Thermodynamics I
Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am 10:00am & Monday 8:00am - 9:00am MEP 261 Class ZA Dr. Walid A. Aissa Associate Professor, Mech. Engg. Dept. Faculty of Engineering at Rabigh, KAU, KSA Chapter #4 October XX, 2011
Announcements: Dr. Walid’s e-mail and Office Hours
[email protected] Office hours for Thermo 01 will be every Sunday and Tuesday from 9:00 – 12:00 am in Dr. Walid’s office (Room 5-213). Text book: Thermodynamics An Engineering Approach Yunus A. Cengel & Michael A. Boles 7th Edition, McGraw-Hill Companies, ISBN-978-0-07-352932-5, 2008
Chapter 4
ENERGY ANALYSIS OF CLOSED SYSTEMS
Objectives of CH4: To • Examine the moving boundary work or P dV work. • Identify the first law of thermodynamics as simply a statement of the conservation of energy principle for closed (fixed mass) systems. • Develop the general energy balance applied to closed systems. • Define the specific heat at constant volume and the specific heat at constant pressure.
* Relate the specific heats to the calculation of
the changes in internal energy and enthalpy of ideal gases. *Describe incompressible substances and determine the changes in their internal energy and enthalpy. *Solve energy balance problems for closed (fixed mass) systems that involve heat and work interactions for general pure substances, ideal gases, and incompressible substances
Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS 4–1 ■ MOVING BOUNDARY WORK
One form of mechanical work frequently encountered in practice is associated with the expansion or compression of a gas in a piston–cylinder device.
the expansion and compression work is often called moving boundary work, or simply boundary work. The boundary work is +ve during expansion & -ve during compression
The total boundary work done during the entire process as the piston moves is
So, p = f (V) should be available.
The boundary work done during a process depends on the path followed as well as the end states
Net work done during a cycle is the difference between the work done by the system and the work done on the system.
a) Constant Volume (V = C) process EXAMPLE 4–1 Boundary Work for a Constant-Volume Process. A rigid tank contains air at 500 kPa and 150°C. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65°C and 400 kPa, respectively. Determine the boundary work done during this process
Solution:
V = C , Hence dV = 0
b) For isobaric process (p = C). EXAMPLE 4–2 Boundary Work for a Constant-pressure Process. A frictionless piston–cylinder device contains 4.53 kg of steam at 413.57 kPa and 160°C. ° Heat is now transferred to the steam until the temperature reaches 204.4 °C. If the piston is not attached to a shaft and its mass is constant, determine the work done by the steam during this process.
Solution: p , kPa
p0 = 413.57 kPa.
m = 4.53 kg p = 413.57 kPa.
v, m3/kg
T T 2 = 204.4 °C
2 T1 = 160 °C
1
v1
v2
v
The total boundary work done during the entire process as the piston moves is
But as p = C
Hence, Wb= p *( V2 - V1 ) = p *m* ( v2 - v1 ) kJ Where, Wb is the total boundary work done during the entire process as the piston moves. Evaluation of Wb implies evaluation of v1 & v2 .
From Table A-5, psat corresponding to T = 160°C = 618.23 kPa. As psat > p (= 413.57 kPa). Hence, state 1 is superheated steam.
1) Evaluation of v1 From Table A-6, For p = 0.4 MPa T, °C 150 160 200
v, m3/kg 0.47088 v 0.53434
u, kJ/kg 2564.4 u 2647.2
h, kJ/kg 2752.8 h 2860.9
Hence, vT= 160°C,
p = 0.4 MPa
= 0.4836 m3/kg
From Table A-6, For p = 0.5 MPa
T, °C Tsat = 151.83°C 160 200
v, m3/kg 0.37483
u, kJ/kg h, kJ/kg 2560.7 2748.1
v 0.42503
u 2643.3
h 2855.8
Hence, vT= 160°C,
p = 0.5 MPa
= 0.3833443 m3/kg
Hence by interpolation, For T = 160°C
p, kPa 0.4 0.41357 0.5
v, m3/kg 0.4836 v 0.3833443
Hence, v1= vT= 160°C,
p = 0.41357 MPa
= 0.47 m3/kg
Similarly, 2) Evaluation of v2 For p =0.4 MPa T, °C v, m3/kg u, kJ/kg h, kJ/kg 200 0.53434 2647.2 2860.9 204.4 v u h 250 0.5952 2726.4 2964.5
Hence, vT= 204.4°C,
3/kg = 0.5397 m p = 0.4 MPa
Similarly, From Table A-6, For p = 0.5 MPa
T, °C 200 204.4 250
v, m3/kg 0.42503 v 0.47443
Hence, vT= 204.4°C,
u, kJ/kg 2643.3 u 2723.8
h, kJ/kg 2855.8 h 2961.0
3/kg = 0.42938 m p = 0.5 MPa
Hence by interpolation, For T = 204.4°C
p, kPa 0.4 0.41357 0.5
v, m3/kg 0.5397 v 0.42938
Hence, v2= vT= 204.4°C,
3/kg = 0.52473 m p = 0.41357 MPa
Hence, V1= m * v1 = 4.53 kg * 0.47 m3/kg = 2.129 m3 . & V2= m * v2 = 4.53 kg * 0.52473 m3/kg= 2.377 m3 .
Hence, Wb= p *( V2 - V1 ) = 0.41357 MPa * (2.377 m3 - 2.129 m3)= 0.41357× × 106 Pa * (2.377 - 2.129) m3 = 1.026 × 105 J = 102.6 kJ.
c) For isothermal process T= Constant).
pV = mRT
Hence, for isothermal process : Hence, pV = C Hence, p = C V-1
(#-1)
Hence,
(#-2)
Where, C = p1V 1= p2V 2
(#-3)
EXAMPLE 4–3 Isothermal Compression of an Ideal Gas. A piston–cylinder device initially contains 0.4 m3 of air at 100 kPa and 80°C. The air is now compressed to 0.1 m3 in such a way that the temperature inside the cylinder remains constant. Determine the work done during this process.
Solution: V 1 = 0.4 m3 ,p1 = 100 kPa and T = 80°C. V 2 = 0.1 m3. Hence, From Eq. (#-3) Where, C = p1V 1= 100 kPa * 0.4 m3 = 40 kJ Hence, From Eq. (#-2) Wb= 40 kJ * ln (0.1 m3 /0.4 m3) = -55.45kJ
The negative sign indicates that the work is done on the system.
d) For polytropic process ; n pV = Constant).
n As, pV = C
Hence, p = C V-n
(#-4)
Hence,
n But, C = p2V2
Hence,
=
n p1V1
(4-9) But, From Equation of State pV = mRT Hence, p1V1 = mRT1& p2V2 = mRT2
(#-5)
By substituting by p2V2 & p1V1 from Eq. (#-4) in Eq. (4-9) to get
(4-10)
EXAMPLE 4–4 Expansion of a Gas against a Spring. A piston–cylinder device contains 0.05 m 3 of a gas initially at 200 kPa. At this state, a linear spring that has a spring constant of 150 kN/m is touching the piston but exerting no force on it. Now heat is transferred to the gas, causing the piston to rise and to compress the spring until the
volume inside the cylinder doubles. If the cross-sectional area of the piston is 0.25 m 2, determine (a) the final pressure inside the cylinder, (b) the total work done by the gas, and (c) the fraction of this work done against the spring to compress it.
Fig. (EXAMPLE 4–4)
V1 = 0.05 m 3 , p1 = 200 kPa, k = 150 kN/m , V2 = 2*V1 , A = 0.25 m 2 Required? p2 = ?, Wb = ?, Fraction of work done against the spring to compress it? Solution: V2 = 2 *V1 = 2 * 0.05 m3 = 0.1 m3
Displacement of the spring; x
i.e.;
Force applied by the linear spring at the final state; F is F = k x = (150 kN/m)*(0.2 m)= 30 kN Additional pressure applied by the spring on the gas at this state; p is
Without the spring, the pressure of the gas would remain constant at 200 kPa while the piston is rising. But under the effect of the spring, the pressure rises linearly from 200 kPa to 200 + 120 = 320 kPa at the final state. i.e. p2 = 320 kPa
(b) One way for finding the work done is to plot the process on a P-V diagram and find the area under the process curve. From Fig. (Example 4–4) the area under the process curve (a trapezoid) is
Hence; W = 13 kJ +ve sign indicates that the work is done by the system.
( c) Total work consists of two portions: I) Work represented by the rectangular area (region I) is done against the piston and the atmosphere, and II) the work represented by the triangular area (region II) is done against the spring.
Thus, - Work done against the piston and the atmosphere (region I), - Fraction of work done against the spring to compress it; (region II),
4–2 ■ ENERGY BALANCE FOR CLOSED SYSTEMS Energy balance for any system undergoing any kind of process (as shown in Chap. 2) was expressed as: (4-11) or, in the rate form, as (4-12)
For constant rates, the total quantities during a time interval ∆t are related to the quantities per unit time as (4-13)
The energy balance can be expressed on a per unit mass basis as (4-14)
For a closed system undergoing a cycle, the initial and final states are identical, and thus,
For a cycle ∆E =0, thus Q = W.
For a closed system undergoing a cycle, the initial and final states are identical, Hence, Eq. (4-11) can be re-written as
∆Esystem = E2 – E1 = 0
(#-6)
Ein – Eout = 0 i.e. Eout = Ein
&
(#-7)
Energy balance for a cycle can be expressed in terms of heat and work interactions as
Wnet, out = Qnet, in i.e.
(4-16)
Energy balance (first-law) relation for a closed system becomes Q- W =∆E = ∆U+∆PE+∆KE (4-17) where, Qnet, in= Qin - Qout is the net heat input & Wnet, out= Wout - Win is the net work output
Various forms of the first-law relation for closed systems
EXAMPLE 4–5 Electric Heating of a Gas at Constant Pressure.
A piston–cylinder device contains 25 g of saturated water vapor that is maintained at a constant pressure of 300 kPa. A resistance heater within the cylinder is turned on and passes a current of 0.2 A for 5 min from a 120-V source. At the same time, a heat loss of 3.7 kJ occurs.
(a) Show that for a closed system the boundary work Wb and the change in internal energy ∆U in the first-law relation can be combined into one term, ∆H, for a constant pressure process. (b) Determine the final temperature of the steam.
Solution:
Qin
Schematic and P-v diagram for Example 4–5.
m = 25 g, p2= p1= 300 kPa I = 0.2 A , V = 120 V, t= 5 min
Qout= 3.7 kJ From Eq. (4-17)
Qnet, in- Wnet, out=∆E= ∆U+∆PE+∆KE As system is stationary, hence ∆PE=0,∆KE=0
Hence, Eq. (4-17) can be re-written as 0
0
Qnet, in- Wnet, out =∆E = ∆U+∆PE+∆KE
i.e., Qnet, in- Wnet, out =∆U (#-8) P = I2R = I V= 0.2 A* 120 V= 24 W Qin= P * t = 24 W* (5*60) s= 7200 J Qin= 7.2 kJ i.e.
Wnet, out = Wb
(#-9)
But, from Eq. (4-2)
But, p2 =p1= p0 . Hence, Eq. (4-2) can be re-written as:
Wb = p0 ( V2 – V1)
(#-10)
By substituting by Wb from Eq. (#-10) in Eq. (#-9) to get:
Wnet, out = Wb = p0 ( V2 – V1) (#-11) By substituting from Eq. (#-11) in Eq. (#-8) to get:
Qnet, in= Wnet, out +∆U= = (p2 V2 – p1V1) + (U2 –U1) (#-12)
But, By definition;
pV + U = H
Hence, Eq. (#-12) can be re-written as:
Qnet, in= H2 –H1
(#-13)
Hence,
Qin= m(h2 –h1)
(#-13′′)
From Table (A-5); Pressure table
Saturated
water-
h1 = hg( p = 300 kPa) = 2724.9 kJ/kg Qnet, in= Qin – Qout = 7.2 kJ- 3.7 kJ =3.5 kJ Hence, 3.5 kJ = (25/1000) kg* (h2 2724.9 kJ/kg) Hence, h2 = 2864.9 kJ/kg
It is clear that h2 (= 2864.9 kJ/kg) > hg (p2 = 300 kPa) Hence, state 2 is superheated stream. Hence, from Table (A-6); superheated stream
150 2761.2 T2 2864.9 200 2865.9 Hence, T2 = 199.5224 °C
EXAMPLE 4–6 Unrestrained Expansion of Water.
A rigid tank is divided into two equal parts by a partition. Initially, one side of the tank contains 5 kg of water at 200 kPa and 25°C, and the other side is evacuated. The partition is then removed, and the water expands into the entire tank. The water is allowed to exchange
heat with its surroundings until the temperature in the tank returns to the initial value of 25°C. Determine (a) the volume of the tank, (b) the final pressure, and (c) the heat transfer for this process.
Solution:
Schematic and p-v diagram for Example 4–6
m = 5 kg , p1 = 200 kPa , T1 = 25°C, T2 = T1 = 25°C. V = ?, p2 = ?, Qheat transfer with soroundings= ?, (a) Evaluation of the volume of the tank, Vtank
From Table (A-5) for Saturated waterPressure table Tsat (p =200 kPa ) =120.21 °C It is clear that T1 < [Tsat (p =200 kPa ) =120.21 °C]. Hence, state 1 is compressed liquid.
It is clear that no data for compressed liquid at p =200 kPa. Hence, v1 ≈ vf [T= 25°C]= 0.001003 m3/kg, u1 ≈ uf [T= 25°C]= 104.83 kJ/kg, h1 ≈ hf [T= 25°C]= 104.83 kJ/kg. Initial volume of the water (V1) is
V1 = m* v1 =5 kg * (0.001003 0.005 m3.
3 m /kg)
=
Total volume of the tank (Vtank )is twice the initial volume of the water (V1)
Vtank = 2*V1 = 2* 0.005 m3= 0.01 m3 (b) Evaluation of final pressure, p2
v2 = Vtank / m = 0.01 m3/kg
3 m
/5 kg = 0.002
Specifying conditions at state 2: At state 2, T2 = 25°C, v2 = 0.002 m3/kg From Table (A-4); Temperature table
Saturated
water-
At T2 = 25°C, vf = 0.001003 m3/kg , vg = 43.34 m3/kg Hence, state 2 is saturated liquid-vapor mixture.
i.e., x = 2.30047 From Table (A-4); Temperature table
-5
Saturated
water-
p2 = psat (T = 25°C) = 3.1698 kPa
From Eq.(#-8) 0
i.e., Qnet, in- Wnet, out =∆U =m(u2 – u1 ) From Table (A-4); Temperature table
(#-14) Saturated water-
uf (T = 25°C) = 104.83 kJ/kg ufg (T = 25°C) = 2304.3 kJ/kg u2 = uf + (x* ufg) =104.83 +(2.30047 10 -5 * 2304.3) kJ/kg = 104.883kJ/kg
Hence, from Eq. (#-14)
i.e., Qnet, in =m(u2 – u1 ) Hence, Qnet, in = 5 kg (104.883 –104.83) kJ/kg Qnet, in = 5 kg (104.883 –104.83) kJ/kg = 0.265 kJ The +ve sign indicates that heat is transferred to the water
4–3 ■ SPECIFIC HEATS The specific heat is defined as the energy required to raise the temperature of a unit mass of a substance by one degree. specific heat at specific heat at constant volume constant volume ; cv. ; cp.
specific heat at constant volume ; cv.: the energy required to raise the temperature of the unit mass of a substance by one degree as the volume is maintained constant specific heat at constant pressure ; cp.: the energy required to raise the temperature of the unit mass of a substance by one degree as the pressure is maintained constant
(4-19)
Formal definitions of cv. Cv dT = du
at constant volume
(4-20)
Formal definitions of cp. Cp dT = dh
at constant pressure
Eqs. (4-20) & (4-20) are properties relations and as such are independent of the type of processes. A common unit for specific heats is kJ/kg ·°C or kJ/kg·K. Notice that these two units are identical since T(°C) T(K), and 1°C change in temperature is equivalent to a change of 1 K.
4–4 ■ INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF IDEAL GASES
Ideal gas is a gas whose temperature, pressure, and specific volume are related by p v = RT (3-10) u = u(T) only (4-21) But by definition; enthalpy and internal energy of an ideal gas r related by:
h= u + p v
(#-15)
By substitution by p v from Eq. (3-10) in Eq. (#-15) to get:
h= u + RT
(#-16)
Since R is constant and u = u(T). Hence, enthalpy of an ideal gas is also a function of temperature only: i.e. (4-22) h = h(T) only
As u = u(T) only; Eq. (4-21) & h = h(T) only; Eq. (4-22) . Hence, the partial derivatives; (4-19) and (4-20) turn to ordinary derivatives; i.e. (4-23) du = Cv (T) dT (4-24) dh = Cp (T) dT The change in internal energy or enthalpy for an ideal gas during a process from state 1 to state 2 is determined by integrating these equations:
(4-25) and (4-26) To carry out these integrations, we need to have relations for Cv and Cp as functions of temperature.
The variation of specific heats with T over Small temperature intervals (a few hundred degrees or less)may be approximated as linear. Therefore the specific heat functions in Eqs. 4–25 and 4–26 can be replaced by the constant average specific heat values. Then the integrations in these equations can be performed, yielding:
(4-27)
(4-28)
The specific heat values for some common gases are listed as a function of temperature in Table A–2b. The average specific heats Cp,avg, and Cv,avg are evaluated from this table at the average temperature (T1 + T2)/2, as shown in Fig. 4–26.
FIGURE 4–26 For small T intervals, the specific heats may be assumed to vary linearly with T.
If the final temperature T2 is not known, the specific heats may be evaluated at T1 or at the anticipated average temperature. Then T2 can be determined by using these specific heat values. The value of T2 can be refined, if necessary, by evaluating the specific heats at the new average temperature.
Specific Heat Relations of Ideal Gases Differentiating the relation (#-16) h = u + RT to get dh = du + RdT
(#-16) (#-17)
By substituting by du & dh from Eqs. (4-23)&(4-24) in Eq. (#-17) to get Cp dT= Cv dT + R dT
Hence, Cp = Cv + R
(kJ/kg.K) (4–29)
In addition, Cp & Cv are related by idealgas property called specific heat ratio k, defined as; k = Cp / Cv
(4–31)
In molar basis; (4–31) where; ,
& (4–30)
To summarize
Three ways of calculating u
EXAMPLE 4–7 Evaluation of the ∆u of an Ideal Gas.
Air at 300 K and 200 kPa is heated at constant pressure to 600 K. Determine the change in internal energy of air per unit mass, using (a) data from the air table (Table A–17), (b) the functional form of the specific heat (Table A–2c), and (c) the average specific heat value (Table A–2b).
Solution:
(a) Evaluation ∆u using Table A–17 u1 = u (T=300 K) =214.07 kJ/kg u2 = u (T=600 K) =434.787 kJ/kg Hence, ∆u = u2 –u1 = 434.79 kJ/kg 214.07 kJ/kg = 220.72 kJ/kg
(b) Evaluation of ∆u using functional form of the specific heat (Table A–2c) kJ/kmol.K -2 a = 28.11, b = 0.1967 -5 c = 0.4802 d = -1.966
Ru = 8.31447 kJ/kmol.K
-9
Cv (T) = Cp (T) - Ru
Cv (T) = (a- Ru) + bT+ cT2+dT3 But, from Eq. (4–30) , From Table (A–1); M (Air) = 28.97 kg/kmol
Hence,
Cv (T) = [(a- Ru) + bT+ cT2+dT3]/M
∆u = (1/M) {[(a- Ru) T+ b(T2/2) + c(T3/3) +d(T4/4)]T=600K - [(a- Ru) T+ b(T2/2) + c(T3/3) +d(T4/4)]T=300K }
∆u = (1/ 28.97) {[(28.11 - 8.31447) -2) [(6002(600-300)]+ (0.1967 -5 3002) /2] + (0.4802 [(6003 -9 3003)/3] +(-1.966 (60043004)/4] } = 223 kJ/kg
(c) Evaluation of ∆u using the average specific heat value (Table A–2b). Cv (T=600 K) = 0.764 kJ/kg.K Cv (T=300 K) = 0.718 kJ/kg.K Cv, average = [Cv(T=600 K) + Cv (T=300 K)]/2= (0.764+0.718)/2 kJ/kg.K= 0.741 kJ/kg.K
Hence, ∆u = Cv, average * [600 – 300] = 0.741 kJ/kg.K * 300 K = 222.3 kJ/kg
EXAMPLE 4–8 Heating of a Gas in a Tank by Stirring.
An insulated rigid tank initially contains 0.5 kg of helium at 30°C and 3 kPa. A paddle wheel with a power rating of 15 W is operated within the tank for 30 min. Determine (a) the final temperature and (b) the final pressure of the helium gas.
Solution: p , kPa
m= 0.5 kg T1= 30°C p1= 3 kPa
m = 0.5 kg, T1= 30°C , p1= 3 kPa, P =15 W , t = 30 min. (a) T2= ? and (b) p2= ? . From Eq. (4-17) (4-17) 0
0
0
Q- W =∆E = ∆U+∆PE+∆KE 0
Wnet,out = Wout - Win Hence,
(#-18)
(#-19)
Win = P *∆t By substituting Eq. (#-18) to get
from
Eq.
P *∆t =∆U=m (u2 – u1) Hence,
∆t =30*60 = 1800 s
(#-19)
in
(#-20)
From Table (A-2) ,C
v=
3.1156 kJ/kg.K
T1= 30+273 K= 303 K Hence, Eq. (#-20) can be re-written as:
P *∆t =∆U=m (u2 – u1) Hence,
(15/1000) kW* 1800 s = 0.5 kg * 3.1156 kJ/kg.K *(T2 - 303 K)
T2 =320.32 K= 47.3 K Treating Helium as an ideal gas. Hence, pv = RT
As v2= v1 Hence, p2/T 2= p1/T 1 p2= p1* (T2/T1 )= 3 kPa * (320.32 K / 303 K)
Hence, p2= 3 kPa * (320.32 K / 303 K)= 3.17 kPa
EXAMPLE 4–9 Heating of a Gas by a Resistance Heater A piston–cylinder device initially contains 0.5 m 3 of Nitrogen gas at 400 kPa and 27°C. An electric heater within the device is turned on and is allowed to pass a current of 2 A for 5 min from a 120-V source. Nitrogen expands at constant pressure, and a heat loss of 2800 J occurs during the process. Determine the final temperature of nitrogen..
Solution:
Schematic and P-V diagram for Example 4–9.
From Eq. (4-17)
0
0
Qnet, in- Wb,out=∆E= ∆U+∆PE+∆KE Qnet, in= Qin –Qout=(IV.∆t)- Qout i.e.
Qnet, in= [2*120 *(5*60)]/1000 kJ2.8 kJ = 69.2 kJ Qnet, in=∆H
V1= 0.5 m 3 , p1= 400 kPa , T1= 27°C. Treating Nitrogen as ideal gas. Hence, pv = RT From Table A-2, RN2 = 0.2968 kJ/Kg.K , Cp,N2 =1.039kJ/Kg.K p1v1 = RT1 Hence, 400 kPa *v1 = 0.2968 kJ/Kg.K *(27+273 K) v1 = 0.55725 m 3 /Kg
v1 = V1 /m 0.2226 m 3 /Kg = 0.5 m 3 /m
Hence, m = 2.212 kg
Qnet, in=∆H= m*∆h= m*(h2-h1) i.e., Qnet, in= m*Cp*(T2-T1)
69.2 kJ = 2.212 kg * 1.039 kJ/Kg.K *(T2-300 K) T2= 330.1 K i.e., t2= 57.1 °C.
Homework 4–4C, 4–5, 4–6, 4–7, 4–8, 4–9, 4–12, 4–13, 4–14, 4–18, 4–21, 4–28, 4–56, 4–61.
