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Cerrena [4.2K]
4 years ago
15

What is the non-linearity error, as a percentage of full range, produced when a 1K Ω potentiometer has a load of 10KΩ & is a

t one-third of its maximum displacement?
Engineering
1 answer:
Darina [25.2K]4 years ago
7 0

Answer:

2.28%

Explanation:

Being at one third of its maximum range a potentiometer should output V0/3.

However if this 1kΩ potentiometer has a 10kΩ load:

(1) I1 = I2 + I3

(2) V0 = I1 * 2/3*Rp + I3 * 1/*Rp

(3) Vp = I2 * Rl

(4) Vp = I3 * 1/3 * Rp

Where

I1: current entering the potentiometer

I2: current going to the load

I3: current going to the other leg of the potentiometer

V0: supply voltage

Vp: output voltage of the potentiometer

Rp: total resistance of the potentiometer

Rl: load resistance

First we determine the intensity of I3 in function of supply power

I3 = 3 * Vp / Rp  = 3 * Vp / 1000 = 0.003*Vp

Then the load current

I2 = Vp / Rl  = Vp / 10000 = 0.0001*Vp

With these we determine I1

I1 = 0.003*Vp + 0.0001*Vp = 0.0031*Vp

Then

V0 = 0.0031 * Vp * 2/3*Rp + 0.003*Vp * 1/3*Rp

V0 = 0.00207 * Vp * Rp + 0.001 * Vp * Rp

V0 = 0.00307 * Vp * 1000

V0 = 3.07 * Vp

Vp = V0 / 3.07

Vp = 0.3257 * V0

Now the percentage error is:

(100 * (0.3333 - 0.3257)) / 0.3333 = 2.28 %

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A refrigerated space is maintained at -15℃, and cooling water is available at 30℃, the refrigerant is ammonia. The refrigeration
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Answer:

(1) 5.74

(2) 5.09

(3) 3.05×10⁻⁵ kg/s

(4) 0.00573 kW

Explanation:

The parameters given are;

Working temperature, T_C  = -15°C = 258.15 K

Temperature of the cooling water, T_H = 30°C = 303.15 K

(1) The Carnot coefficient of performance is given as follows;

\gamma_{Max} = \dfrac{T_C}{T_H - T_C}  =  \dfrac{258.15}{303.15 - 258.15}   = 5.74

(2) For ammonia refrigerant, we have;

h_2 = h_g = 1466.3 \ kJ/kg

h_3 = h_f = 322.42 \ kJ/kg

h_4 = h_3 = h_f = 322.42 \ kJ/kg

s₂ = s₁ = 4.9738 kJ/(kg·K)

0.4538 + x₁ × (5.5397 - 0.4538) = 4.9738

∴ x₁ = (4.9738 - 0.4538)/(5.5397 - 0.4538) = 0.89

h_1 = h_{f1} + x_1 \times h_{gf}

h₁ = 111.66 + 0.89 × (1424.6 - 111.66) = 1278.5 kJ/kg

\gamma = \dfrac{h_1 - h_4}{h_2 - h_1}

\gamma = \dfrac{1278.5 - 322.42}{1466.3 - 1278.5} = 5.09

(3) The circulation rate is given by the mass flow rate, \dot m as follows

\dot m = \dfrac{Refrigeration \ capacity}{Refrigeration \ effect \ per \ unit \ mass}

The refrigeration capacity = 105 kJ/h

The refrigeration effect, Q = (h₁ - h₄) = (1278.5 - 322.42) = 956.08 kJ/kg

Therefore;

\dot m = \dfrac{105}{956.08}  = 0.1098 \ kg/h

\dot m = 0.1098 kg/h = 0.1098/(60*60) = 3.05×10⁻⁵ kg/s

(4) The work done, W = (h₂ - h₁) = (1466.3 - 1278.5) = 187.8 kJ/kg

The rating power = Work done per second = W×\dot m

∴ The rating power = 187.8 × 3.05×10⁻⁵ = 0.00573 kW.

6 0
3 years ago
A spherical balloon with a diameter of 9 m is filled with helium at 20°C and 200 kPa. Determine the mole number and the mass of
SIZIF [17.4K]

Answer:

<em>number of mole is 31342.36 moles</em>

<em>mass is 125.369 kg</em>

<em></em>

Explanation:

Diameter of the spherical balloon d = 9 m

radius r = d/2 = 9/2 = 4.5 m

The volume pf the sphere balloon ca be calculated from

V = \frac{4}{3} \pi r^3

V = \frac{4}{3}* 3.142* 4.5^3 = 381.75 m^3

Temperature of the gas T = 20 °C = 20 + 273 = 293 K

Pressure of the helium gas = 200 kPa = 200 x 10^3 Pa

number of moles n = ?

Using

PV = nRT

where

P is the pressure of the gas

V is the volume of the gas

n is the mole number of the gas

R is the gas constant = 8.314 m^3⋅Pa⋅K^−1⋅mol^−1

T is the temperature of the gas (must be converted to kelvin K)

substituting values, we have

200 x 10^3 x 381.75 = n x 8.314 x 293

number of moles n  = 76350000/2436 = <em>31342.36 moles</em>

We recall that n = m/MM

or m = n x MM

where

n is the number of moles

m is the mass of the gas

MM is the molar mass of the gas

For helium, the molar mass = 4 g/mol

substituting values, we have

m = 31342.36 x 4

m = 125369.44 g

m =<em> 125.369 kg</em>

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Answer:

Explanation:

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