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Papessa [141]
3 years ago
15

Your boss asks you to classify some of the components in a temperature measurement system. The system consists of a thermocouple

and an analog-digital conversion (ADC) device. The thermocouple has a linear response from 0 to 70°C. Over this range, its minimum output voltage is 0 V and its maximum output voltage is 3.649 mV. The ADC device has 8 bits and two possible input ranges +5 mV and ±5 V. You can assume that the ADC rounding process operates in the same manner as we discussed in class.a. Determine the resolution voltage and its related uncertainty of the ADC for both input ranges b. Determine the sensitivity of the thermocouple c. You set a water bath to a 57°C temperature and place the thermocouple in the bath.i. What do you expect the thermocouple voltage to read? ii. Based off both ADC input ranges, what do you expect the ADC output voltage to read? iii. How can you explain the difference in values? iv. Which input range is optimal for this measurement scenario? d. Based on the previous step and the lower ADC input range: If you didn't know the bath temperature and only knew the ADC voltage, what temperature would you say the bath is set to?
Engineering
1 answer:
lianna [129]3 years ago
8 0

Answer:

Explanation:

1) Resolution and uncertainty of both ADC ranges

a) Resolution (for +/- 5 mV range) = 5 mV / 2n where n = number of bits of ADC

Resolution = 5 mV / 28 = 5 mV / 256 = 0.0195 mV    ..................1

uncertainty (for +/- 5 mV range) = +/- 0.5*resolution = +/- 0.5*0.0195 mV = 0.00975 mV       ..........2

b) Resolution (for +/- 5 V range) = 5 V / 2n where n = number of bits of ADC

Resolution = 5 mV / 28 = 5 V / 256 = 0.0195 V = 19.5 mV    ..................3

uncertainty (for +/- 5 mV range) = +/- 0.5*resolution = +/- 0.5*19.5 mV = 9.75 mV       ..........4

2)Thermocouple sensitivity = ( Maximum output voltage - Minimum Output voltage) / (Maximum Temperature - Minimum Temperature)

Thermocouple sensitivity = (3.649mv - 0 ) / (70 - 0) = 0.0521 mV / Deg.C           ............5

This is the required Thermocouple sensitivity

3) Water bath temperature is given as 57 deg.C

Hence voltage read by Thermocouple = Sensitivity*57 = 0.0521*57 mV = 2.9697 mV    ........6

4)We need to use ADC with a range of +/- 5 mV range as ADC with +/- 5 V range can not do measurement as it's resolution is higher than output voltage.

ADC will measure voltage as 2.9695 mV                    ......................7

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The phasor technique makes it pretty easy to combine several sinusoidal functions into a single sinusoidal expression without us
devlian [24]

Answer:

The phasor technique can't be applied directly in the following cases:

a) 45 sin(2500t – 50°) + 20 cos(1500t +20°)

b) 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°)

c)  -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)

d)  75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)

Explanation:

For a) and c), it is not possible to use the phasor technique, due this technique is only possible when the sinusoidal signals to be combined are all of the same frequency.

This is due to the vector representing a signal is showed as a fixed vector in the graph( which magnitude is equal to the amplitude of the sinusoid and his angle is the phase angle with respect to cos (ωt)), which is rotating at an angular speed equal to the angular frequency of the sinusoidal signal that represents, like a radius that shows a point rotating in a circular uniform movement.

This rotating vector represents a sinusoidal signal, in the form of a cosine (as the real part of the complex function e^{j(wt+\alpha)}), so it is not possible to combine with functions expressed as a sine, even though both  have  the same frequency.

If we look at the graphs of cos (ωt) and sin (ωt) we can say that the sin lags the cos in 90º, so we can say the following:

sin (ωt) = cos (ωt-90º)

This means that in order to be able to represent a sine function  as a cosine, we need to rotate it 90º in the plane clockwise.

This is the reason why before doing this transformation, it is not possible to use the phasor technique for b) and d).

8 0
3 years ago
the AADT for a section of suburban freeway is 150000 veh/day. Assuming this is an urban radial facility, what range of direction
igomit [66]

Answer:

design hour volumes will be 4000 to 6000

Explanation:

given data

AADT  = 150000 veh/day

solution

we get here design hour volumes that is express as

design hour volumes  = AADT × k × D    ..............1

here k is factor and its  range is 8 to 12 % for urban

and D is directional distribution i.e traffic equal divided by the direction

so here design hour volumes will be 4000 to 6000

7 0
3 years ago
A Capacitor is a circuit component that stores energy and can be charged when current flows through it. A current of 3A flows th
Neko [114]

Answer:

8 μC

Explanation:

By definition, current is the rate of change of charge, so we can write the following equation for current I:

I = ΔQ / Δt

As charge must be conserved, all the charge carried by the current must add to the charge on the plates of the capacitor, so we can finf this incremental charge as follows:

ΔQ = I* Δt (assuming that current remains constant during the charging process)

⇒ ΔQ = 3 A* 2 μsec = 3 coul/sec*2 μsec = 6 μC

As the initial charge must be conserved also, the magnitude of the net electric charge of the capacitor must be as follows:

Qnet = Q₀+ ΔQ = 2 μC + 6 μC = 8 μC

5 0
3 years ago
Water is the working fluid in an ideal Rankine cycle. Steam enters the turbine at 1400 lbf/in2 and 1000°F. The net power output
Alex17521 [72]

Answer:

(a) The mass flow rate of the steam is approximately 1.803×10⁶ lb/h

(b) The rate of heat transfer is approximately 2.52×10⁹ BTU/h

(c) The thermal efficiency  is approximately 39.68%

(d) The mass flow rate of cooling water is approximately 9.478 × 10⁷ lb/h

Explanation:

(a) The parameters are;

T₁ = 1000 F, P₁ = 1400 psi

By using an online application, we have;

h₁ = 1494 BTU/lb = 3,475 kJ/kg

s₁ = 1.61 BTU/(lb·R) =  6.741 kJ/(kg·K)

Therefore, due to isentropic expansion from state 1 to state 2, we have;

s₁ = s₂ = 1.61 BTU/(lb·R)

P₂ = 2 psi

T₂ =

s_f = 0.17498 BTU/(lb·R)

h_{f2} = 94.02 BTU/(lb)

h_g = 1116 BTU/lb

s_g = 1.919 BTU/(lb·R)

We have;

x₂ = (1.61 - 0.17498)/(1.919 - 0.17498) ≈ 0.823

h₂ = h_f + x₂×(h_g -

P₃ = P₂ = 2 psi

h₃ = h_{f2} = 94.02 BTU/(lb)

v₃ = 0.01605 ft³/lb

h₄ = h₃ + v₃ × (P₄ - P₃)

h₄ = 94.02 + 0.01605 × (1400 - 2) ×144/778 = 98.17 BTU/lb

The mass flow rate of the steam, \dot m = \dot W/((h₁ - h₂) - (h₄ - h₃)) = 1 * 10^9/((1494 -935.11) - (98.17 -94.02)) ≈ 1.803×10⁶ lb/h

\dot m ≈ 1.803×10⁶ lb/h

The mass flow rate of the steam ≈ 1.803×10⁶ lb/h

(b)The rate of heat transfer, \dot Q_{in} = \dot m × (h₁ - h₄) = 1.803×10⁶×(1494 -98.17) ≈ 2.52×10⁹ BTU/h

\dot Q_{in}  ≈ 2.52×10⁹ BTU/h

(c) The thermal efficiency =  \dot W_{cyc}/\dot Q_{in} = 1×10⁹/(2.52×10⁹) = 0.3968 ≈ 39.68%

The thermal efficiency ≈ 39.68%

(d) The mass flow rate of cooling water \dot m_w = \dot m(h₂- h₃)/(c_w \Delta T)

c_w = 1 BTU/(lb·°F)

\dot m_w =  1.803×10⁶ (935.11- 94.02)/(1 * (76 - 60)) ≈ 9.478 × 10⁷ lb/h

\dot m_w ≈ 9.478 × 10⁷ lb/h

The mass flow rate of cooling water ≈ 9.478 × 10⁷ lb/h.

5 0
3 years ago
Once a design is final engineer needs a plan for product
Makovka662 [10]

Answer:

Correct

Explanation:

7 0
3 years ago
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