EG-210 EG- 210 Tutor i al Shee heet N o. 2 (2013) (2013) An swer wer s
The Second Law of Thermodynamics and Cyclic Processes 1) (a) Steam enters a horizontal pipe operating at steady state with a specific enthalpy of 3000 kJ/kg and a mass flow rate of 0.5 kg/s. At the exit, the specific enthalpy is 1700 kJ/kg. If there is no significant change in kinetic energy from inlet to exit, determine the rate of heat transfer between the pipe and its surroundings, in kW. Answer Diagram of system: h1 = 3000 kJ/kg m = 0.5 kg/s
h2 = 1700 kJ/kg m = 0.5 kg/s
Starting with the general energy balance for open systems: 0 Q W S m (h1 h2 ) g ( z 1 z 2 )
1
2
2
(v1
v22 )
No work is done (W (W S = 0) and ignoring potential and kinetic energy, 0 Q m(h1 h2 ) Q m(h2 h1 ) Q 0.5(1700 3000) 650 650 kJ/kg [5 marks] (b) Steam enters a well-insulated turbine operating at steady state at 4 MPa with a specific enthalpy of 3015.4 kJ/kg and a velocity of 10 m/s. The steam expands to the turbine exit where the pressure is 0.07 MPa, specific enthalpy is 2431.7 kJ/kg, and the velocity is 90 m/s. The mass flow rate is 11.95 kg/s. Neglecting potential energy effects, determine the power developed by the turbine, in kW. Answer
Starting with the general energy balance for open systems: 1
0 Q W S m (h1 h2 ) g ( z 1 z 2 )
1
2
(v12
v22 )
We are told the turbine is well insulated, therefore Q = 0 and we also assume PE change is zero, giving: W S
m (h1 h2 ) 12 (v12 v22 )
We know m = 11.95 kg/s, then; W S m (3015.4 2431.7) 1 2 (10
2
90 2 ) / 1000 6927 kW [5 marks]
(c) Steam enters a turbine operating at steady state with a mass flow of 10 kg/min, a specific enthalpy of 3100 kJ/kg, and a velocity of 30 m/s. At the exit, the specific enthalpy is 2300 kJ/kg and the velocity is 45 m/s. The elevation of the inlet is 3 m higher than at the exit. Heat transfer from the turbine to its surroundings occurs at a rate of 1.1 kJ per kg of steam flowing. Let g Let g = 9.81 = 9.81 m/s2. Determine the power developed by the turbine, in kW. Answer
Starting with the general energy balance for open systems: 0 Q W S m (h1 h2 ) g ( z 1 z 2 )
1
2
(v12
v22 )
Re-arranging gives: W S m
Q m
(h1 h2 ) g ( z 1 z 2 ) 1 2 (v12 v22 )
Therefore: W S m
1.1 (3100 2300) (9.81 3 / 1000) 1 2 (30 2 452 ) / 1000 798.37 kJ/kg
Now m = 10 kg/min = 0.1667 kg/s, so: W S S = 0.1667 × 798.37 = 133.09 kJ/s = 133.09 kW [5 marks] 2
(d) Air enters a compressor operating at steady state at 1 atm with a specific enthalpy of 290 kJ/kg and exits at a higher pressure with a specific enthalpy of 1023 kJ/kg. The mass flow rate is 0.1 kg/s. If the compressor power input is 77 kW, determine the rate of heat transfer between the compressor and its surroundings, in kW. Neglect kinetic and potential energy effects and assume the ideal gas model. Answer
Starting with the general energy balance for open systems: 0 Q W S m (h1 h2 ) g ( z 1 z 2 )
1
2
2
(v1
v22 )
Ignoring potential and kinetic energy, and re-arranging gives; Q W S m(h2
h1 )
Q 77 0.1 (1023 290 290) 3.7 kW [5 marks] (e) A pump delivers water through a hose terminated by a nozzle. The exit of the nozzle has a diameter of 2.5 cm and is located 4 m above the pump inlet pipe, which has a diameter of 5.0 cm. The pressure is equal to 1 bar at both the inlet and the exit, and the temperature is constant at 20 C. The magnitude of the power input required by the pump is 8.6 kW and the acceleration due to gravity, g = 9.81 m/s 2. The specific volume of water at 20 C and 1 bar is 1.0018 × 10 -3 m3/kg. Determine the mass flow rate delivered by the pump in kg/s. Answer
3
Starting with the general energy balance for open systems: 0 Q W S m (h1 h2 ) g ( z 1 z 2 )
1
2
(v12
v22 )
Assuming that heat transfer is negligible (Q (Q = 0) and as the input and output water conditions are the same (h (h1 = h2), then; 0 W S
m g ( z 1 z 2 ) 1 2 (v12 v 22 )
Now assuming that the water is incompressible the velocity of water can be calculated using: v m spec specific ific volume / A So v1 v2
m 1.0018 10 3 /( 0.052 / 4) 0.5102m
m 1.0018 10 3 /( 0.0252 / 4) 2.04085m
0 (8.6) m 9.81 (4) / 1000
1
2
((0.5102m) 2
0 8.6 1.9524 103 m 3
(2.04085m) 2 ) / 1000
0.03924m
Solving for m gives; m = 15.98 kg/s [5 marks]
4
2) (i) The data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1000 K and 300 K, respectively. For each case, determi ne whether the cycle operates reversibly, operates irreversibly, or is impossible. impossible. (a) Qh = 600 kJ, W eng eng = 300 kJ, Qc = 300 kJ (b) Qh = 400 kJ, W eng eng = 280 kJ, Qc = 120 kJ (c) Qh = 700 kJ, W eng eng = 300 kJ, Qc = 500 kJ (d) Qh = 800 kJ, W eng eng = 600 kJ, Qc = 200 kJ Answer For each process first check the first law balance: W eng eng = Qh - Qc and if OK then check that: Qh
For a reversible process:
T h Qh
For an irreversible process:
T h Qh
For an impossible process:
T h
Qc
T c Qc T c Qc T c
a) First Law: W eng eng = Qh - Qc 300 = 600 – 300 Qh T h
600 1000
0.6
Qc T c
300 300
1 therefore
b) First Law: W eng eng = Qh - Qc 280 = 400 – 120 Qh T h
400 1000
0.4
Qc T c
120 300
0.4
First law OK so, Qh T h
Qc T c
so process is irreversible
First law OK so,
therefore
Qh T h
Qc T c
so process is reversible
c) First Law: W eng eng = Qh - Qc 300 ≠ 700 – 500
First law violated so process impossible
d) First Law: W eng eng = Qh - Qc 600 = 800 – 200
First law OK so,
Qh T h
800 1000
0.8
Qc T c
200 300
0.66
therefore
Qh T h
Qc T c
so process is impossible [4 marks]
(ii) A reversible power cycle operating between hot and cold reservoirs at 1000 K and 300 K, respectively, receives 100 kJ by heat transfer from the hot reservoir for each cycle of operation. Determine the net work developed in 10 cycles of operation, in kJ. Answer First calculate the efficiency of the heat engine using;
5
1
T c T h
1
300 1000
0.7
The efficiency can also be calculated using:
net work output heat input
W eng Qh
So, W eng eng = Qh = 0.7 × 100 = 70 kJ For 10 cycles power generated is: Net work = no cycles × W eng eng = 10 × 70 = 700 kJ [3 marks] (iii) In a heat-treating process, a 1-kg metal part, initially at 1075 K, is quenched in a tank containing 100 kg of water, initially at 295 K. There is negligible heat transfer between the contents of the tank and their surroundings. The metal part and water can be modeled as incompressible with specific heats 0.5 kJ/kg K and 4.2 kJ/kg K, respectively. Determine (a) the final equilibrium temperature after quenching, in K, and (b) the amount of entropy produced within the tank, in kJ/K. Answer First write the equation describing the energy change of the metal in cooling from 1075 K to T . Qmet = mC p T = = 1 × 0.5 × (1075 – T ) = 0.5 (1075 – T ) Next, write the equation describing the energy change of the water in being heated from 295 °C to T . Qwat = mC p T = = 100 × 4.2 × (T ( T – 295) = 420 (T ( T – 295) – 295) – 295) Applying first law heat balance: Qwat = Qmet – 295) 420 (T (T – 295) = 0.5 (1075 – T )
T = = 295.93 K Calculate the entropy change of the copper,
ΔS met
T 295.93 mC p ln 2 1 0.5 ln 0.645 kJ / K T 1075 1 6
Calculate the entropy change of the water,
ΔS wat
T 295.93 mC p ln 2 100 4.2 ln 1.322 kJ / K T 295 1
Calculate the overall entropy change,
S T S met ΔS wat wat 0.645 1.322 0.677 kJ / K [6 marks] (iv) Propane at 0.1 MPa, 20 C enters an insulated compressor operating at steady state and exits at 0.4 MPa, 90 C Neglecting kinetic and potential energy effects, determine (a) the power required by the compressor, in kJ per kg of propane flowing. (b) the rate of entropy production within the compressor, in kJ/K per kg of propane flowing. Data supplied: at 0.1 MPa, 20 C h propane = 517.6 kJ/kg and s propane = 2.194 kJ/kg K, at 0.4 MPa, 90 C h propane = 639.2 kJ/kg and s and s propane = 2.311 kJ/kg K, Answer
Starting with the general energy balance for open s ystems: 0 Q W S m (h1 h2 ) g ( z 1 z 2 )
1
2
(v12
v22 )
As the compressor is insulated Q = 0 and neglecting PE and KE, gives; W S m W S m
h1 h2
517.6 639.2 121.6 kJ / kg
Starting with the entropy balance: 0
Qk
T m s m s i i
j j
S gen
k
As the mass flow is constant and Q is zero then;
7
0 m( s1 S gen m
s 2 ) S gen
s 2 s1 2.311 2.194 0.117 kJ/kg K [3 marks]
(v) The pressure – volume volume diagram of a Carnot power cycle executed by an ideal gas with constant specific heat ratio is shown below. Show that V 4V 2 = V 1V 3
Answer As step 1 – 2 2 occurs at a constant temperature which is higher than step 3 to 4 then, T 1 = T 2 = T h For an isothermal change:
and
T 3 = T 4 = T c
V 2 therefore; V 1
Q W RT ln
V 2 V 1
For step 1 to 2:
W 12 RT h
ln
For step 3 to 4:
W 34 RT c
ln
V 4 V 3
The cycle efficiency is given by:
net work W 12 energy in
W 34
Qin
1
T c T h
8
V 4 V 2 RT l n c V V 1 3 1 T c T h V RT h ln 2 V 1
RT h ln
V 3 V 4 T 1 c 1 T h V RT h ln 2 V 1 RT c
ln
V 3 V 4 T c V T h T h ln 2 V 1 T c
ln
Therefore,
V 3 V 2 l n V 4 V 1
ln
V 3 V 2 V 4 V 1 V 3 V 1 = V 4 V 2 [6 marks]
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3) A heat engine cylinder contains 0.1 kg of air. The engine is ass umed to operate on a Carnot cycle comprising the following four processes: Step A-B: Isothermal expansion at 1000 K during which the volume of the air is doubled. Step B-C: Adiabatic expansion until the temperature of the air reaches 350 K and the pressure 1 bar. Step C-D: Isothermal compression of the air at 350 K. Step D-A: Adiabatic compression until the air temperature reaches 1000 K and the cycle is complete.
Assuming that air behaves as an ideal gas and given the following data (specific heat capacity of air in a constant volume process C V 0.7175 kJ kg -1 K -1; Gas constant for air R = 0.2867 V = kJ kg-1 K -1), then a) b) c) d) e) f) g)
Sketch the cycle on a P-V diagram. What is the maximum pressure achieved in the c ycle? Show that the volume of air is halved during the isothermal compression. Calculate the work done during each of the adiabatic processes. Calculate the heat and thus work effects for each of the isothermal processes. Determine the net work done per cycle. Confirm the cycle efficiency matches the prediction given by = 1 – (T (T c/T h) [25 marks]
Answer: a)
[4 marks] b) Step B-C is an adiabatic expansion from T B =1000 K to T C K, P C C = 350 K, P C = 1 bar, therefore;
T C P C T B P B
1
10
Re-arranging
P B
T P C B T C
1
Now C P = C V + R = = 0.7175 + 0.2867 = 1.0042 kJ kg -1 K -1 V + R So,
C P C V
1.0042 0.7175
1.3996
Resulting in: 1.3996
P B
1000 1 350
1.3996 1
39.53 bar bar
As air behaves as an ideal gas: P AV A = P = P BV B Also, V B = 2V 2V A, so; P AV A = 39.53 × 2 × V A P A = 79.06 bar [6 marks] c) For adiabatic compression from T D = 350 K to T A = 1000 K, P K, P A = 79.06 bar, then
1
P D 79.06
T D P D T A P A 350 1000
1.3996 1
1.3996
P D
350 1.3996 1
79.06 1000
2.00 bar bar
For an ideal gas: P DV D = P = P C CV C C V D = P = P C 0.5 V C CV C C / P / P D = 1 × V C C / 2 = 0.5V C [5 marks] 11
d) Step B-C: Adiabatic expansion; W BC = m C V (T B – T C 350) = 46.64 kJ V (T C) = 0.1 × 0.7175 × (1000 – 350) Step D-A: Adiabatic compression; W DA = m C V ( 1000) = -46.64 kJ V (T T D – T A) = 0.1 × 0.7175 × (350 – 1000) [3 marks] e) Step A-B: Isothermal expansion at 1000 K
Q AB
P 79.06 W AB mRT h ln A 0.1 0.2867 1000 ln 19.87 kJ P 39 . 53 B
Step C-D: Isothermal compression at 350 K
QCD
P 1 W CD mRT c ln C 0.1 0.2867 350 ln 6.96 kJ 2 P D [3 marks]
f) Net work; W net 6.96 – 46.64 46.64 = 12.91 kJ net = W AB + W BC + W CD CD + W DA = 19.87 + 46.64 – 6.96 [2 marks] g) Efficiency is given by
W net Qin
W net Q AB
12.91 19.87
0.65
Check;
1
T c T h
1
350 1000
0.65 [2 marks]
12
4) At the Bear Point refinery complex on the St Lawrence Seaway between Canada and the USA, the process waste gas is being considered as a thermal energy source for raising steam to generate electrical power in an energy park associated with the refinery. The preliminary design of a Rankine Cycle steam turbine power plant is being considered to make use of this thermal energy. In the proposed design, steam is to be supplied from a boiler to the turbine inlet at a pressure of 15 MPa and temperature of 600 C whilst the exhaust pressure from the turbine is 10 kPa. The exhaust mixture is passed through a condenser and then pumped back to the boiler. The power plant is assumed to be a closed system. a) Draw a block flow diagram of the proposed power plant and a general TemperatureEntropy diagram of the cycle, labelling the appropriate points. [6 marks] b) Calculate the operational data for the steam/water after each step of the cycle, namely: temperature, pressure, quality, specific enthalpy, specific entropy. State any assumptions you make. [12 marks] c) Calculate the thermal efficiency of the cycle. [2 marks] d) Determine the flow rate of water/steam required to produce 60 MW. [2 marks] e) What methods could be employed to improve the thermal efficienc y of the cycle? [3 marks] Answer a) Diagram:
Q h P2 = 15 MPa
2
3
P3 = 15 MPa T3 = 600C
Boiler
Win
Wout
Compressor (pump)
Turbine Heat 4
P1 = 10 kPa 1
P4 = 10 kPa
Q c
13
T T3 = 600oC 3
P= 15 MPa
2 P = 10 kPa
4
1
s [6 marks] b) Assumptions: Pump and turbine are isentropic P2 = P3 = 15 MPa T3 = 600C P4 = P1 = 10 kPa The working fluid is completely condensed to a liquid in the condenser Kinetic and potential energy changes are zero Known Data:
State 1 2 3 4
T ( C) ? ? 600 ?
P (kPa) 10 15000 15000 10
h (kJ/kg) ? ? ? ?
s (kJ/kg K) ? ? ? ?
x ? ? ? ?
At State 3:
P3 = 15 MPa = 150 bar and T 3 = 600 C. From the Steam Tables we can see that this situation corresponds to superheated steam (i.e. the water is all vapour and the quality of the steam, x, is equal to 1) so: hg3 = h3 = 3581 kJ/kg
sg3 = s3 = 6.677 kJ/kg K
At State 4:
The turbine is isentropic so the water/steam mixture at State 4 has an entropy of: s4 = s3 = 6.677 kJ/kg K
14
The steam has been condensed in the turbine to a mixture of water and steam at a pressure of P4 = 10 kPa = 0.1 bar and a temperature of T 4 = 45.8 C (this temperature is from the given pressure and Steam Tables). At these conditions: sf4 = 0.649 kJ/kg K hf4 = 192 kJ/kg
s fg4 = 7.5 kJ/kg K hfg4 = 2392 kJ/kg
sg4 = 8.149 kJ/kg K hg4 = 2584 kJ/kg
If we let the quality of steam (fraction of water as steam) = x, then: smix = x sg + (1 – x) x) sf Now: s4 = x sg4 + (1 – x) x) sf4 6.677 = 8.149 x + 0.649 (1 – x) x) x = (6.677 – 0.649) 0.649) / (8.149 – 0.649) 0.649) = 0.804 Also hmix = x hg + (1 – x) x) hf so, h4 = x hg4 + (1 – x) x) hf4 h4 = (0.804 2584) + [(1 – 0.804) 0.804) (192)] h4 = 2115.2 kJ/kg At State 1:
The pressure is P1 = 10 kPa = 0.1 bar and the temperature T 1 = 45.8 C, therefore from Steam Tables: hf1 = 192 kJ/kg sf1 = 0.649 kJ/kg K
hfg1 = 2392 kJ/kg s fg1 = 7.5 kJ/kg K
hg1 = 2584 kJ/kg sg1 = 8.149 kJ/kg K
Here we need to make an assumption that steam/water mixture from State 4 is condensed down to just water at State 1 (so quality of steam, x, at State 1 = 0). This is a common assumption in Rankine cycles as it makes pumping of the fluid easier. Therefore: h1 = 192 kJ/kg
s1 = 0.649 kJ/kg K
We are also given the specific volume of liquid water at 10 kPa = 0.00101 kg/m 3 in the question. Therefore: State 1 2 3 4
T ( C) 45.8 ? 600 45.8
P (kPa) 10 15000 15000 10
h (kJ/kg) 192 ? 3581 2115.2 2115. 2
s (kJ/kg K) 0.649 ? 6.677 6.677
x 0 ? 1 0.804
15
At State 2:
The pump work is given by: Win
Win
v( P2
P1 ) h 2 h1
0.00101 (15000000 10000) 15139.9 J/kg 15.1 kJ/kg
Therefore the enthalpy at pump outlet: h2 = Win + h1 h2 = 15.1 + 192 = 207.1 kJ/kg This process is also isentropic so: s2 = s1 = 0.649 kJ/kg From the steam tables, saturated liquid water at P = 15 MPa has T S = 342.1 C, hf = 1610 kJ / kg, sf = 3.685 kJ / kg K, which are greater than conditions at position [2] so water is subcooled, giving: T2 = 45.8 C
P2 = 15 MPa
h2 = 207.1 kJ / kg
s 2 = 0.649 kJ / kg K
x=0
Giving: State 1 2 3 4
T ( C) 45.8 45.8 600 45.8
P (kPa) 10 15000 15000 10
h (kJ/kg) 192 207.1 3581 2115.2 2115. 2
s (kJ/kg K) 0.649 0.649 6.677 6.677
x 0 0 1 0.804
[12 marks] c) We now have all the information we need to calculate the cycle efficiency. H eat in to Boil er:
Qh = h3 – h h2 = 3581 – 207.1 207.1 = 3373.9 kJ/kg H eat re r emoved by condense condenserr :
Qc = h4 – h h1 = 2115.2 – 192 192 = 1923.2 kJ/kg Tur bine work: work:
Wout = h3 – h h4 = 3581 – 2115.2 2115.2 = 1465.8 kJ/kg Pump work (alr eady calcul ated): ated):
Win = 15.1 kJ/kg Cycle Cycle ef ef fi ciency: ciency:
Wnet Qh
Wout
Win
Qh
16
Wnet Qh
1465.8 15.1 3373 .9
0.42998
Therefore the cycle efficiency is: = 43 %
[2 marks] d) Net work output = 60 MW: Net work output
= mass flow rate (Wout - Win) = mass flow rate ((h3 - h4) - (h2 - h1)) 60 MW = mass flow rate ((h3 - h4) - (h2 - h1))
mass flow rate = 60000/((3581 – 2115.2) 2115.2) - (207.1 - 192)) = 41.36 kg/s [2 marks] e) Thermal efficiency can be improved by: (a) Lowering the condensing pressure (lower condensing temperature, lower T c) (b) Superheating the steam to higher temperature (c) Increasing the boiler pressure (increase boiler temperature, increase Th) [3 marks]
17
5) Refrigerant 134a is the working fluid in an ideal vapour-compression refrigeration cycle that communicates thermally with a cold region at -4C (minus four) and a warm region at 24C. Saturated vapour enters the compressor at -4 C and saturated liquid leaves the condenser at 24C. The mass flow rate of the refrigerant is 0.2 kg/s. a) Draw a block flow diagram of the proposed refrigeration plant and explain how the refrigeration cycle works. [5 marks] b) Draw a Temperature-Entropy diagram of the cycle, cycle, labelling the appropriate points. [2 marks] c) Calculate the operational data for the refrigerant after each step of the cycle, namely: temperature, pressure, quality, specific enthalpy and specific entropy. State any assumptions you make. [10 marks] d) Calculate the compressor power, in kW. [2 marks] e) Calculate the refrigeration capacity, in kW. [2 marks] f) Determine the coefficient of performance of the ideal vapour-compression refrigeration cycle. [2 marks] g) Determine the coefficient of performance of an ideal Carnot refrigeration cycle operating at the same conditions. [2 marks] Answer: a) Diagram: a) Diagram:
18
– 2: Pr ocess ocess 1 2: Isentropic compression of the refrigerant from state 1 (usually from a saturated vapour) to the condenser pressure at state 2 (usuall y a superheated vapour). – 3: Pr ocess ocess 2 3: Heat transfer from the refrigerant as it flows at constant pressure through the condenser. The refrigerant exits at state 3 (usually as a saturated liquid). – 4: Throttling process from state 3 (usually as a saturated liquid) to a two-phase Pr ocess ocess 3 4: liquid – vapour vapour mixture at state 4. This occurs at consta nt enthalpy. – 1: Pr ocess ocess 4 1: Heat transfer to the refrigerant as it flows at constant pressure through the evaporator to complete the cycle. [5 marks] b)
[2 marks] c) Assumptions: The compressor is isentropic. The condenser and evaporator operate at constant pressure. The working fluid is completely condensed to a liquid in the condenser The compressor and expansion valve operate adiabatically. Kinetic and potential energy changes are negligible. T 3 = 24°C T 1 = T 4 = -4C Known Data:
State 1 2 3 4
T ( C)
P (bar)
h (kJ/kg)
s (kJ/kg K)
x
-4 ? 24 -4
? ? ? ?
? ? ? ?
? ? ? ?
1 1 0 ?
At State 1:
We have a saturated vapour at T 1 = -4 C. To get the other properties at this condition is a straightforward look-up in the Saturated Refrigerant 134a Temperature table: h1 = h g @ -4°C = 244.90 kJ/kg
19
s1 = s = s g @ -4°C = 0.9213 kJ/kg K P 1 = P = P @ -4°C = 2.5274 bar At State 2:
At state 2 we have a superheated vapour. The pressure of this vapour is the pressure according to the temperature of the liquid coming from the condenser i.e. the pressure when T = 24°C. From the Saturated Refrigerant 134a Temperature table this pressure is: P 2 = P = P @ 24°C = 6.4566 bar The compressor is isentropic so the refrigerant at State 2 has an entropy of: s2 = s = s1 = 0.9213 kJ/kg K In order to calculate the enthalpy and temperature at state 2 we now need to use the tables for the properties of the Superheated Refrigerant 134a. However, we have a problem as there is no exact table for a pressure of 6.4566 bar, hence we need to use the tables at 6 and 7 bar and perform a double interpolation as follows. First we need to generate a new T -h- s table s table for superheated vapour at a pressure of 6.4566 bar. We will do this by interpolating h and s s data at each temperature data point i.e. T sat , 30°C, 40°C, 50°C etc. as far as is necessary. We already know that when at P at P = = 6.4566 bar, T sat = 24°C, so now we can interpolate h and s and s as follows: h@ 6.4566bar T sat
h@ 6bar T sat
h@ 6bar T ) ( P 6.4566 bar P 6bar ) ( P 7 bar P 6bar )
(h@ 7 bar T sat
sat
From tables: h g @ 7 bar Tsat = 261.85 kJ/kg h g @ 6 bar Tsat = 259.19 kJ/kg so h@ 6.4566 bar T sat
259.19
( 261.85 259.19) ( 7 6)
(6.4566 6) 260.40 kJ / kg
Doing the same for the enthalpy; s@ 6.4566 bar T sat
s@ 6bar T sat
s@ 6bar T ( P 7 bar P 6bar )
( s@ 7 bar T sat
sat
)
( P 6.4566 bar P 6bar )
From tables: s g @ 7 bar Tsat = 0.9080 kJ/kg 20
s g @ 6 bar Tsat = 0.9097 kJ/kg so s @ 6.4566 bar T sat
0.9097
(0.9080 0.9097 ) ( 7 6)
(6.4566 6) 0.9089 kJ / kg K
Repeat the calculations for the higher temperatures until the value of s of
[email protected] bar exceeds s exceeds s2. So the next calculation is at T = = 30°C. h@ 6.4566bar 30C
h@ 6bar 30C
h@ 6bar 30C ) ( P 6.4566 bar P 6bar ) ( P 7 bar P 6bar )
(h@ 7 bar 30C
From tables: h g @ 7 bar 30°C = 265.37 kJ/kg h g @ 6 bar 30°C = 267.89 kJ/kg so h@ 6.4566 bar 30 C
267.89
( 265.37 267.89) ( 7 6)
(6.4566 6) 266.74 kJ / kg
Doing the same for the enthalpy; s@ 6.4566 bar 30C
s@ 6bar 30C
s@ 6bar 30C ) ( P 6.4566 bar P 6bar ) ( P 7 bar P 6bar )
( s@ 7 bar 30C
From tables: s g @ 7 bar 30°C = 0.9197 kJ/kg s g @ 6 bar 30°C = 0.9388 kJ/kg so s @ 6.4566 bar 30
0.9388
(0.9197 0.9388) ( 7 6)
(6.4566 6) 0.9301 kJ / kg K
Hence we have already exceeded s exceeded s2 and our superheated refrigerant table at P = = 6.4566 bar reads: Enthalpy -1 kJ kg
Entropy -1 -1 kJ kg K
T
h g g
s g g
°C 24 30
260.40 266.74
0.9089 0.9301
21
We now move onto the second part of the interpolation to calculate T 2 and h2 at s2 = 0.9213 kJ/kg K. Using the table generated we get: T 2 @ s2
T 2 @ s2
T @ s0.9089
24
T @ s 0.9089 ) ( s 0.9213 s 0.9089 ) ( s0.9301 s0.9089 )
(T @ s 0.9301
(30 24) (0.9301 0.9089 )
(0.9213 0.9089 ) 27.5C
and h2 @ s2
h2 @ s2
h@ s 0.9089
260.40
h@ s0.9089 ) ( s 0.9213 s 0.9089 ) ( s0.9301 s0.9089 )
(h@ s 0.9301
(266.74 260.40) (0.9301 0.9089 )
(0.9213 0.9089 ) 264.11 kJ / kg
Summarizing h2 = 264.11 kJ/kg s2 = 0.9213 kJ/kg K P 2 = 6.4566 bar T 2 = 27.5 °C At State 3:
We have a saturated liquid at T 3 = 24 °C. Again this becomes a straightforward read off from the Saturated Refrigerant 134a Temperature table: h3 = h f @ 24°C = 82.90 kJ/kg s3 = s f @ 24°C = 0.3113 kJ/kg K P 3 = P = P @ 24°C = 6.4566 bar At State 4:
The throttle valve is isenthalpic, hence: h4 = h3 = 82.90 kJ/kg T 4 = T 1 = -4°C P 4 = P = P @-4°C @-4°C = 2.5274 bar The refrigerant is a two-phase liquid – vapour vapour mixture at a temperature of -4°C. At these conditions: h f 4 = h = h f @ -4°C = 44.75 kJ/kg
h g 4 = h g @ -4°C = 244.90 kJ/kg
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If we let the quality of refrigerant ref rigerant (fraction of refrigerant as vapour) v apour) = x = x,, then: hmix = x = x h g + (1 – x) x) h f Now: = x h g 4 + (1 – x) h4 = x x) h f 4 82.90 = 244.90 x 244.90 x + + 44.75 (1 – x) x)
x = x = (82.90 – 44.75) 44.75) / (244.90 – 44.75) 44.75) = 0.1906 We can now calculate the entropy using: s f 4 = s = s f @ -4°C = 0.1777 kJ/kg K and
s g 4 = s = s g @ -4°C = 0.9213 kJ/kg K
smix = x = x s s g + (1 – x) x) s f s4 = x = x s s g4 + (1 – x) x) s f4 s4 = (0.1906 0.9213) + [(1 – 0.1906) 0.1906) (0.1777)] s4 = 0.3194 kJ/kg
Results: State 1 2 3 4
T ( ( C)
P (bar) (bar)
h (kJ/kg) (kJ/kg)
s (kJ/kg (kJ/kg K)
x
-4 27.5 24 -4
2.5274 6.4566 6.4566 2.5274
244.90 264.11 82.90 82.90
0.9213 0.9213 0.3113 0.3194
1 1 0 0.1906 [10 marks]
d) Calculate the compressor power W in (h2 – h1) = 0.2 (264.11 – 244.90) 244.90) = 3.842 kW in = m (h [2 marks] e) The refrigeration capacity Qc = m (h (h1 – h4) = 0.2 (244.90 – 82.90) 82.90) = 32.4 kW [2 marks] f) The coefficient of performance COP COP
Qc W in
32.4 3.842
8.43 [2 marks]
23
g) Carnot COP COP COP
T C T H
T C
[273 (4)] (273 24) [273 (4)]
9.61 [2 marks]
24
