FUND OF ENG THERMODYN(LLF)+WILEYPLUS
9th Edition
ISBN: 9781119391777
Author: MORAN
Publisher: WILEY
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A fluid expands reversibly according to a linear law from 3.8 bar to 1.1 bar, the initial volume is 0.006 m3. 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 m3. 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
In an air conditioning system running at steady-state, m ̇ = 0.7 kg/s of refrigerant 3
134a in saturated liquid state at 48◦C flow through a throttling valve reducing its pressure
to a value of p4 = 4 bars. The system is shown in Fig. 1. Then the refrigerant flows through
the internal side of a heat exchanger exiting at saturated vapor with p5 = p4. Air enters the
external side of the heat exchanger at T1 = 300 K and exits at T2 = 295 K moved by a fan ̇
Figure 1: Problem 1
that consumes WCV = 0.15 kW. Determine the mass flow rate of the air, in kg/s
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- Liquid water flows isothermally at 20°C through a one-inlet, one-exit duct operating at steady state. The duct's inlet and exit P2 = 4.8 bar T = 320°C diameters are 0.02 m and 0.04 m, Water vapor (AV)2 = (AV)3 respectively. At the inlet, the velocity is 50 m/s and the pressure is 1 bar. At the exit, determine the mass flow rate, in kg/s, and V, T A1 = 0.2 m? P1 = 5 bar 3 velocity, in m/s. P3= 4.8 bar T3 = 320°Carrow_forwardQ1: 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 diagramarrow_forwardA 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_forward
- 8Q Prove that, whenever a system undergoes a cycle, p<0 (b) T.arrow_forward4arrow_forwardA fluid at 0.6 bar occupying 0.1 m° is compressed reversibly to a pressure of 11.1 bar and specific volume of 0.4 m/kg according to the law pv =c. The fluid then expands reversibly according to the law pv = c to 2.1 bar. A reversible cooling at constant volume then restores the fluid back to initial state. Calculate the value of n in the procesS.arrow_forward
- 2. An ideal gas undergoes a process from state 1 ( T1= 300 K, P1= 100 kPa) to state 2 ( 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 from state 1 to state 2 (in kJ/kg-K) is (correct to two decimal places).arrow_forward3. A thermodynamic system operates under steady flow conditions, the fluid entering at 2 bar and leaving at 10 bar. The entry velocity is 30 m/s and exit velocity is 10 m/s. During the process 25 MJ/hr of heat from an external source is supplied and the increase in enthalpy is 5kJ/kg. The exit point is 20 m above the entry point. Determine flow work from the system if the fluid flow rate is 45 kg/min.arrow_forwardThe entropy of an ideal gas depends on both T and P. The function s° represents only the temperature-dependent part of entropy.arrow_forward
- Q2: Steam enters a converging-diverging nozzle operating at steady state with Pi=40 bar, T-400C, and a velocity of 10 m/s. The steam flows through the nozzle adiabatically and no significant change in elevation. At the exit, p:=1.5 MPa, and the velocity is 665 m/s. The mass flow rate is 2 kg/s. Determine the exit area of the nozzle, in (m²). also, drive the (T-V) diagram for the steam. י2kg-מ Insulation -15 bar -665 ms 40 bar - 400 "C -10 m's Control volume boundaryarrow_forwardAn 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_forwardArgon gas flows through a well-insulated nozzle at steady state. The temperature and velocity at the inlet are 590°R and 150 ft/s, respectively. At the exit, the temperature is 440°R and the pressure is 40 Ibę/in?. The area of the exit is 0.0085 ft². Use the ideal gas model with k = 1.67, and neglect potential energy effects. Determine the velocity at the exit, in ft/s, and the mass flow rate, in Ib/s.arrow_forward
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