5 kg
p₁ = 20 bar
T₁ = 360°C
v₁ = 0.141147 m³/kg
Process 1 to 2 = Constant pressure process
p₂ = 20 bar
Process 2 to 3 = Constant volume process
p₃ = 5 bar
Process 3 to 4 = Constant pressure process
Process 4 to 1 = Polytropic process pv = Constant
For Stage 1 to 2, we have;
p₂ = 20 bar
From the steam tables for superheated steam, we have;
T₂ = 212.385°C
v₂ = 0.0995805 m³/kg
Work done = p₂×(v₂ - v₁) = 2×10⁶ × (0.0995805 - 0.141147 ) = -83133 J/kg
For the 5 kg, we have;
Stage 2 to 3: Constant volume cooling
v₂ = v₃ = 0.0995805 m³/kg
p₃ = 5 bar
T₃ = 151.836°C
(0.0995805 - 0.00109256)/(0.374804 - 0.00109256) = 0.2635 liquid vapor mixture
Work done,
= 0
Stage 3 to 4: Constant pressure heating
p₃ = p₄ = 5 bar
v₄/T₄ = v₃/T₃
v₄ = 0.374804 m³/kg
T₄ = v₄×T₃/v₃ = 0.374804*(273.15 + 151.836)/0.0995805 = 1599.6 K = 1326.4 °C
Work done = p₄×(v₄ - v₃) = 5×10⁵ × (0.374804 - 0.0995805 ) = 137611.75 J/kg
For the 5 kg, we have;
Stage 4 to 1: Polytropic process
Which gives;
n = log(5/20) ÷log(0.141147/0.374804) = 1.42
Work done,
, is given as follows;
Which gives;
For the 5 kg, we have;
The net work done,
, is therefore;
-415,665 + 688,058.75 -11.2967 = 272,382.45 J = 272.38 kJ.
Ответ: