A fluid at 0.7 bar occupying 0.09m³ is compressed reversibly to a pressure of(a)3.5 bar according to a law pv" = constant. The fluid is then heated reversibly at constantvolume until the pressure is 4 bar: the specific volume is then 0.5m³/kg. A reversibleexpansion according to a law pv = constant restores the fluid to its initial state. Sketch thecycle on a p-v diagram and calculate:i) the mass of the fluid present;ii) the value of n in the first process;1.iii) the net work of the cycle. Table6.5 Properties ofsome common gasesat 300KRGas(kJ/kg K)(kJ/kmol K)(kg/kmol)(kJ/kg K)DiatomicCarbon monoxide (CO)Hydrogen (H2)Nitrogen (N2)Oxygen (O2)0.29681.041014.32301.04000.918220.84520.56120.81921.06828.0102.01628.01331.9990.74421.39929.15828.87529.13429.3824.12430.29680.259810.19871.4041.3991.3950.74320.6584MonatomicArgon (Ar)Helium (He)0.520312.47212.47339.9500.20812.07710.31221.6661.66620.78620.7885.19303.11594.003TriatomicCarbon dioxide (CO2)Sulphur dioxide (SO2)0.84570.656837.2191.2881.25228.90644.0100.18890.64480.515041.30632.99164.0600.1298HydrocarbonsEthane (C,H6)Methane (CH,)Propane (C,Hg)1.49031.71321.76681.18653.12835.79574.57844.81327.48066.26330.0700.27652.23161.69151.3031.12616.0400.51840.18861.502944.090

Given: The pressure at state 1, P1 = 0.7 bar The volume at state 1, V1 = 0.09 m^3 The pressure at…

1. (a) A fluid at 0.7 bar occupying 0.09m3 is compressed reversibly to a pressure of 3.5 bar according to a law pv" = constant. The fluid is then heated reversibly at constant volume until the pressure is 4 bar: the specific volume is then 0.5m³/kg. A reversible expansion according to a law pv = constant restores the fluid to its initial state. Sketch the cycle on a p-v diagram and calculate: i) the mass of the fluid present; ii) the value of n in the first process; iii) the net work of the cycle. %3D

1. (a) A fluid at 0.7 bar occupying 0.09m3 is compressed reversibly to a pressure of 3.5 bar according to a law pv" = constant. The fluid is then heated reversibly at constant volume until the pressure is 4 bar: the specific volume is then 0.5m³/kg. A reversible expansion according to a law pv = constant restores the fluid to its initial state. Sketch the cycle on a p-v diagram and calculate: i) the mass of the fluid present; ii) the value of n in the first process; iii) the net work of the cycle. %3D

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Work and Heat Transfer

It is generally related to thermal engineering which tells about the use, creation, conversion and transfer of heat energy between the systems. It is also considered that to transfer heat from one source to another, transfer of masses is also required. There are various mechanisms by which, transfer of heat takes place i.e. thermal conduction, convection, radiation or by change of phase. All these mechanisms have their own specific features which sometimes occur simultaneously within the same system.

Question

A fluid at 0.7 bar occupying 0.09m³ is compressed reversibly to a pressure of
(a)
3.5 bar according to a law pv" = constant. The fluid is then heated reversibly at constant
volume until the pressure is 4 bar: the specific volume is then 0.5m³/kg. A reversible
expansion according to a law pv = constant restores the fluid to its initial state. Sketch the
cycle on a p-v diagram and calculate:
i) the mass of the fluid present;
ii) the value of n in the first process;
1.
iii) the net work of the cycle.

Table
6.5 Properties of
some common gases
at 300K
R
Gas
(kJ/kg K)
(kJ/kmol K)
(kg/kmol)
(kJ/kg K)
Diatomic
Carbon monoxide (CO)
Hydrogen (H2)
Nitrogen (N2)
Oxygen (O2)
0.2968
1.0410
14.3230
1.0400
0.9182
20.845
20.561
20.819
21.068
28.010
2.016
28.013
31.999
0.7442
1.399
29.158
28.875
29.134
29.382
4.1243
0.2968
0.2598
10.1987
1.404
1.399
1.395
0.7432
0.6584
Monatomic
Argon (Ar)
Helium (He)
0.5203
12.472
12.473
39.950
0.2081
2.0771
0.3122
1.666
1.666
20.786
20.788
5.1930
3.1159
4.003
Triatomic
Carbon dioxide (CO2)
Sulphur dioxide (SO2)
0.8457
0.6568
37.219
1.288
1.252
28.906
44.010
0.1889
0.6448
0.5150
41.306
32.991
64.060
0.1298
Hydrocarbons
Ethane (C,H6)
Methane (CH,)
Propane (C,Hg)
1.4903
1.7132
1.7668
1.186
53.128
35.795
74.578
44.813
27.480
66.263
30.070
0.2765
2.2316
1.6915
1.303
1.126
16.040
0.5184
0.1886
1.5029
44.090

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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
Q2/ A power cycle operating between two reservoirs receives energy Qi by heat transfer from a hot reservoir at T 1000K and rejects energy Q2 by heat transfer to a cold reservoir at Tz-200K For each of the following cases, determine whether the cycle operates reversibly, irreversibly, or is impossible. (a) QI = 500KJ, W 350KJ (b) Q1 = 1000KJ, Q2 200KJ (c) W= 800KJ, Q2 = 500KJ (d) 7 = 95 % %3D %3D
4.30 Refrigerant 134a enters a heat exchanger operating ai steady state as a superheated vapour at 10 bars. 60°C. where it is cooled and condensed to saturated liquid at 10 bars. The mass flow rate of the refrigerant is 10 kg/min. A separate stream of air enters the heat exchanger at 37°C with a mass flow rate of 80 kg/min. Ignoring heat transfer from the outside of the heat exchanger and neglecting kinetic and potential energy effects, determine the exit air temperature, in °C.
A 2.00 mol sample of an ideal diatomic gas is taken through a reversible cycle A → B → C → A as shown in the figure below. Process A → B is isothermal expansion. For the cycle, calculate: a) the work done by the gas, b) the heat energy added to the gas, c) the heat energy exhausted by the gas, d) the thermal efficiency, e) the change in entropy of the gas. f) Determine the efficiency of a Carnot engine operating between the same temperature extremes, R = 8.314 J/mol K Data at vertices for the above PV diagram are as follow: PA = 5.00 x 105 Pa, VA = 1.00 x 10 -2 m3, PB = 1.25 x 105 Pa, VB = 4.00 x 10 -2 m3 PC = 1.25 x 105 Pa, VC = 1.00 x 10 -2 m3
The manufacturer of the device claims that it accepts a heat transfer at a rate of QH = 3.0 kW from a low-grade source of geothermal heat at TH = 80ºC and rejects heat at rate QC to the atmosphere at TC = 20ºC. The device operates at steady state and produced work at a rate of W= 2.5 kW. There are no other heat or work transfers from the device. a.) What is the rate at which the device rejects heat to the atmosphere, QC ? b.) Is this device possible? Justify your answer using an entropy balance.
Thermodynamics
The last 3 parts B,D,E