Practice Problems (KEY)

Practice Problems (KEY)#

Answer the following questions about the devices shown in the following table

Device	Current	Voltage
A	3 A rms	190 V rms
B	8 A rms	190 V rms
C	3 A rms	190 V rms
X	10 A rms	40 V rms
Y	4 A rms	40 V rms
Z	5 A rms	40 V rms

a. Draw the block diagram to represent this set of devices

b. Label circuit breaker locations and values

c. Calculate total power supplied

P gen = I svs = 18*190 = 3.42kW

Calculate the following for the power distribution system below:

a. turns ratio of the transformer

b. circuit breaker values for circuit breakers A, B, C

c. Power the source produces

\[a = \frac{V_{1}}{V_{2}} = \frac{35}{205} = 0.171\]

\[P_{flaps,out} = \frac{15hp}{1}*\frac{1492W}{2hp} = 11.19kW\]

\[P_{flaps,in} = \frac{11,190W}{0.62} = 18kW\]

\[I_{S} = \frac{18000W}{205V} = 88A\]

I~C~ = I~Flaps~ + I~AVIONICS~ + I~NAV~ = 88A + 12A + 12A = 112A ; CB I~C~ = 112*1.1 = 123A, round up to nearest 0 or 5, CB I~C~ = 125A

P~C~ = V~C~I~C~ = 205$\ V_{RMS}$*112$A$ = 22.97kW

\[P_{B} = \frac{22970W}{0.8} = 28.7kW\]

$I_{B} = \frac{P_{B}}{V_{B}} = \frac{28700W}{32V} = 897A;\ $CB I~B~ = 897*1.1 = 986.7A, round up to nearest 0 or 5, CB I~B~ = 1kA

I~A~ = I~comms~ + I~cam~ + I~B~ = 897A + 5A + 2.3A = 904A; CB I~A~ = 904*1.1 = 994.4A, round up to nearest 0 or 5, CB I~A~ = 1kA

P~S~ = V~S~ I~S~ = 32*904 = 28.9kW

T / F Buses work by delivering power to devices connected in series.
T / F A circuit breaker works the same as a fuse in that it opens when current exceeds the circuit breaker’s rating.
The electrical system for a small reconnaissance aircraft is shown below. The aircraft has two busses, one at 160 V~RMS~ and one at 50 V~RMS~, powered by a 160 V~RMS~ alternator.

What transformer turns ratio is required to drive the 50 V~RMS~ bus?

\[a = \frac{V_{1}}{V_{2}} = \frac{160}{50} = 3.2\]

Given the following table of required currents, what circuit breaker ratings would you choose for the circuit breakers labeled A, B, C, and D?

Equipment	current (A rms)
Flaps	2.5
Avionics	4
Navigation	6
Comm	3.5
Camera	4.5

\[I~D~ = I~Comm~ = 3.5A; CB I~D~ = 3.5*1.1 = 3.85A\]

round up to nearest 0.0 or 0.5 $$CB I~D~ = 4A$$

\[I~C~ = I~comm~ + I~cam~ = 3.5A + 4.5A = 8 A; CB I~C~ = 8*1.1 = 8.8A\]

round up to nearest 0.0 or 0.5 $$CB I~C~ = 9A$$

\[a = \frac{I_{C}}{I_{B}};I_{B} = \frac{I_{C}}{a}\ = \frac{8A}{3.2} = 2.5A;\ CB I~B~ = 2.5*1.1 = 2.75A\]

round up to nearest 0.0 or 0 $$CB I~B~ = 3A$$

\[I~A~ = I~FLAPS~ + I~AVIONICS~ + I~NAV~ + I~B~\]

\[= 2.5 + 4 + 6 + 2.5 = 15;\]

\[CB I~A~ = 15*1.1 = 16.5A\]

round up to nearest 0.0 or 0.5 $$CB I~A~ = 16.5A$$

The electrical system for a fighter is shown below. The aircraft has two busses, one at 80 V~RMS~ and one at 20 V~RMS~, powered by a 115-V~RMS~ alternator.

a. What transformer turns ratios are required to drive the two buses?

\[a_{XFMR1} = \frac{V_{1}}{V_{2}} = \frac{115}{80} = 1.4375\]

\[a_{XFMR2} = \frac{V_{1}}{V_{2}} = \frac{115}{20} = 5.75\]

b. Given the following table of required currents, how much power must the alternator provide?

Equipment (A rms)	Current
Flaps	5
Landing Gear	8
Radar	4
Computer	2
IRS	5

$I~20VRMS~ = I~Comp~ + I~IRS~ = 2A + 5A = 7A$

$a_{XFMR2} = \frac{I_{20VRMS}}{I_{1,XFMR2}};I_{1,XFMR2} = \frac{I_{20VRMS}}{a_{XFMR2}} = \frac{7}{5.75} = 1.21A$

I~80VRMS~ = I~FLAPS~ + I~LG~ + I~RADAR~ = 5A + 8A + 4A = 17A

$a_{XFMR1} = \frac{I_{80VRMS}}{I_{1,XFMR1}};I_{1,XFMR1} = \frac{I_{80VRMS}}{a_{XFMR1}} = \frac{17}{1.4375} = 11.8A$

I~S~ = I~1,XFMR2~ + I~1,XFMR1~ = 1.21A + 11.8 A = 13A

P~ALT~ = I~S~V~S~ = 13A*115V = 1.495kW

Or….

Since there are no losses in the system,

\[P_{ALT} = P_{useful} = P_{load} = 80V_{RMS}*(5 + 8 + 4)A_{RMS} + 20V_{RMS}*(2 + 5)A_{RMS} = 1.5\ kW\]

The electrical system for a fighter is shown below. The aircraft has two busses, one at 80 V~RMS~ and one at 20 V~RMS~, powered by a 115-V~RMS~ alternator.

a. What transformer turns ratios are required to drive the two buses?

\[a_{XFMR2} = \frac{V_{1}}{V_{2}} = \frac{80}{20} = 4\]

\[a_{XFMR1} = \frac{V_{1}}{V_{2}} = \frac{115}{80} = 1.4375\]

b. Given the following table of required currents, how much power must the alternator provide?

Equipment (A rms)	Current
Flaps	5
Landing Gear	8
Radar	4
Computer	2
IRS	5

$I~20VRMS~ = I~Comp~ + I~IRS~ = 2A + 5A = 7A$

$a_{XFMR2} = \frac{I_{20VRMS}}{I_{1,XFMR2}};I_{1,XFMR2} = \frac{I_{20VRMS}}{a_{XFMR2}} = \frac{7}{4} = 1.75A$

I~80VRMS~ = I~FLAPS~ + I~LG~ + I~RADAR~ + I~1,XFMR2~= 5A + 8A + 4A + 1.75= 18.75A

$a_{XFMR1} = \frac{I_{80VRMS}}{I_{1,XFMR1}};I_{1,XFMR1} = \frac{I_{80VRMS}}{a_{XFMR1}} = \frac{18.75}{1.4375} = 13A$ = I~s~

P~ALT~ = I~S~V~S~ = 13A*115V = 1.495kW

Or ….

Since there are no losses in this system,

\[P_{ALT} = P_{useful} = P_{load} = 80V_{RMS}*(5 + 8 + 4)A_{RMS} + 20V_{RMS}*(2 + 5)A_{RMS} = 1.5\ kW\]

An aircraft requires a 205-V~RMS~ bus and a 32-V~DC~ bus as shown below. Calculate the turns ratio of the transformer and the circuit breaker values for circuit breakers A and B. The Comm equipment draws 4 A of current and the Camera draws 5 A. The efficiency of the AC-to-DC Converter is 80%.

Solution: First, we will calculate the turns ratio for the transformer. Note the input voltage for the AC-DC converter is given in the diagram. Since this is in RMS, we can use the turns ratio in conjunction with the source’s RMS value:

\[a = \frac{V_{1}}{V_{2}} = \frac{205}{35} = 5.86\]

Next, we can calculate the circuit breaker values. First, we are given the Comm and Camera current draws. Thus, the circuit breaker at location B must be larger than 4A+5A = 9 A. Therefore, we choose a 10 A breaker.

Now, we find the power used by the equipment attached to the 32 VDC bus is:

P~comm~ = (32V~DC~)(4A) = 128W

P~Camera~ = (32V~DC~)(5A) = 160W

The total power is 128 + 160 = 288 W. This is the power on the right side of the AC-DC converter. Since we know the AC-DC converter is 80% efficient, we know the power on the left side of the AC-DC converter is:

\[P_{in,ACDC} = \frac{288W}{0.8} = 360W\]

The power above is the power in the secondary of the transformer. Since we assume lossless transformers, the power in the primary is also 360 W. We know the voltage drop across the primary is 205 VRMS . Therefore, we can calculate the current:

\[I_{XFMR} = \frac{360W}{205V_{RMS}} = 1.75A\]

CB I = 1.75 * 1.1=1.925A.

The circuit breaker should be rated around 2A.

You have previously designed everything shaded gray in the block diagram below for an end user. Now, your user has decided to give you a new requirement to drive a 5 V DC bus for the satellite’s computer and camera. Since you really don’t want to have to start all over, you decided to retrofit your system as below. Is this a good design? Why or why not?

Solution: No, this is not a good design. The added transformer is attached to a DC bus. Transformers only work with AC power. Therefore, there will be no current or voltage on the right side of the transformer, meaning the 5V bus will not receive any power.

The electrical distribution system shown below is for a building’s security system. Each component is vital to some aspect of the security of the complex. How many breakers would you add to this diagram? Where would you put them?

Solution: Since each component is vital to the system, you would not want any one item to knock out an entire bus, or worse yet, the entire system. Therefore we definitely need to put a breaker on each component from the 60V and 25V buses (8 total). After that you want breakers before the buses to account for if the `combined’ current being drawn from the bus becomes too high (2 more at that point). Finally you may want to put extra breakers in-between the XFMR2 and ADC, and on either side of the XFMRs, just to be safe if something shorts or malfunctions inside of the ADC or XFMRs (up to 3 more here). Therefore, in total, you should have between 10 and 13 (optimal) breakers in your diagram.