Solution manual for renewable and efficient electric power system reeps2 chapter 2 problems (gilbert masters)

Solution Manual for Renewable and Efficient electric Power System REEPS2 (Gilbert Masters)Solution Manual for Renewable and Efficient electric Power System REEPS2 (Gilbert Masters)Solution Manual for Renewable and Efficient electric Power System REEPS2 (Gilbert Masters)Solution Manual for Renewable and Efficient electric Power System REEPS2 (Gilbert Masters)Solution Manual for Renewable and Efficient electric Power System REEPS2 (Gilbert Masters)Solution Manual for Renewable and Efficient electric Power System REEPS2 (Gilbert Masters)

CHAPTER 2: PROBLEM SOLUTIONS

2.1 A source is connected through a switch to a load that is either a resistor, a capacitor or an inductor At t = 0, the switch is closed and current delivered to the load results in the voltage shown below What is the circuit element and what is its magnitude?

Figure P2.1 SOLN:

The relationship follows the form i = C dv/dt, so it is a CAPACITOR

C = i

dv dt = 0.1

100 10 = 0.01F

2.2 A voltage source produces the square wave shown below The load, which is either

an ideal resistor, capacitor or inductor, draws current as shown below

a. Is the “Load” a resistor, a capacitor or an inductor ?

SOLN: Since v = L di

dt fits the graph, while v = Ri and i = C dv

dt do not, is an INDUCTOR

b. Sketch the power delivered to the load versus time

SOLN:

c. What is the average power delivered to the load?

?"

V"

Source"

SOLN: From the graph of power, average power = 0

2.3 A source supplies voltage v(volts) = 10 2 cosωt to a 5-Ω resistive load

a Write an expression for the current (amps) delivered to the load

R =10 2 cosωt

5 = 2 2 cosωt

b Write an expression for the power delivered to the load

SOLN: p = vi = 10 2 cosωt ⋅ 2 2 cosωt = 40 cos2(ωt)

c Sketch a graph of power versus time What is the average power (W) delivered to the load?

SOLN: From the sketch, easy to see average value of cos2x = 1/2

P avg = avg 40 cos( 2ωt)= 40avg cos( 2ωt)= 40x12 =20W

2.4 Suppose your toaster has 14-gage wire inside and, to simplify the analysis, suppose

we approximate the normal 60-Hz sinusoidal current flowing through that wire with

a square wave carrying ± 10A Using a drift-velocity calculation, find the average back-and-forth distance those electrons travel

SOLN: From Table 2.3, 14-ga wire has diameter 0.0641 in, so its cross-sectional area is

A=π

4

D2 =π 4 0.0641 in x 2.54 cm/in x 10-2m/cm

= 2.08x10−6m2 From (2.2) the average drift velocity (m/s) is

Source'

R = 5Ω

i= ?

1"

0"

$1"

1"

0"

1/2"

10A$

%10A$

60$cycles/s$ +/%$10A$

14$ga$

i!

t!

d$=$?$

Drift = i

8.48x1028elec/m3x 1.602x10-19coul/elec x 2.08x10-6m2 = 0.000354 m/s = 3.54 x 10−4 m/s

In one half of a 60-Hz cycle the electrons go one way, then turn around and go back

Distance = 3.54x10−4 m/s x 1

2⋅ s

60 cycle = 2.94x10−6 m

So, literally, the electrons never leave the toaster as asserted in the text

2.5 As is the case for all metals, the resistance of copper wire increases with temperature in

an approximately linear manner that can be expressed as

where α = 0.00393/oC How hot do copper wires have to get to cause their resistance to increase by 10% over their value at 20oC?

SOLN:

2.6 A 52-gallon electric water heater is designed to deliver 4800 watts to an

electric-resistance heating element in the tank when it is supplied with 240V (it doesn’t

matter if this is ac or dc)

a. What is the resistance of the heating element?

SOLN:

b. How many watts would be delivered if the element is supplied with 208V instead of 240V?

SOLN:

R T2 =R T1[1+ α T( 2−T1) ]

R

T2 = R T11 + α T2 − T

1

R

T2

R T1 = 1.10 = 1 + 0.00393/o C ⋅ T

2 −20

( )o C T

2 = 20 + 25.4 = 45.4o C = 114 o F

240V

4800W

52 gal

R=V

2

P = 240

2

4800 = 12 Ω

P = V

2

R =208

2

12 = 3605 W = 3.605 kW

c. Neglecting any losses from the tank, how long would it take for 4800 W to heat the 52 gallons of water from 60oF to 120oF? The conversion between kilowatts of electricity and Btu/hr of heat is given by 3412 Btu/h = 1 kW Also, one Btu heats 1 lb of water by

1oF and 1 gallon of water weighs 8.34 lbs

SOLN:

d If electricity costs $0.12/kWh, what is the cost of a 15-gallon, 110oF shower if the cold-water supply temperature is 60oF?

SOLN:

Cost = 15 g x 8.34 lb/g x 1Btu/lb

o

F x (110-60)oF x $0.12/kWh

3412 Btu/kWh = $0.22

2.7 Suppose an automobile battery is modeled as an ideal 12-V battery in series with an internal resistance of 0.01 Ω as shown in (a) below

a. What current will be delivered when the battery powers a 0.03-Ω starter motor, as in (b)? What will the battery output voltage be?

SOLN:

a While driving the starter motor:

b Compare the power delivered by the battery to the starter with the power lost in the battery's internal resistance What percentage is lost in the internal resistance

SOLN: Delivered to the starter

p delivered( )=i2R = 300( )2

x0.03 = 2736W p(battery loss) = 300( )2

x0.01 = 900W

Heat needed = 52 gal x 8.34 lb/gal x 1 Btu/lb o

F x (120− 60)o

F = 26,020 Btu

ΔT hr( )= 26,020 Btu

4.8 kW x 3412 Btu/kWhr = 1.59 hr

Starter&

I = V

R = 12

0.01 + 0.03= 300A

V b = 0.03 Ω x 300 A = 9V or with the voltage divider V b = 12⋅ 0.03

0.01 + 0.01= 9V

c. To recharge the battery, what voltage must be applied to the battery in order to deliver a 20A charging current as in (c)?

SOLN: While being charged with 20A:

d. Suppose the battery needs another 480 Wh of energy to be fully charged, which could be achieved with a quick-charge of 80A for 0.5 hour (80A x 0.5hr x 12V = 480 Wh) or a trickle charge of 10A for 4 hours Compare the Wh of energy lost in the internal

resistance of the battery for each charging scheme

SOLN: @ 80A I2

R x Δt loss = (80)2 x 0.01 x 0.5 hr = 32 Wh (32/480 = 6.7%) @ 10A I2

R x Δt loss = (10)2 x 0.01 x 4 hr = 4 Wh (4/480 = 0.9%)

e Automobile batteries are often rated in terms of their cold-cranking amperes (CCA), which is the number of amps they can provide for 30 seconds at 0oF while maintaining

an output voltage of at least 1.2V per cell (7.2V for a 12-V battery) What would be the CCA for the above battery (assuming the idealized 12-V source still holds)?

SOLN: CCA= ΔV

R =(12− 7.2)V

0.01Ω = 480A

2.8 A photovoltaic (PV) system is delivering 15A of current through 12-gage wire to a battery 80 feet away

Figure P2.8

a Find the voltage drop in the wires

SOLN: From Table 2.3, 12-ga wire has 0.1588 Ω per 100-ft With 160-ft of wire (to-and-from) the voltage drop would be

ΔV = RI = 15A x 160 ft x(0.1588Ω /100ft) = 15A x 0.254Ω = 3.81V

b. What fraction of the power delivered by the PVs is lost in the connecting wires?

SOLN: Power delivered by the PVs = I V = 15A x (12+3.81)V = 237W

Power lost in wires = I2R = (15)2 x 0.254 = 57.2W

% loss in wire = 57.2W

237W = 24%

c Using Table 2.3 as a guide, what wire size would be needed to keep wire losses to less than 5% of the PV power output (assume the PVs will continue to keep the current at 15A, which, by the way is realistic)?

V

b = 12V + 0.01Ω x 20A = 12.2V

12#V#

15A#

80#)#

AWG#12##

SOLN:

5%PV-power = R-power

0.05x15x 12 + 15R( )=152R 0.6 + 0.75R = 15R

R = 0.6

14.25=0.0421Ω /160 ft

R≤100

160x 0.0421 = 0.0263Ω /100 ft

Since 4-ga wire has 0.0249 Ω/100 ft, it will just be good enough

2.9 Consider the problem of using a low-voltage system to power your little cabin Suppose a 12-V system powers a pair of 60-W lightbulbs (wired in parallel) The distance between these loads and the battery pack is 50 ft

a. Since these bulbs are designed to use 60 W at 12 V, what would be the (filament) resistance of each bulb?

2

R so R = V

2

P =( )12 2

60 = 2.4Ω ea

b. What would be the current drawn by two such bulbs if each receives a full 12 V?

SOLN: I each bulb( )=V

R= 12V 2.4Ω= 5A both bulbs 10A

c. Of the gages shown in Table 2.3, what gage wire should be used if it is the minimum size that will carry the current?

SOLN: 14-ga wire will handle 15A, so that is adequate

d. Find the equivalent resistance of the two bulbs plus the wire resistance to and from the battery Both lamps are turned on (in this and subsequent parts)

SOLN: First draw the circuit The wire resistance is 0.2525 Ω/100ft, so

15#

R##5%#loss#

12V#

15A#

PV#

12#+#ΔV#

ΔV#

12#V#

50#'#

60)W#ea.#

@#12#V#

The two lamps in parallel have a total of 1.2 Ω, which added to the 100-ft of wire

gives RTotal = 1.2 + 0.2525 = 1.4525 Ω

e Find the current delivered by the battery with both lamps turned on

0.2525 + 1.2= 8.26A

f Find the power delivered by the battery

SOLN: P Battery = VI =12V x 8.26A = 99.1 W

g Find the power lost in the connecting wires in watts and as a percentage of battery power

SOLN: P wire = I2R= 8.26( )2

x0.2525= 17.2 W

99.1 W

= 17.4%

h Find the power delivered to the lamps in watts and as a percentage of their rated

power

SOLN: P Lamps = P Battery − P Wires = 99.1 − 17.2 = 81.9W

% of full power to lamps = 81.9

60 + 60= 68%

… so they're going to be much dimmer than full brightness !

2.10 Suppose the system in Problem 2.9 is redesigned to work at 24 volts with 12-gage wire and two 24-V 60-W bulbs What percentage of the battery power is now lost in the wires?

SOLN: 100-ft of 12-ga wire has 0.1588 Ω of resistance Each bulb has

R = V2/P = (24)2

/60 = 9.6 Ω, so two in parallel have 9.6/2 = 4.8 Ω The circuit is now

Power lost in the wires

0.2525/2&Ω&

0.2525/2&Ω&

2.4&Ω&&&&&&&&&&&&&&&&&&&&&&2.4&Ω&

12&V&

24#V#

0.1588#Ω#

4.8#Ω# V bulbs= 24 4.8

4.8 + 0.1588

⎛

⎝⎜ ⎞⎠⎟ =23.231V

vbulbs

P wire =(ΔV)2

R wire

=(24 − 23.231)2

0.1588 = 3.72 W Power from the battery and % losses

P battery = V

2

R tot = 24

2

0.1588 + 4.8 = 116.16 W

P loss

P battery = 3.72

116.16= 0.032 = 3.2%

2.11 Suppose the lighting system in a building draws 20 A (ac or dc; it doesn't matter) and the lamps are, on the average, 100 ft from the electrical panel Table 2.3 suggests that 12-ga wire meets code, but you want to consider the financial merits of wiring the circuit with bigger 10-ga wire Suppose the lights are on 2500 hours per year and electricity costs $0.10 per kWh

a. Find the energy savings per year (kWhr/yr) that would result from using 10 ga instead

of 12 ga wire

SOLN:

12 ga wire @ 0.1588 Ω/100 ft x 200 ft = 0.3176 Ω

Pwire (12 ga) = I2 R = 202 x 0.3176 = 127.04 watts

10 ga wire @ 0.0999 Ω/100 ft x 200 ft = 0.1998 Ω

Pwire (10 ga) = I2 R = 202 x 0.1998 = 79.92 watts

Energy savings = (127.04 – 79.92)W x 2.500 khr/yr = 117.8 kWh/yr

b. Suppose 12-ga Romex (2 conductors, each 100-ft long, plus a ground wire that carries

no current, in a tough insulating sheath) costs $50 per 100 ft and 10 ga costs $70 per

100 ft What would be the “simple payback” period (1st cost divided by annual

savings) when utility electricity costs $0.10/kWh

SOLN: Payback = Extra cost

Annual savings= $70-$50

117.8 kWh/yr x $0.10/kWh=

$20

$14.14/yr=1.7 yr

c. An effective way to evaluate energy efficiency projects is by calculating the annual cost associated with conservation and dividing it by the annual energy saved This is

the Cost of Conserved Energy (CCE ) and is described more carefully in Appendix

A CCE is defined as follows

15#20A# Lights

100#'#

Panel

Hot

Neutral Ground Romex

CCE = annual cost of saved electricity $/yr( )

annual electricity saved (kWh/yr) =

ΔP ⋅ CRF i, n( ) kWh/yr

where ΔP is the extra cost of the conservation feature (heavier duty wire in this case),

and CRF is the capital recovery factor (which means your annual loan payment on $1

borrowed for n years at interest rate i

What would be the “cost of conserved energy” CCE (¢/kWh) if the building (and

wiring) is being paid for with a 7%, 20-yr loan with CRF = 0.0944/yr How does

that compare with the cost of electricity that you don’t have to purchase from the

utility at 10¢/kWhr?

SOLN: With CRF(7%, 20yr) = 0.07(1 + 0.07)

20

(1 + 0.07)20− 1 = 0.0944 / yr CCE = ΔP ⋅ CRF

ΔkWh/yr=

$ 70-50( )⋅ 0.0944/yr

117.8kWh/yr =$0.016/kWh = 1.6¢/kWh way cheaper than 10¢

2.12 Thevenin's theorem says that the output of any circuit consisting of resistors and ideal voltage sources can be modeled as a voltage source in series with a resistance Suppose the Thevenin equivalent of a circuit consists of a 12-V source in series with a 6-Ω resistance

a What is output voltage with an infinite load so no current flows (called the open-circuit

voltage, VOC)?

SOLN: V OC = 12 V

b. What is the output current when the terminals are shorted together (called the short-circuit

current, ISC)

SOLN: I SC =12V6

Ω = 2A

c Write an equation for the output current I as a function of the output voltage Vout Draw a

graph of I versus Vout (as the load changes)

SOLN:

d Using the equation found in (c), determine the location (I, V) on the graph at which the maximum power will be delivered to a load This is called the maximum power point and

you'll see it a lot in the chapters on photovoltaics Show that point on your I-V graph from part (c)

SOLN:

P = IV = I 12 − 6I( )= 12I − 6I2

Max Power Point MPP : dP

dI = 12 − 12I = 0

so Imax = 1A

and Vmax = 12 - 6x1= 6V

d What load resistance will result in the circuit delivering maximum power to the load? How much power would that be?

SOLN: The maximum power point occurs when the circuit delivers 1 A at 6V to a load

R=V

I = 6V

1A = 6Ω

P

max= I2R= 1( )2

x6= 6 W

2.13 When circuits involve a source and a load, the same current flows through each one and the same voltage appears across both A graphical solution can therefore be obtained by

simply plotting the current-voltage (I-V) relationship for the source onto the same axes that the I-V relationship for the load is plotted, and then finding the crossover point

where both are satisfied simultaneously This is an especially powerful technique when the relationships are nonlinear as will be the case for the analysis of photovoltaic

systems

V OUT%

I!

I SC = 2A!

V OC = 12V!

V out = 12 − 6I

I=12 − V out

6 = 2 −V out

6

V OUT%

I!

I SC = 2A!

V OC = 12V!

MPP#at#1A,#6V#

Consider the following I-V curve for a source delivering power to a load For the

following loads, plot their I-V curves onto the I-V curve for the source shown and at the

crossover points note the current, voltage and power delivered to the load

a The load is a simple 2-Ω resistor Find I, V and P

b. The load is an ideal battery that is always at 12V no matter what current

c. The source is charging a battery that is modeled as an ideal 12-V battery in series with a 2-Ω internal series resistance

SOLN:

a. 2-Ω : v = 8 V, i = 4A, P = 32 W

b. 12V battery: v = 12 V, i = 3 A, P = 36 W

c. 12V with 2Ω: v = 16 V, i = 2A, P = 32 W

2.14 Suppose a photovoltaic (PV) module consists of 40 individual cells wired in series, (a)

In some circumstances, when all cells are exposed to the sun it can be modeled as a series combination of forty 0.5-V ideal batteries, (b) The resulting graph of current versus voltage would be a straight, vertical 20-V line as shown in (c)

0"

1"

2"

3"

4"

5"

6"

7"

8"

Voltage"(V)"

I!

Source"

Load

+

-

0"

1"

2"

3"

4"

5"

6"

7"

8"

Voltage"(V)"

2Ω# ba&ery#12V,#2Ω#

Ideal##

12V#ba&ery#

Source#I"V$

R= v 2

12V battery: v = 12

12V battery with 2Ω resistance:

v = 12 + 2R