Presentasi.docx

SLIDE 1 UNCERTAINTY OF RESISTANCE OF A COPPER WIRE The resistance of a certain size of copper wire is given as 𝑅 = 𝑅0 [1 + 𝛼(𝑇 − 20)] where 𝑅0 = 6 Ω ± 0.3% is the resistance at 20℃, a = 0.004℃−1 ± 1% is the temperature coefficient of resistance, and the temperature of the wire is T = 30 ± 1℃. Calculate the resistance of the wire and its uncertainty!

SLIDE 3 Solution The nominal resistance is R = (6) [1 + (0.004) (30-20)] = 6.24 Ω The uncertainty in this value is calculated by applying Eq. (3.2). The various terms are 𝜕𝑅 = 1 + 𝛼(𝑇 − 20) = 1 + (0.004)(30 − 20) = 1.04 𝜕𝑅0 𝜕𝑅 = 𝑅0 (𝑇 − 20) = (6)(30 − 20) = 60 𝜕𝛼 𝜕𝑅 = 𝑅0 𝛼 = (6)(0.004) = 0.024 𝜕𝑇 𝑤𝑅0 = (6)(0.003) = 0.018 Ω 𝑤𝛼 = (0.004)(0.01) = 4 × 10−5 ℃−1 𝑤𝑟 = 1 ℃ Thus, the uncertainty in the resistance is 𝑊𝑅 = [(1.04)2 (0.018)2 + (60)2 (4 × 10−5 )2 + (0.024)2 (1)2 ]1/2 = 0.035 Ω 𝑜𝑟 0.49%

SLIDE 4 UNCERTAINTY IN POWER MEASUREMENT. The two resistors R and Rs are connected in series as shown in the accompanying figure. The voltage drops across each resistor are measured as E = 10 V ± 0.1 V (1%) Es = 1.2 V ± 0.005 V (0.467%) Along with a value of Rs = 0.0066 Ω ± ¼% From these measurements determine the power dissipated in resistor R and its uncertainty.

SLIDE 5 The power dissipated in resistor R is P = EI The current through both resistors is I = Es/Rs so that 𝐸 𝐸𝑠 𝑅𝑠

𝑃=

(𝑎)

The nominal value of the power is therefore P = (10)(1.2)/(0.0066) = 1818.2 W The relationship for the power given in Eq. (a) is a product function, so the fractional uncertainty in the power may be determined from Eq. (3.2a). We have 𝑎𝐸 = 1

𝑎𝐸𝑠 = 1

𝑎𝑅𝑠 = −1

So that 1/2

𝑎𝐸 𝑤𝐸 2 𝑎𝑅 𝑤𝑅 2 𝑤𝑝 𝑎𝐸 𝑤𝐸 2 = [( ) + ( 𝑆 𝑆) + ( 𝑆 𝑆) ] 𝑃 𝐸 𝐸𝑠 𝑅𝑠

1/2

0.1 2 0.005 2 = [(1) ( ) + (1)2 ( ) + (−1)2 (0.0025)2 ] 10 1.2 2

Then 𝑤𝑝 = (0.0111)(1818.2) = 20.18 𝑊

SLIDE 6 SELECTION OF MEASUREMENT METHOD. A resistor has a nominal stated value of 10 Ω ± 1%. A voltage is impressed on the resistor, and the power dissipation is to be calculated in two different ways: (1) from P = E2/R and (2) from P = EI. In (1) only a voltage measurement will be made, while both current and voltage will be measured in (2). Calculate the uncertainty in the power determination in each case when the measured values of E and I are E = 100 V ± 1%

(for both cases)

I = 10 A ± 1 %

SLIDE 7 The chematic is shown in the accompanying figure. For the first case we have 𝜕𝑃 2𝐸 = 𝜕𝐸 𝑅 And we apply Eq. (3.2) to give

𝜕𝑃 𝐸2 = − 2 𝜕𝑅 𝑅

2

1/2

2𝐸 2 2 𝐸2 𝑤𝑝 = [( ) 𝑤𝐸 + (− 2 ) 𝑤𝑅2 ] 𝑅 𝑅

(𝑎)

Dividing by P = E2/R gives 𝑤𝑝 𝑤𝐸 2 𝑤𝑅 2 1/2 = [4 ( ) + ( ) ] 𝑃 𝐸 𝑅

(𝑏)

Inserting the numerical values for uncertainty gives 𝑤𝑝 = [4(0.01)2 + (0.01)2 ]1/2 = 2.236 % 𝑃 SLIDE 8 For the second case we have 𝜕𝑃 =𝐼 𝜕𝐸

𝜕𝑃 =𝐸 𝜕𝐼

And after similar algebric manipulation we obtain 𝑤𝑝 𝑤𝐸 2 𝑤𝐼 2 1/2 = [( ) + ( ) ] 𝑃 𝐸 𝐼

(𝑐)

Inserting the numerical values of uncertainty yields 𝑤𝑝 = [(0.01)2 + (0.01)2 ]1/2 = 1.414 % 𝑃

SLIDE 9 INSTRUMENT SELECTION. The power management in Example 3.2 is to be conducted by measuring voltage and current across the resistor with the circuit shown in the accompanying figure. The voltmeter has an internal resistance Rm, and the value of R is known only approximately. Calculate the nominal value of the power dissipated in R and the uncertainty for the fllowing conditions: R

= 100 Ω

Rm

= 1000 Ω

I

= 5 A ± 1%

E

= 50 V ± 1%

SLIDE 10 A current balance on the circuit yields 𝐼1 + 𝐼2 = 𝐼

𝐸 𝐸 + =𝐼 𝑅 𝑅𝑚 𝐼1 = 1 −

𝐸 𝑅𝑚

(𝑎)

The power dissipated in the resistor is 𝑃 = 𝐸𝐼1 = 𝐸𝐼 −

𝐸2 𝑅𝑚

(𝑏)

The nominal value of the power is thus calculated as 𝑃 = (500)(5) −

5002 = 2250 𝑊 1000

In terms of known quantities the power has the functional from 𝑃 = 𝑓(𝐸, 𝐼, 𝑅𝑚 ), and so we form the derivatives 𝜕𝑃 2𝐸 =𝐼− 𝜕𝐸 𝑅𝑚

𝜕𝑃 =𝐸 𝜕𝐼

𝜕𝑃 𝐸2 = 𝜕𝑅𝑚 𝑅𝑚 2 SLIDE 11 The uncertainty for the power is now written as 2

1/2

2𝐸 2 2 𝐸2 𝑤𝑝 = [(1 − ) 𝑤𝐸 + 𝐸 2 𝑤12 + ( 2 ) 𝑤𝑅2𝑚 ] 𝑅𝑚 𝑅𝑚

(𝑐)

Inserting the appropiate numerical values gives 2

1/2

1000 2 2 104 𝑤𝑝 = [(5 − ) 5 + (25 × 104 )(25 × 10−4 ) + (25 × 6 ) (2500)] 1000 10 = [16 + 25 + 6.25]1/2 (5) = 34.4 𝑊 Or 𝑤𝑝 34.4 = = 1.53% 𝑃 2250 SLIDE 12 WAYS TO REDUCE UNCERTAINTIES

A certain obstruction-type flowmeter (orifice, venturi, nozzle), shown in the accompnying figure, is used to measure the flow of air at low velocities. The relation describing the flow rate is 𝑚̇ = 𝐶𝐴 [

2𝑔𝑐 𝑝1 (𝑝1 𝑅𝑇1

1/2

− 𝑝2 )]

(𝑎)

where

C = empirical-discharge coefficient A = flow area 𝑝1 and 𝑝2 = upstream and downstream pressure, respectively T1 = upstream temperature R = gas constant for air

Calculate the percent uncertainty in the mass flow rate for the following conditions: C = 0.92 ± 0.005

(from calibration data)

P1 = 25 psia ± 0.5 psia T1 = 70°F ± 2°

T1 = 530°R

Δp = p1 – p2 = 1.4 psia ± 0.005 psia

(measured directly)

A = 1.0 in2 ± 0.001 in2

SLIDE 13 In this example the flow rate is a function of several variables, ech subject to an uncertainty. ṁ = 𝑓(𝐶, 𝐴, 𝑝1 , Δ𝑝, 𝑇1 )

(𝑏)

Thus, we form the derivatives 1/2 𝜕ṁ 2𝑔𝑐 𝑝1 = 𝐴( Δ𝑝) 𝜕𝐶 𝑅𝑇1 1/2 𝜕ṁ 2𝑔𝑐 𝑝1 = 𝐶( Δ𝑝) 𝜕𝐴 𝑅𝑇1 1/2 𝜕ṁ 2𝑔𝑐 = 0.5𝐶𝐴 ( Δ𝑝) 𝑝1 −1/2 𝜕𝑝1 𝑅𝑇1

(𝑐)

𝜕ṁ 2𝑔𝑐 𝑝1 1/2 = 0.5𝐶𝐴 ( ) Δ𝑝−1/2 𝜕Δ𝑝 𝑅𝑇1 1/2 𝜕ṁ 2𝑔𝑐 𝑝1 = −0.5𝐶𝐴 ( Δ𝑝) 𝑇1 −3/2 𝜕𝑇1 𝑅

The uncertainty in the mass flow rate may now be calculated by assembling these derivatives in accordance with Eq. (3.2). Designating this assembly as Eq. (c) and then dividing by Eq. (a) gives 1/2

𝑤𝑚̇ 𝑤𝐶 2 𝑤𝐴 2 1 𝑤𝑝 2 1 𝑤Δ𝑝 2 1 𝑤𝑇1 2 = [( ) + ( ) + ( 1 ) + ( ) + ( ) ] 𝑚̇ 𝐶 𝐴 4 𝑝1 4 Δ𝑝 4 𝑇1

(𝑑)

SLIDE 15 We may now insert the numerical values for the quantities to obtain the percent uncertainty in the mass flow rate.

1/2

𝑤𝑚̇ 0.005 2 0.001 2 1 0.5 2 1 0.005 2 1 2 2 = [( ) + ( ) + ( ) + ( ) + ( ) ] 𝑚̇ 0.92 1.0 4 25 4 1.4 4 530

= [29.5 × 10−6 + 1.0 × 10−6 + 1.0 × 10−4 + 3.19 × 10−6 + 3.57 × 10−6 ]1/2 = [1.373 × 10−4 ]1/2 = 1.172 %