Introduction to engineering experimentation 3rd edition wheeler solutions manual

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CHAPTER 7 7.1 (a) Series R = 1, R

R = R +R 1 2 w

R

= [(

1

R =1 R 2

R R 2 2 1/ 2 w ) +( w ) ] R R1 R R 2 1

1

2 2 1/ 2 = [( 0.2 * 1) + ( 0.1 * 1) ] = 0.22  w

R,max

=

R R w + w R R1 R R 2 1

1

= 0.2 + 0.1 = 0.3 

(b) parallel R1R2 R= R1 + R2

R R22 R R12 = , = R1 (R1 + R2 )2 R2 (R1 + R2 )2 50 2 = (150)2 = 0.11

100 2 = 150 2 = 0.44

wR = [(0.11 * 0.2)2 + (0.44 * 0.1)2 ]1/ 2 = 0.05  wR,max = 0.11*.2 + 0.44 * 0.1 = 0.07 

7.1

7.2 K=

M = [tA (2gh)1/2 ]

393.00 [600 * (

0.0127 2 ) * 1000(2 * 9.81 * 3.66)1/2 ] 4

= 0.610 =

w k = [(

kg = K (dimensionless) [sec. m . kg/m3 (m/sec 2 * m)1/2 ] 2

1/ 2 K K K K w m )2 + ( w t )2 + ( w  )2 + ( w n ) 2 ] M T  n

K 1 K = = = 155 * 10 − 4 kg −1 1/ 2 M tA (2gh) M K K 0.610 =− =− = −10.16 * 10 − 4 sec −1 t t 600 K −K 0.610 = =− = −4814 .40m −2 −4 A A 1.266 * 10 K K 0.610 =− =− = −6.10 * 10 − 4 m 3 / kg   1000 K K 0.610 =− =− = −833 .33 * 10 − 4 m −1 ( h ) 2 h 2 * 3.66 A A d  0.0127 wd , = = = 0.01995 m d d 2 2 w A = 0.01995 * 0.000025 = .005 * 10 − 4 m 2 wA =

w K = [(1.55 * 10 − 4 * 0.03 ) 2 + (10.16 * 10 − 4 * 1) 2 + ( 4815 .40 * 0.005 * 10 − 4 ) 2 + (6.10 * 10 − 4 * 0.003 ) 2 ]1 / 2 w K = 0.0026

So,

K= 0.610  0.003

The maximum error for this case is larger: n K w K max =  w x i = 10 − 4 [1.55 * 0.03 + 10.16 * 1 + 4815 .4 * 0.005 + 6.10 * 0.001 + 833 .33 * 0.00366 ]  x n =1 i = 0.037  0.004

7.2

7.3 K=

865 M = −3 1/ 2 . * 10 ) * 63.36 * (2 * 32.17 * 12.02)1/ 2 [tA(2gh ) ] 600 * (136

= 0.601

lbm  K dim ensionless ft lbm 1/ 2 2 * ft ) sec* ft * 3 * ( sec 2 ft K K K K w )]1/ 2 wM )2 + ( wt )2 + ( w  )2 + ( w K = [( h h  t M = 10 −4 [(6.94 * 0.005)2 + (10 * 1)2 + (440 * 10 4 * 5.45 * 10 −6 )2 + (249.58 * 0.01)2 ]1/ 2 = 26.10 * 10 −4 where K 1 k = 6.94 * 10 −4 lbm −1 = = 1/ 2 M M tA(2gh ) ft 3 C C K K K K = − = −96.22 * 10 −4 = − = −440 ft −2 , = − = −10 * 10 −4 sec −1 , lbm   A A t t K K = 249.58 * 10 −4 ft −1 =− 2h  ( h ) A d  *.5 0.001 A = 5.45 * 10 −6 ft 2 * = = wd , wA = 2 * 12 12 2 d d So K = 0.601+ 0.003 The maximum error for this case is larger: n K w K ,max =  w x 1 = 10 −4 [6.49 * 0.05 + 10 * 1 + (440 * 10 +4 * 5.45 * 10 −6 ) + (249.58 * 0.01)]  x i =1 1 = 0.0037

7.3

7.4 The general formula for calculation of uncertainty (Eq. 7.4) will be applied to n R wR = [ (w xi 1 )2 ]1/ 2 individual cases. x i i =1 R 1 R R (a) R1 = ax 1 + bx 2 + cx 3 = a, 1 = b, 1 = C x 1 x 2 x 3 w R1 = [a 2w x21 + b 2 w x22 + c 2w x23 ]1 / 2

For the general case of summation of, R =  ai x i

wR w2 = [ ai2 xi2 ]1/ 2 R R R 2 R (b)R 2 = d ( x 1 )( x 2 )( x 3 ), ,= d , ( x 2 )( x 3 ), 2 = d ( x 1 )( x 3 ) x 1 x 2 wR = [ ai2w xi2 ]1/ 2 and

R 2 = d ( x 1 )( x 2 ) x 3 w R 2 = [d 2 ( x 2 ) 2 ( x 3 ) 2 w x21 + d 2 ( x 1 ) 2 ( x 3 ) 2 w x22 + d 2 ( x 1 ) 2 ( x 2 ) 2 w x23 ]1 / 2 For the general case of multiplication: n n wR w xi2 1 / 2 R 2 2 1/ 2 w R = d [ 2 w xi ] and = [ 2 ] R i =1 x i i =1 x i (c)R 3 = e( x 1 )( x 2 ) / x 3 ,

R 3 x R x R x x = e 2 , 3 = e 1 , 3 = −e 1 2 2 x 1 x 3 x 2 x 3 x 3 x3

x2 2 2 x1 2 2 x1 x 2 2 2 1/ 2 w x21 w x22 w x23 1 / 2 w R 3 = e[( ) w x1 + ( ) w x 2 + ( 2 ) w x 3 ] and w R 3 / R 3 = [ 2 + 2 + 2 ] x3 x3 x3 x1 x2 x3 (d) R 4 = f ( x 1 ) g ( x 2 ) h ( x 3 ) i ,

R 4 R = fg ( x 1 ) g −1 ( x 2 ) h ( x 3 ) i , 4 = fh( x 1 ) g ( x 2 ) h −1 ( x 3 ) i , x 1 x 2

R 4 = fi ( x 1 ) g ( x 2 ) h ( x 3 ) i −1 x 3 w R 4 = f [g 2 ( x 1 ) 2( g −1) ( x 2 ) 2h ( x 3 ) 2i w x21 + h 2 ( x 1 ) 2g ( x 2 ) 2( h −1) ( x 3 ) 2i w x22 + i 2 ( x 1 ) 2g ( x 2 ) 2h ( x 3 ) 2( i −1) ]1 / 2 w w R4 w w = [g 2 ( x1 ) 2 + h 2 ( x 2 ) 2 + i 2 ( x 3 ) 2 ]1 / 2 R4 x1 x2 x3 n

For the general case of R = f  x iai  = multiplication i =1

n

w R = f [ ai2 R 2 ( i =1

n w x1 2 1 / 2 w w ) ] and R = [ ai2 ( x1 ) 2 ]1/ 2 x1 R x1 i =1

7.4

7.5 F = kx (w F )max = w k

F F + wx k x

F = k = 700N / cm x F = x = 12.5cm k (w F )max = (18 * 12.5 + 1.25 * 700 ) (w F )max = 1,100N F 2 F 2 1/ 2 ) + (w x ) ] K X = [(18 * 12.5) 2 + (1.25 * 700 )2 ]1/ 2

w F = [(w k

= 903N F

= 700 * 12.5 = 8,750N (absolute value)

wF 903 * 100 = * 100 = 10.3%( relative value) F 8750

7.6

F F + wx k x F 2 F 2 1/ 2 w F = [(w k ) + (w x ) ] k x F F = x = 5 in, = 20 lbf / in k x

F = kx

(w F )max = w k

(w F )max 12.5 = =.125 F 5 * 20 w w F = [(0.5 * 5)2 + (0.5 * 20)2 ]1/ 2 = 10.3, F =.103, 10.3% F (w F )max = 0.5 * 5 + 0.5 * 20 = 12.5 lbf ,

7.5

7.7 F = kx ( wF ) max = wk

F F + wx k x

F = k = 800 N / cm x F = x = 20.0cm k ( wF ) max = (16 * 20.0 + 1.25 * 800) ( wF ) max = 1,320 N F 2 F 2 1 / 2 ) + ( wx ) ] K X = [(16 * 20.0) 2 + (1.25 * 800) 2 ]1 / 2

wF = [( wk

= 1,050 N F

= 800 * 20.0 = 16,000 N ( absolute value)

wF 903 *100 = *100 = 10.3%(relativevalue) F 8750

7.8 F 1/ 2 2 ) , = 250,000, = 500rad / sec mr Eq.7.6

 =( w

w w 1 wm 2 [( ) + ( r ) 2 + ( F ) 2 ]1 / 2  2 m r F w 1 .5 2 .02 2 .5 2 1 / 2 = [( ) +( ) +( ) ]  2 100 20 500 1 = [25 * 10 −6 + 10 − 6 + 10 −6 ]1 / 2 2 w = +2.60 * 10 −3 ,0.26%(95%confidencelevel ) =



w  = +1.30rad / sec (95%confidencelevel )

7.6

7.9 F 1/ 2 2 ) , = 200,000,  = 447.21rad / sec mr Eq.7.6

=(

wr wF 2 1 / 2 1 wm 2 [( ) + ( ) 2 + ( ) ]  2 m r F w 1 .5 2 .02 1.2 2 1 / 2 = [( ) + ( )2 + ( ) ]  2 120 25 600 1 = [4.691  10 − 3 ] 2 w = + 2.345 * 10 − 3 ,0.23% (95%confidencelevel) w

=



w = +1.03rad / sec (95%confidencelevel)

7.10 E = (f / A) /(

L L

)=

F .L A.L

Eq.7.6 w wE w w w = [( F ) 2 + ( L ) 2 + ( A ) 2 + ( L ) 2 ]1 / 2 E F L A L .5 2 1 2 1.5 2 5 2 1/ 2 = [( ) +( ) +( ) +( ) ] 100 100 100 100 1 = [.25 + 1 + 2.25 + 25]1 / 2 = 5.3 / 100or 5.3% 100 To reduce the uncertainty in E by 50%, we have to reduce the uncertainty in measurement of L (to less than 1/2 of the present value.)

7.7

7.11

wR = [(

R R wd )2 + ( wT )2 ]1/ 2 d T

We have assumed that R0 and T0 are known values with high degrees of accuracy.

R = [R0 (1 +  (T − T0 )]1/ 2

R R = R0 (T − T0 ) and = R0 , so  T wR = R0 [((T − T0 )w )2 + (wT )2 ]1/ 2 0.1 * 0.0048 = + 0.0048 * 10 −3 1 / C 100 wT = [BT2 + (tST )2 ]1/ 2 t = 2 (for l arg e (  30) samples ) w = +

= [0.12 + (2 * 0.1)2 ]1/ 2 = 0.22 C wR = 100[((25 − 0)*.0048 * 10 −3 )2 + (.0048 * 0.22)2 ]1/ 2 = 0.11 ohm

while R25 = 100.00(1 + 0.0048 * 25) = 112.00 ohms.

7.8

7.12

f m 2 N 5.0 * 10− 4 (kg / sec) = = 1.06 * 10−8 kg / J 1 2 * 50 * 150( .N .m.) sec kg / hr = 0.38 * 10 − 4 W Wbsfc Wm f 2 W W (b) = [( ) + ( N ) 2 + (  ) 2 ]1 / 2 f bsfc m N  (a) bsfc =

w bsfc at 1%, the maximum tolerable uncertainty in bsfc m f , N and  is 1%. This implies that if we have 1% uncertainty in one, the other two must have a much less uncertainty (negligible value compared to 1%.)

To have the value of

(c) If all measured parameters have the same uncertainty, then, W W bsfc Wm f 2 1 / 2 W 2 = [3( ) ] = [3( N ) 2 ] = [3( ) ] bsfc m f N  = 0.01 Wm f WN W = = = 7.6 *10 −3 or 0.8% m f N  d − Wm f = 7.6 *10 −3 * 5.0 *10 − 4 = 3.8 *10 −6 kg / sec = 14.4 g / hr!

WN = 22.8RPM W = 1.14N .m

7.9

7.13

f m 2 N 2.5 *10− 4 ( kg / sec) = = 1.06 *10− 9 kg / J 1 2 * 50 * 750( .N .m.) sec kg / hr = 0.38 *10− 5 W Wbsfc Wm f 2 W W (b) = [( ) + ( N ) 2 + (  ) 2 ]1 / 2 f bsfc m N  (a) bsfc =

w bsfc at 1.5%, the maximum tolerable uncertainty in bsfc m f , N and  is 1.5%. This implies that if we have 1.5% uncertainty in one, the other two must have a much less uncertainty (negligible value compared to 1.5%.)

To have the value of

(c) If all measured parameters have the same uncertainty, then, W W bsfc Wm f 2 1 / 2 W 2 = [3( ) ] = [3( N ) 2 ] = [3( ) ] bsfc m f N  = 0.015 Wm f WN W = = = 8.7 *10 −3 or 0.87%  mf N  d − Wm f = 8.7 *10 −3 * 2.5 *10 − 4 = 2.2 *10 −6 kg / sec = 7.8g / hr!

WN = 26.1RPM W = 1.31N .m

7.10

7.14 hysteresis Lineariz.error Resolution error zero off set repeatability

error type systematic systematic random systematic random

+ 0.1C 0.2% of reading 0.05C 0.1C +0.2C

B = (0.12 + [(.002)(120)]2 + 0.12 )1 / 2 = 0.28C Assuming that the random errors have been determined with samples>30, P = (.05 2 + .2 2 )1 / 2 = 0.21C So total uncertainty w = [ B 2 + p 2 ]1 / 2 = [ B 2 + (tS ) 2 ]1 / 2 w = [(0.28) 2 + (0.21) 2 ]1 / 2 w = 0.35C

7.15 hysteresis Lineariz.error Resolution error zero off set repeatability

error type systematic systematic random systematic random

+ 0.1C 0.2% of reading 0.10C 0.1C + 0.1C

B = (0.12 + [(.002)(100)]2 + 0.12 )1 / 2 = 0.24C Assuming that the random errors have been determined with samples>30, P = (.10 2 + .12 )1 / 2 = 0.14C So total uncertainty w = [ B 2 + p 2 ]1/ 2 = [ B 2 + (tS ) 2 ]1/ 2 w = [(0.24) 2 + (0.14) 2 ]1/ 2 w = 0.28C

7.11

7.16 Range: 0 → 2 kg Calib. Accuracy: 2% of reading (Systematic) A/D Converter Uncertainty: 0.5 Range 0.5  2 = 12 = 0.00024 kg 2N 2 2 wm = [ wCalib + w102 ]1 / 2 = [(0.02 *1.25) 2 + (0.00024) 2 ]1 / 2 = 0.025 kg

7.17 Range: 0 → 5 kg Calib. Accuracy: 2% of reading (Systematic) A/D Converter Uncertainty: 0.5 Range 0.5  5 = 12 = 0.00061 kg 2N 2 2 wm = [ wCalib + w102 ]1 / 2 = [(0.02 * 3.20) 2 + (0.00061) 2 ]1 / 2 = 0.064 kg

7.12

7.18 accuracy resolution linearity temp. stab.

error type systematic random systematic random

+0.5% range =5 kPa 1 kPa 4 kPa + 2 kPa (0 − 50C )

Systematic Errors: Normally, accuracy includes the error of linearity, so the only systematic error is the accuracy. Then, B = 5 kPa.

Random Errors: The random errors are those due to resolution and temperature stability.

Assuming more than 30 samples, t = 2 P = (12 + (2)2 )1/ 2 = 2.2 kPa Overall Uncertainty w = [B2 + P 2 ]1/ 2 = (5 2 + 2.22 )1/ 2 = 5.5 kPa or 5.5 / 500 = 0.011or 11% . of measured value of 500 kPa 7.19 accuracy resolution temp. stab.

error type systematic random random

+0.3%range =1.5 kPa 1 kPa + 1kPa (0 − 50 c)

Random Errors: The random errors are those due to resolution and temperature stability.

Assuming more than 30 samples, t = 2 P = (12 + (1) 2 )1 / 2 = 1.4kPa Overall Uncertainty w = [B 2 + P 2 ]1/ 2 = (1.5 2 + 1.4 2 )1/ 2 = 2.05kPa or 2.05 / 250 = 0.0082 0.8% of measured value

7.13

7.20 Average Temperature n

T

T = i =1 n T = 248.3 C

Random Uncertainty of Single Measurement n

Sx = [  ( x i − x )2 / (n − 1)]1/ 2 i =1

Sx = 0.55 C Table 6.6

10 − 1 = 9  t = 2.262 95% conf .

Pi = tSx = (2.262)(0.55) Pi = 12 . C

Random Uncertainty of Mean Px = tSx / n 1/ 2 = (2.262)(0.55) / 10 Px =.39 C

7.14

7.21 (a) mav =

 m = 1.93 + 1.95 + 1.96 + 1.93 + 1.95 + 1.94 + 1.96 + 1.97 + 1.92 + 1.93 i

n

10

= 1.94 kg

 (m − m ) ] =[ 2

(b)

S samp

i

av

1/ 2

n −1 ( 0.01) + ( 0.01) 2 + ( 0.02) 2 + .......... 1 / 2 =[ ] 9 = 1.65 *10 − 2 kg S Samp 1.70 *10 − 2 S mean = = = 0.52  10 − 2 kg n 10 2

(c)

PSingle = tS samp = 3.73 * 10 −2 kg (95% Confidence ) t = 2.262 is for  = n - 1 = 9 degrees of freedom from Student - t Table 6.6 Bsingle = 0.015  Range = 0.015  5kg = 0.075kg 2 1/ 2 Total uncertainty : wsingle = ( Psin2 gle + Bsin = 0.084 (95%confidence) gle )

(d) Pmean = tS mean = 1.18  10 −2 kg (95% Confidence) B mean = Bsingle = 0.075kg 2 2 Total uncertainty : wmean = ( Pmean + Bmean )1 / 2 = 0.076(95%confidence) As can be seen, the dominant factor is the systematic uncertainty.

7.15

7.22 (a)

(b)

mav =

i

n 10 = 2.94 lb (mi − mav ) 2 1/ 2  Ssamp = [ ] n −1 ( 0.04) 2 + ( 0.01) 2 + ( 0.02) 2 + .......... 1 / 2 =[ ] 9 = 2.72 *10 − 4 lb

S mean =

(c)

 m = 2.90 + 2.95 + 2.96 + 2.92 + 2.95 + 2.94 + 2.96 + 2.97 + 2.93 + 2.91

2.72 *10 − 4 = = 0.86 *10 − 4 lb n 10

S Samp

PSingle = tS samp = 6.15  10 −4 lb (95% Confidence ) t = 2.262 is for  = n - 1 = 9 degrees of freedom from Student - t Table 6.6 Bsingle = 0.015  Range = 0.015  10lb = 0.15lb 2 1/ 2 Total uncertainty : wsingle = ( Psin2 gle + Bsin = 0.15 (95%confidence) gle )

(d) Pmean = tS mean = 1.95  10 −4 kg (95% Confidence) B mean = Bsingle = 0.15kg 2 2 Total uncertainty : wmean = ( Pmean + Bmean )1 / 2 = 0.15(95%confidence) As can be seen, the dominant factor is the systematic uncertainty.

7.16

7.23 (a) If measurement is performed on over 50 pieces of pipe. Based on problem 7.13, for the sample of 10  mi = 1.94kg mav = n (b) Based on Problem 7.13, (mi − mav ) 2 1/ 2  Ssamp = [ ] = 1.65*10-2 kg n −1 But now, n=50 so the standard deviation of the mean is: SSamp 1.65 * 10 −2 Smean = = = 0.23 * 10 −2 kg n 50 (c)

PSingle = tS samp = 2 * 1.65 * 10 −2 kg = 3.3 * 10 −2 kg (95% Confidence ) t = 2 is for   30 degrees of freedom from Student - t Table 6.6 B single = 0.015 * Range = 0.015 * 5kg = 0.075 kg 2 2 1/ 2 Total uncertainty : w single = (Psin = 0.082 (95% confidence ) gle + Bsin gle )

(d) Pmean = tSmean = 2 * 0.23  10 −2 = 0.46  0 −2 kg (95% Confidence) B mean = Bsingle = 0.075 kg 2 2 Total uncertainty : w mean = (Pmean + Bmean )1/ 2 = 0.075 (95% confidence ) As can be seen, the dominant factor is the systematic uncertainty.

7.24 S = 0.1

P = 2S

(Large Sample, t = 2)

= 2 * 0.10 = 0.20 kg (With 95% Confidence Level) 1.5 B = * 12 = 0.18 kg 100 w Total = [B 2 + P 2 ]1/ 2

= [0.18 2 + 0.2 2 ] = 0.27 kg (With 95% confidence level)

7.17

7.25 1 .5 * 4.5 = 0.07kg 100 Calib : B1

w Total =

= 0.01 * 4.5 = 0.045 kg

Calibration will be the only systematic uncertainty. w 2 Total = B 2 + P 2 P 2 = w 2 Total - B 2 w total = .015  4.5 = 0.068 = 0.068 2 - 0.045 2 P = 0.05 kg P = 0.027 2 This the acceptableS for the cheese blocks.

The A/D converter contributes to this precision uncertainty but there may be other sources of precision error.

For a large sample : S Total =

7.18

7.26

 .t , where m = mass, m  = mass flow rate and t = filling time m=m m 1 (a) t = = = 4 sec  m 0.25 (b) The flowrate uncertainty is almost certainly random since it depends on factors that vary somewhat randomly (like viscosity( It is most likely that the time uncertainty is also random. If it were systematic, it would likely be adjusted to close to zero by adjusting the speed of the cam. We will treat it as random.

This means there are no significant bias errors to consider. (c) 2 2  m   m   Pm =  Pm  +  Pt      t    m

0.5

  = (4.0  0.0025 ) + (0.25  0.1)   Pt ) = (tPm ) + (m 2

2 0 .5

2 0 .5

2

= 0.0001 + 0.00063 

0 .5

= 0.027 kg Since we have not considered any systematic uncertainties, this is the total uncertainty, wm = P (d) The uncertainty in time has a much larger effect than the uncertainty in flowrate (2.5% vs. 1%) therefore the approach should be to make the timing more accurate.

Note: When the total uncertainty is computed in this problem, it doesn’t matter whether uncertainties were classified as systematic or random since they were combined with RSS.

7.19

7.27  t , where m = mass, m  = mass flow ratio and t = filling time m=m m 2 (a) t = = = 4 sec  0. 5 m (b) The flowrate uncertainty is almost certainly random since it depends on factors that vary somewhat randomly (like viscosity( It is most likely that the time uncertainty is also random. If it were systematic, it would likely be adjusted to close to zero by adjusting the speed of the cam. We will treat it as random.

This means there are no significant bias errors to consider. (c) 2 2  m   m   Pm =  P  + P   m   t t    m 

0.5

  = (4.0  0.005 ) + (0.5  0.1)  2  Pt )2 = (tPm ) + (m

0 .5

2 0 .5

2

= 0.0004 + 0.0025 

0 .5

= 0.054 lbm

Since we have not considered any systematic uncertainties, this is the total uncertainty, wm.= P. (d) The uncertainty in time has a much larger effect than the uncertainty in flowrate (2.5% vs. 1%) therefore the approach should be to make the timing more accurate.

Note: When the total uncertainty is computed in this problem, it doesn’t matter whether uncertainties were classified as systematic or random since they were combined with RSS.

7.20

7.28 standard deviation:

S=450 kJ/kg

systematic error:

B=.01*100000=100 kJ/kg

for 95% confidence level and degrees of freedom 15-1 , Pi = tSx = 2145 . * 450 kJ / kg = 965

kJ / kg

Pi 965.25 = = 249 kJ / kg n 15 a) w x = (B 2 + P 2 ) y 2 = (1000 2 + 249 2 )1/ 2 Px =

w x = 1030 kJ / kg b) w i = (B 2 + Pi 2 )1/ 2 = (1000 2 + 965 2 )1/ 2 w i = 1390

kJ / kg

7.21

t=2.145

(Table 6.6)

7.29 (a) Output ( mA) is linear with input(I mH 2 O) in the range; At Depth = 0 mH 2O I = 4 mA At Depth = 20 mH 2O I = 20 mA Input span is 20 m, Output span is 16 mA At depth = 15 m Output = 4 + (15/20)16 = 16 mA (b) Acccuracy 0.2% of span (systematic) In m H 2O :

0.002  20 = 0.04 m H 2O

In mA :

0.002  16 = 0.032 mA

This is 0.032/16 = 0.002 = 0.2% of reading Zero Balance :

2% Span (systematic) = 0.02  20 = 0.40 m H 2O 0.02  16 = 0.3 2mA

In mA :

This is 0.32/16 = 0.02 = 2% of reading Thermal Effects :

1.5% of span (random)

In m H 2O :

0.015  20 = 0.3 m H 2O

In mA :

0.015  16 = 0.24 mA

This is 0.24/16 = 0.015 = 1.5% of reading

The following table shows the uncertainties in m, mA and % of reading: Uncertainty Accuracy Zero Balance Thermal Effect Total Uncertainty

Uncertainty Category Systematic Systematic Random

Meter Water 0.04 0.40 0.30

0.03 0.30 0.24

% reading at 15 m 0.26 2.6 2.0

0.50

0.40

3.3

(c) B Depth = [  Bi2 ]1 / 2 = [0.042 + 0.4 2 ]1 / 2 = 0.40m water

wDepth = [ B 2 + P 2 ]1 / 2 = [0.402 + 0.30 2 ]1 / 2 = 0.50 m water or 0.50  16/20 = 0.40 mA This will be 0.4/16 = 2.5%

7.22

mA

7.30 (a) Output ( mA) is linear with input(I mH 2 O) in the range; At Depth = 0 mH 2O I = 4 mA At Depth = 25 mH 2O I = 20 mA Input span is 25 m, Output span is 16 mA At depth = 15 m Output = 4 + (15/25)16 = 13.6 mA (b) Acccuracy 0.1% of span (systematic) In m H 2O :

0.001  25 = 0.025 m H 2O

In mA :

0.001  16 = 0.016 mA

This is 0.016/16 = 0.001 = 0.1% of reading Zero Balance :

1.5% Span (systematic) = 0.015  25 = 0.38 m H 2O 0.015  16 = 0.24 mA

In mA :

This is 0.24/16 = 0.015 = 1.5% of reading Thermal Effects :

1% of span (random)

In m H 2O :

0.01  25 = 0.25 m H 2O

In mA :

0.01  16 = 0.16 mA

This is 0.16/16 = 0.01 = 1% of reading

The following table shows the uncertainties in m, mA and % of reading: Uncertainty Accuracy Zero Balance Thermal Effect Total Uncertainty

Uncertainty Category Systematic Systematic Random

Meter Water 0.025 0.38 0.25

0.016 0.24 0.16

% reading at 15 m 0.12 1.76 1.18

0.46

0.29

2.13

(c) B Depth = [  Bi2 ]1 / 2 = [0.0252 + 0.38 2 ]1 / 2 = 0.38m water wDepth = [ B 2 + P 2 ]1 / 2 = [0.382 + 0.252 ]1 / 2 = 0.45 m water or 0.45  16/20 = 0.36 mA This will be 0.36/16 = 2.3%

7.23

mA

7.31 (a) For an un-amplified output of 20mV, considering that the device is linear, assuming no zero offset, F (N ) mV VOutput = * 3( ) * VExitation FRange V VOutput =

F (N ) 500N

* 3(

mV ) * 10V V

Vout 20 = = 333 N 0.060 0.060 0.1 (b) Linearity Uncertainty : 0.1% FS = * 500 = 0.5 N 100 or 0.060 * 0.50 = 0.030 mV or 0.03/20 =0.15% Similarly for Hysteresis and Repeatability (see table below) 0.002 Max. Temperature effect: 0.002 % FS/ oC = * 500 * (40 − 20) 100 = 0.20N Vout

= 0.060 F (N)

 F=

Or 0.06 * 0.2 = 0.012 mV Other uncertainties are shown in the table below: Uncertainty Linearity Hysteresis Repeatability Temp. Effect Total (c)

Category Systematic Systematic Random Random

Total Uncertaint y :

mV 0.030 0.025 0.009 0.012 0.042

N 0.50 0.40 0.15 0.20 0.70

% output 0.15 0.12 0.05 0.06 0.21

2 2 w Total = [B Force + PForce ]1 / 2

2 2 B Force = [ Bi2 ]1 / 2 = [B Linl + BHys ]1 / 2 = 0.039 mV

or

0.65N

Assuming that all random uncertainties are based on samples greater than 30 : 2 2 1/ 2 PForcel = [ P 2 i ]1 / 2 = (PRe = (0.009 2 + 0.012 2 )1 / 2 = 0.015 mV p + PTem )

or

0.25N

2 2 w Total = (BForce + PForce )1 / 2 = (0.039 2 + 0.015 2 )1 / 2 = 0.042 mV

or

0.70N

which is about 0.21% of output.

7.24

7.32 (a) Uncertainty due to accuracy (considered systematic) 6 For 2 lpm : 6% of flow = * 2 = 0.12 lpm 100 6 For 10 lpm : 6% of flow = * 10 = 0.6 lpm 100 Uncertainty due to repeatability (considered random) 1 For 2 lpm : 1% of flow = * 2 = 0.02 lpm 100 1 For 10 lpm : 1% of flow = * 10 = 0.10 lpm 100 (b)

Total Uncertainty: w Total = [B 2 + P 2 ]1/ 2

For 2 lpm

wTotal = [0.12 2 + 0.02 2 ]1 / 2 = 0.12 lpm

For 10 lpm

wTotal = [0.6 2 + 0.12 ]1/2

= 0.61 lpm

In both the cases, the uncertainty due to accuracyis dominant

7.33 (a) Uncertainty due to accuracy (considered systematic) 5 For 5 lpm : 5% of flow = * 5 = 0.25 lpm 100 5 For 15 lpm : 5% of flow = *15 = 0.75 lpm 100 Uncertainty due to repeatability (considered random) 1 For 5 lpm : 1% of flow = * 5 = 0.05 lpm 100 1 For 15 lpm : 1% of flow = *15 = 0.15 lpm 100 (b)

Total Uncertainty:

wTotal = [B2 + P 2 ]1 / 2

For 5 lpm

wTotal = [0.25 2 + 0.05 2 ]1 / 2 = 0.25 lpm

For 15 lpm

wTotal = [0.75 2 + 0.152 ]1/2

= 0.76 lpm

In both the cases, the uncertainty due to accuracyis dominant

7.25

7.34 Accuracy: average: STD Dev:

0.2 C 250 C 0.2 C

(systematic error)

(a) Random Uncertainty of Mean Value degrees of freedom 15-1 = 14, 95% confidence.

Table 6.6  t = 2145 .

PX = tSx / n 1/ 2 = (2145 . )(0.2) / 15 = 0.11 C w = (B2 + P 2 ) y 2 = (0.22 + 0.112 )1/ 2 w = 0.23

C

overall accuracy of the mean value

(b) Random Uncertainty - Single Reading Pi = tSx = (2145 . )(0.2) =.43 w = (B 2 + P 2 )1/ 2 = (.22 +.43 2 )1/ 2 = 0.5 C

accuracy of any sin gle measurement

(c) It will have a greater effect on mean since random error is about the same, but for single reading - the effect will not be as much because random error is greater.

7.35 Tcorr =



(Ts4 − Tw4 ) =

h Tcorr = 135.8 K

5.669 * 10 −8 * 0.9 [(600 + 273)4 − (545 + 273)4 ] 50

w + 2 w n 2 1/ 2 w +T 0.1 2 10 2 1/ 2 b− = [( ) + ( ) ] = [( ) + ( ) ] = 0.23 T  n 0.9 50 + w T = 0.23 * 135.8 = 312 . K w− w −T w 0.2 2 10 2 1/ 2 = [(  )2 + ( n )2 ]1/ 2 = [( ) + ( ) ] = 0.30 T  n 0.9 50 − w T = 0.30 * 135.8 = 40.8 K  Tg = 600 + 135.8 Tg = 735.8 + 312 . / −40.8 K

7.26

7.36 10 measurements of strain ave: 80 mV S = 0.5 mV (Standard deviation) t= 2.262 (deg. of freedom 10 - 1 = 9, 95% confidence) Px = ts = 2.262 * 0.5 = 1.13mV (dueto strain gage ) 15 measurements of transmitted voltage STD = 1 mV (Standard deviation) t = 2.145 (deg. of freedom 15 – 1 =14, 95% confidence) Pz = ts = 2145 . * 1 = 214 . mV (due to transmission ) Overall standard deviation of voltage: m

S x = [ s i2 ]1 / 2 = [.5 2 + 12 ]1 / 2 = 1.1mV i =1 m

[ s i2 ] 2

[.5 2 + 12 ] 2 = 20 .5 4 14 4 si / i +  9 14 i =1 For 95%conf .fromTable6.6,t = 2.086,Px = ts x = 2.3mV

 x = mi =1

=

7.37 10 measurements of strain ave: 100 mV S = 0.8 mV (Standard deviation) t= 2.262 (deg. of freedom 10 - 1 = 9, 95% confidence) Px = ts = 2.262 * 0.8 = 1.81mV (duetostrain gage) 10 measurements of transmitted voltage STD = 1 mV (Standard deviation) t = 2.262 (deg. of freedom 10 - 1 = 9, 95% confidence) Pz = ts = 2.262 *1 = 2.26mV (duetotransmissi on) Overall standard deviation of voltage: m

S x = [  si2 ]1 / 2 = [.8 2 + 12 ]1 / 2 = 1.3mV i =1

m

[  si2 ]2

[.8 2 + 12 ]2 = 17 .8 4 14 4  si /  i 9 + 9 i =1 For95%conf . fromTable6.6,t = 2.110,Px = ts x = 2.7mV

 x = mi=1

=

7.27

7. 38 15 measurements - degrees of freedom = 15 - 1 = 14 at 95% confidence. t = 2145 .

(Table 6.6) Pr ecision index Sx = 5 kP P = ts x = 2145 . * 5 = 10.7 kPa In Problem 7.11, there are two systematic errors, accuracy and linearity.

However, the linearity is normally included in the accuracy so the only systematic error is the accuracy. As a result: B=5 kPa

Overall uncertainty of the pressure measurements: total: w x = (Bx2 + Px2 )1/ 2 = (5 2 + 10.7 2 )1/ 2 w x = 118 . kPa

7.39 Yes the uncertainty will change because it is dependent on t which is determined from the sample size. If n = 30, t = 2.042. t would be smaller so the uncertainty would also be smaller.

7.28

7.40 (a) V =

 V = 459.9 i

n

 (V − V ) ] S =[ 2

i

I=

= 1.25 V

1/ 2

n −1

V

 I = 31.0 amps i

n

 (I − I ) ] S =[ 2

1/ 2 = 0.63 amp n −1  PFi = 0.79 PF = n  (PFi − PF )2 ]1 / 2 = 0.02 SPF = [ n −1  Pi =  Vi * Ii * PFi 3 = 19,629 Watts P = n n i

I

 (P − P ) ] S =[ 2

i

1/ 2

n −1 PV = tSV = 2.95 V , P

= 787 Watts PI = tSI = 1.49 amp,

PPF = tSPF = 0.05, and

PP = tSP = 1,861Watts

where " t" has been obtained from Student - t test Table 6.6 for

 = n - 1 = 7 for 95% confidence (

(b)

P

PP = t

(c)

=

SP n

 2

= 0.0025) to be 2.365

 P = V * I * PF 3 = 19,629 Watts i

i

i

n

= 2.365 *

i

n

787 8

= 658 Watts

95% confidence

B v = B v = 1% of V = 0.01*459.9 = 4.60 V

B I = B I = 1% of I = 0.01 * 31.0 = 0.31 amp B PF = B PF = 1% of PF = 0.01 * 0.79 = 0.79 * 10 − 2

Systematic uncertainty of power (applying Eq. 7.24 and utilizing Eq. 7.6 for simplification), 2

2

2

 P   P   P  BP = [ BV  +  BI  +  BPF  ]1/ 2  V   I   PF 

7.29

B BP B2 B = [( V ) 2 + ( I ) 2 + ( PF ) 2 ]1/ 2 = (0.012 + 0.012 + 0.012 )1/ 2 = 0.017 P V PF I B P = 0.017 * 19629 . = 334 Watts

(d) w P = [ B P2 + PP2 ]1/2 = (334 2 + 658 2 )1/2 = 738 Watts

(With 95% confidence level)

most of the uncertainty is due to random effects.

7.41 El Error freedom Calibration Data Acquisition repeatedly spatial installation transmission Data Reduction linearity

Systematic Uncertainty .

Standard deviation Degrees of

2

---

---

--2 1 ---

.5 ----.3

14 ----29

1

---

---

Systematic Uncertainty : Bx = (.22 + 22 + 12 + 12 )1/ 2 = 2.5 C (.5 2 +.3 2 )2 = 4 = 24 [.5 / 14 +.3 4 / 29] Standard Deviation : from 6.6  t = 2.064 Sx = (.5 2 +.3 2 )1/ 2 =.58 C P = tSx = (2.064)(.58) = 12 . C

Total Uncertainty w x = (P 2 + B 2 )1/ 2 = (12 . 2 + 2.5 2 )1/ 2 = 2.8 C

7.30

7.42 w V = uncerta int y of volume = +1 cc ( precision ) w t = uncerta int y of time = + 0.3

sec

V 100 = = 4cc / sec t 25 Eq. 7.6

Q=

wQ w w = [( V )2 + ( t )2 ]1/ 2 Q V t 1 2 .3 = [( ) + ( )2 ]1/ 2 100 25 = 0.016, 16% . wQ = 4 * 0.016 = 0.06 cc / sec

7.31

( precision )

7.43 Power as measured: P = V x I x PF x 3 1/2 = 460*36*0.81*3 ½ = 23,233 Watts (a) We will take the uncertainty originated from the measurement device as systematic uncertainty For Voltage : B V = 1 % of reading = 0.01 * 460 = 4.60 V For Current : B I = 1% of reading = 0.01 * 36 = 0.36 A For PF

B PF = 2% of reading = 0.02 * 0.81 = 1.62 * 10 -2

:

The total uncertainty due to measurement device applying Eq. 7.24 and utilizing Eq. 7.6 for simplification): 2

2

2

 P   P   P  B P = [ BV  +  BI  +  BPF  ]1 / 2  V   I   PF 

or 2

2

2

BP B  B  B  = [ V  +  I  +  PF  ]1 / 2 = (0.012 + 0.012 + 0.02 2 )1 / 2 = 0.0245 P V   I   PF  B P = 23,233 * 0.0245 = 569 Watts

(b) Assuming that a large number of tests has been performed for calculation (n>30)of the standard deviation, S, the random uncertainties with 95% confidence level will be: PV = 2 S V = 6.00 Volts P1 = 2 SI = 4.00 Amp PPF = 2 SPF = 0.06

The actual power is Pwr = 460  35  0.81  3 0.5 = 22,587 and then B=0.0245x22587=553. (c) To Calculate the total uncertainty of power measurement, we calculate the systematic and random uncertainties separately:

need to

Systematic uncertainty, BP was calculated in part a. The random uncertainty PP will be calculated below.

7.32

2

2

2

 P   P   P  Sp = [  SV  +  SI  +  S PF  ]1 / 2  V   I   PF  or 2

2

2

S  S  S  = [  V  +  I  +  PF  ] 1 / 2 P V   I   PF  2 2 2 Sp  3.00   2.00   0.03  1 / 2 =[  +  +  ] = 0.067 P  460   35.00   0.81  S p = 1,556 Watts with 95% confidence Sp

Assuming a large number of measurements (n>30), PP = 2SP = 3,112 Watts Total uncertainty for measurement of power with 95% confidence,

w P = [ BP2 + PP2 ]1/2 = (553 2 + 3,112 2 )1/2 = 3,160 Watts This is equivalent to 14% of the reading!

7.33

7.44 Power as measured: P = V x I x PF x 3 1/2 = 460*50*0.81*3 ½ = 32,268 Watts (a) We will take the uncertainty originated from the measurement device as systematic uncertainty For Voltage : BV = 1 % of reading = 0.01* 460 = 4.60 V For Current : BI = 1% of reading = 0.01* 50 = 0.5 A For PF

BPF = 1% of reading = 0.01* 0.81 = 0.81

:

The total uncertainty due to measurement device applying Eq. 7.24 and utilizing Eq. 7.6 for simplification):  P   P   P  BP = [ BV  +  BI  +  BPF  ]1 / 2  V   I   PF  2

2

2

or BP  BV   BI   BPF  1 / 2  ] = (0.012 + 0.012 + 0.012 )1 / 2 = 0.0173 = [  +   +   V   I   PF  P BP = 32,268 * 0.0173 = 558 Watts 2

2

2

(b) Assuming that a large number of tests has been performed for calculation (n>30)of the standard deviation, S, the random uncertainties with 95% confidence level will be: PV = 2 S V = 6.00 Volts P1 = 2 S I = 2.00 Amp PPF = 2 S PF = 0.06

We will take the uncertainty originated from the measurement device as systematic uncertainty For Voltage : B V = 1 % of reading = 0.01* 460 = 4.60 V For Current : B I = 1% of reading = 0.01* 49.50 = 0.495 A For PF

:

B PF = 1% of reading = 0.01* 0.80 = 0.80

The actual power is Pwr = 460  49.5  0.8  3 0.5 = 31,551 an then the systematic uncertainty is 0.0173x31551 = 546. W (c) To Calculate the total uncertainty of power measurement, we need to calculate the systematic and random uncertainties separately:

7.34

Systematic uncertainty, BP was calculated in part b. The random uncertainty PP will be calculated below.

2 2 2  P   P   P  1/ 2 Sp = [  S  + S  + S  ]  V V   I I   PF PF  or S  S  S  = [  V  +  I  +  PF  ]1 / 2  V   I   PF  P

Sp

2

2

2

 3.00   1.00   0.03  1 / 2  +  +  ] = 0.043 =[  460   49.50   0.80  P

Sp

2

2

2

Sp = 1,357 Watts with 95% confidence

Assuming a large number of measurements (n>30), PP = 2SP = 2,714Watts Total uncertainty for measurement of power with 95% confidence,

w P = [ BP2 + PP2 ]1/2 = (546 2 + 2,714 2 )1/2 = 2,768 W This is equivalent to 9% of the reading!

7.35

7.45 (a) Although, some of these uncertainties (repeatability and temperature errors specifically) are usually classified as random, because they are related to the accuracy of an instrument, we will combine them as systematic uncertainty.

As an elemental error if they are considered as random uncertainty based on a large sample, and the uncertainty rather than the standard deviation is given, the effect on the total uncertainty of the measurement will be the same. If we are looking at the uncertainty of the mean, this approach will not work and we need to separate systematic and random uncertainties.

0.5 * 100 = 3.5 kPa 100 0.1 BHys = 0.1% of span = * 700 = 0.7 kPa 100 0.1 BRe p = 0.1% of span = * 700 = 0.7 kPa 100 0.3 BStab = 0.3% of span = * 700 = 2.1 kPa 100 0.040 Bzero = 0.040% of span/ C = * 700 * (20 − 15 ) = 1.4 kPa 100 0.030 BSpan = 0.030% of span/ C = * 700(20 − 15 ) = 1.05 kPa 100 4.55 BPr ess = [ B 2i ]1 / 2 =4.55kPa or * 100 = 0.65% of span 700 Blin = 0.5% of span =

(b)

Assuming that a large number of tests has been performed for calculation of the standard deviation, S.

PP = 2S p = 2 * 3.5 = 7.0 kPa with 95% Confidence w P = [ B 2P + PP2 ]1/ 2 = [ 4.55 2 + 7.0 2 ]1/ 2 = 8.3 kPa with 95% Confidence The total uncertainty is more due to pressure fluctuations, i.e random unceratinty. (c)

Quantization uncertainty, PQ , which can be considered a part of the transmission uncertainty

0.5Span 0.5  1400 = = 0.17 kPa 2N 212 0.17 This value is * 100 = 2.0% of the sensor uncertainty 8.3 Note : A unipolar A/D was assumed, For bi - polar system, the PQ will be 0.34 kPa PQ =

(d) Uncertainty due to noise, Pn , which can be considered a part of the transmission uncertainty Pn = 0.02 * Signal value = 0.02 * 550 = 11 kPa Obviously, the uncertainty due to noise is a significant part of the uncertainty of the final result

7.36

7.46 (a) Although, some of these uncertainties (repeatability and temperature errors specifically) are usually classified as random, because they are related to the accuracy of an instrument, we will combine them as systematic uncertainty.

As an elemental error if they are considered as random uncertainty based on a large sample, and the uncertainty rather than S is given the effect on the total uncertainty of the measurement will be the same. If we are looking for the total uncertainty in a mean, this approach will not work and we need to separate systematic and random uncertainties.

0.5 Blin = 0.5% of span = * 100 = 0.5 psi 100 0.1 BHys = 0.1% of span = * 100 = 0.1 psi 100 0.1 BRe p = 0.1% of span = * 100 = 0.1 psi 100 0.3 BStab = 0.3% of span = * 100 = 0.3 psi 100 0.040 Bzero = 0.022% of span/  F = * 100 * (70 − 60 ) = 0.22 psi 100 0.016 BSpan = 0.016% of span/  F = * 100 * (70 − 60 ) = 0.16 psi 100 0.66 BPr ess = [ B i2 ]1/ 2 =0.66 psi or * 100 = 0.66% of span 100 (b)

Assuming that a large number of tests has been performed for calculation of the standard deviation, S,

PP = 2S p = 2 * 0.50 = 1.00 psi with 95% Confidence w P = [ B 2P + PP2 ]1/ 2 = [0.66 2 + 1.0 2 ]1/ 2 = 1.20 psi with 95% Confidence The total uncertainty is more due to pressure fluctuations, i.e random unceratinty. (c)

Quantization uncertainty, PQ , which can be considered a part of the transmission uncertainty

0.5Span 200 = 12 = 2.44 * 10 -2 psi N 2 2 2.44 * 10 -2 This value is * 100 = 3.7% of the sensor uncertainty 0.66 Note : A unipolar DAS was assumed, For bi - polar system, the PQ will be 4.88 * 10 -2 psi PQ =

(d) Uncertainty due to noise, Pn , which can be considered a part of the transmission uncertainty Pn = 0.02 * Signal value = 0.02 * 80 = 1.60 psi Obviously, the uncertainty due to noise is a significant part in the uncertainty of the final result

7.37

7.47 P =  

(a)

3000 * 165 60 = 51,836 N.m/sec (watt) = 2 N.  = 2 *

(b)

The Standard deviation of power : 2

2

 P   P  S P = [ S  +  S  ]1 / 2       P P = , =   S P = [( .S ) + ( .S ) ]1 / 2 2

2

= [(2 * 50 * 4 ) + (165 * 2 2

5 2 1/ 2 ) ] 60

= 1,260 Watts

(c)

The degrees of freedom for power (Eq. 7.28)

P =

(S )

2 2 P 2

1  P  1  [ x Si  ] 2 i  i  n

Where   = n − 1 = 20 − 1 = 19 and

  = n − 1 = 10 − 1 = 9 P =

=

S P4 4

1  P 1  P   S  +  S            

4

(1,260) 4 4

1  5  1 3000  * 4 165 * 2 *  +  2 * 19  60  9 60 

The random uncertainty of power:

7.38

4

=9

 P = 9  t P = 2.262 (Table 6.6) for 95% confidence

PP = t P SP = 2.262 * 1260 = 2,850 Watts

The systematic uncertainty of power : 2

2

 P   P  1/ 2 B P = [ B  +  B  ]       = [(B ) + (B ) ]1/ 2 2

2

= 236 Watts (d)

Total uncertainty of power :

w P = [B P2 + PP2 ]1/ 2 = [(2850 ) + (236 ) ]1/ 2 2

2

= 2,860 Watts (95% confidence)

7.39

7.48 (a)

P =   3000 * 165 60 = 51,836 N.m/sec (Watts) = 2 N.  = 2 *

(b) The Standard deviation of power : 2

2

 P   P  S P = [ . S   +  . S   ]1 / 2       = [(S ) + (S ) ]1/ 2 2

2

= [(2 * 50 * 4 ) + (165 * 2 * 5160 ) ]1/ 2 2

2

= 1,260 Watts

(c)

Random uncertainty of power, considering the large number of measuremen ts :

PP = 2S P = 2,520 Watts The systematic uncertainty of power : 2

2

 P   P  1/ 2 B P = [ B  +  B  ]       = [( B  ) + (B ) ]1/ 2 2

2

2

2

5   3000   = [165 * 2 * * 0 .7  ] 1 / 2  +  2 60   60   = 236 Watts (d) Total Uncertainty of power : w P = [BP2 + PP2 ]1/ 2 = [(2,520 ) + (236 ) ]1/ 2 2

2

= 2,531 Watts

7.40

7.49 (a) The systematic uncertainty of the measurement will remain the same as in the previous two problems: 2

2

 P   P  1/ 2 B P = [ B  +  B  ]       = [( B  ) + (B ) 2 ]1/ 2 2

2

2

5   3000   = [165 * 2 * * 0 .7  ] 1 / 2  +  2 60   60   = 236 Watts

To calculate the random uncertainty: Degrees of freedom,  = n-1 = 10-1 = 9, where n is number of measurements.

Random uncertainty of each power measurement: PP = tSP = 2.262*0.75 = 1.70 kW where t = 2.262 is the value for Student-t value from Table 6.6 for 95% confidence.

For the mean power: tS 2.262  0.75 PP = P = = 0.54 kW n 10 (b) Total Uncertainty of a single measurement : w P = [BP2 + PP2 ]1 / 2 = [(1700 ) + (236 ) ]1 / 2 2

2

= 1,716 watt Total uncertainty of the mean power: Total Uncertainty of the mean power :

w P = [BP2 + PP2 ]1 / 2 = [(236 ) + 540 2 ]1 / 2 2

= 590 watt (c) The systematic uncertainty remains the same but the random uncertainty depends upon the measurement process and number of measurements.

Additionally, the uncertainties in Problems 7.32 and 7.33 are for single measurement experiment although torque and speed have been independently measured multiple times but the averages are used for calculation of the results.

In this problem, power is measured multiple times, and the random uncertainties are based on these measurements.

7.41

7.50 P W 3  340  −3  m   * 700 * 10 3 (Pa )   * 10  60 sec    

 = Q.

(a)

=

5 * 10 3 Watts

= 0.793

(b) Parameter Q

Mean Value 340 lit/m 700 kPa 5 kW

P W

Standard Deviation 5.6 10 0.15

2

   S = [  S i  ]1/ 2  X i  2

Q QP  P  = [ . S Q  + ( .S P ) 2 + ( 2 S ) 2 ]1/ 2 W W W  Applying Eq. 7.6 and 7.26

S



= [(

SQ Q

.)2 + (

S P 2 SW 2 1/ 2 ) +( ) ] P W

5.6 2 10 2 0.15 2 1/ 2 ) +( ) +( ) ] = 0.037 340 700 5 S = 0.793 * 0.037 = 0.029 = [(

7.42

Degree of Freedom 15-1 15-1 15-1

(c) Random uncertainty : P = t S

To find t, we need to calculate degrees of freedom for  (Eq. 7.28 )

(S )

2 2

 =



2

2

2 2 2 1   1    1          SQ   + S P   + SW       Q  Q    P  P      W  

2

(S ) 4



=

4 4 4 4 1  SQ   1  S P   1  Sw       +   +      Q  Q    P  P   W  W     4

 0.029    0.793    = 4 4 4 1  5 .6  1  10  1  0.15  + +       14  340  14  700  14  5 

= 27

Using Table 6.6, for 95% confidence level (

 2

= 0.025 ),

t = 2.052 P = t S = 2.052 * 0.029 = 0.060

with 95% confidence level

Systematic uncertainty of the pump efficiency :  BQ  2  B  2  B  2   +  P  +  W   =    Q   P   W    

B

1/ 2

(Apply Eq. 7.6 and 7.24)

There are two uncertainties associated with the pressure transducer. Accuracy is always treated as systematic. It is hard to predict how thermal stability will affect the uncertainty. It may vary over a long time period and appear constant over a shorter period of time. It is conservative (high estimate of uncertainty) to consider it systematic.

Following Eq. 7.15 : 2 2  0.2   0 .2   BP =  * 1200  +  * 1200     100    100 = 3.39 kPa

1/ 2

7.43

1.5 * 340 ) = 5.10 rpm 100 BW = 0.07 kW BQ = (

 5.10  2  339  2  0.07  2  =   +  +     340   700   5  

B

1/ 2

= 0.021 B = 0.793 * 0.021 = 0.017

(d)

Total uncertainty of the pump efficiency:



w  = B + P 2





2 1/ 2

= 0.017 2 + 0.060 2 = 0.062



1/ 2

With 95% confidence level

 = 0.793  0.062 With 95% confidence level

7.44