FUND OF ENG THERMODYN(LLF)+WILEYPLUS
9th Edition
ISBN: 9781119391777
Author: MORAN
Publisher: WILEY
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When a system undergoes a Isochoric process (P = 5 bar), the change in internal internal is 5 kJ and temperature during the process is 15°C. Then its change in entropy would be approximately 1.736 kJ/kgK
Select one:
True
False
Q1: 0.09 m³ of a fluid at 0.7 bar are compressed reversibly to a pressure of 3.5 bar according to
a law py"=constant. The fluid is then heated reversibly at a constant volume until the pressure
is 4 bar; the specific volume is then 0.5 m³/kg. A reversible expansion according to a law
pv²=constant restores the fluid to its initial state. Calculate the net work done on or by the fluid
in the cycle and sketch the cycle on a p-v diagram
i) Use the thermodynamic identity for a P-V-T system and the equation of state to show
that the entropy change of one mole of an ideal gas of vibrating diatomic molecules,
as volume and temperature are changed, is given by;
7
AS = S(V;,T;) – S(V;, T;) =
;Rln
+ Rln
T
ii) The gas undergoes an isobaric compression from V; to V;/2. Evaluate the change in
the number of microstates of the system that occurs as a result.
ii) Using the equation derived in i), demonstrate that for an ideal gas undergoing an
adiabatic expansion from initial volume V; to final volume V; the change in entropy is
zero.
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- An oxygen gas R = 0.2598 KJ/kg°k and k = 1.395. If 4 kg of oxygen undergo a reversible non flow constant pressure process from initial volume =1.2 cubic meter and initial pressure = 690 kPa to a state where final temperature = 600°C. 1. Determine the Change in Internal Energy. choices: a.200.60 KJ. b.198.45 KJ. c.99.54 KJ. d.200.55 KJ 2. Determine the constant pressure-specific heat. choices: a.0.9865 KJ/kg-°K. b.0.9175 KJ/kg-°K. c.0.8580 KJ/Kg-°K. d.0.7843 KJ/kg-°K need complete solution, cancellation and symbol:)arrow_forwardIdentify the working substance, specify the kind of system and sketch the system boundary. PLEASE ANSWER it in 1hr.arrow_forwardThe following processes occur in a reversible thermodynamic cycle: 1-2: 0.2 kg heating at constant pressure 1.05 bar at specific volume 0.1 m3 3/kg and work done -515 J. 2-3: Isothermal compression to 4.2 bar. 3-4: Expansion according to law pv1./= constant. 4-1: heating at constant volume back to the initial conditions. Calculate the pressure at state four in bar to 3 decimal places?arrow_forward
- When U=f(T,v) for an ideal gas if B=1.4 E -5 /K, and if partial derivative of internal energy with respect to volume at constant temperature 6.16 J/m3 with V=5 m3 then the pressure * :is 7.221 E 4 Pa 3.14 E 5 Pa 1.161 E 5 Pa 6.73 E 6 Pa In Otto cycle if T1=50 C, T2=70 C, * :V1=7.5 m3, y=1.5, then V3 is equal to 7.08 m^3 3.11 m^3 6.65 m^3 18.23 m^3arrow_forwardAn ideal gas undergoes a process from state 1 ( the properties are T₁ = 300 K, p₁ = 100 kPa) to state 2 (the properties are T₂ = 600 K, p₂ = 500 kPa). The specific heats of the ideal gas are: c = 1 kJ/kg-K and c = 0.7 kJ/kg-K.. The change in specific entropy of the ideal gas to two decimal places)from state 1 to state 2 (in kJ/kg-K) is......arrow_forwardFast ,Do not hold. Two heat engines receive heat from a source at temperature of 550◦C. Heat engine "A" receives 200 kJ of heat and rejects the waste heat to a sink at 180◦C. Heat engine "B" receives 180 kJ of heat and rejects the waste heat to a sink at 120◦C.(a) Caclualte the generated entropy, Sgen, in both processes.(b) Based on your answer in part (a), identify the heat transfer that is more irreversible.arrow_forward
- A fluid expands reversibly according to a linear law from 3.8 bar to 1.1 bar, the initial volume is 0.006 mº. The fluid is then cooled reversibly at constant pressure, and finally compressed reversibly according to a law pv = constant back to the initial conditions of 3.8 bar and 0.006 m. Calculate the net work of the cycle if the work done during the constant pressure cooling is given by 5930 N.m. Sketch the cycle on a p-v diagram.arrow_forwardkg 3. The working fluid in a steady flow process flows at a rate of 220 The fluid min rejects 100 - KJ passing through the system. The conditions of the fluid at the inlet and outlet are given: V1 = 320 "; P, = 6.0 bar; U1 = 2,000 ; v1= 0.36 : KJ m3 and kg kg m3 The suffix 1 indicates the kg KJ V2 = 140 4; P2 = 1.2 bar; U2 = 1,400 : ; V2= 1.3 kg condition at the inlet and 2 indicates the outlet of the system. Determine the power capacity of the system in MW.arrow_forwardDetermine the change in specific entropy, in kJ/kg K, of CO2 as an ideal gas undergoing a process from T, = 300 K, p, = 1 bar to T2 = 1420 K. P2 = 5 bar. Additional information g°= 1.70203 KJKG K °2 = 3.37901 kJikg K 1.215 kJ/kg. °C 1.215 kJkg. K 0 1.190 kJ/kg K O 1.373 kJ/kg Karrow_forward
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