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Ksenya-84 [330]
4 years ago
12

Suppose we include the lead resistance in the calculation of temperature for a class A RTD. If R3 = 1000 ohms, Ra = 18 ohms, V0

= 1.80 volts, and V = 5 volts. Including the tolerances, what is the highest temperature you could measure in °C? (Type in a two-decimal number)
Engineering
1 answer:
GenaCL600 [577]4 years ago
5 0

Answer:

The highest temperature that could be measured(in °C) is 14170.27°C

Explanation:

RTDs are thermometers consist metals of which it resistance increase with temperature.

For Class A RTD l,Alpha = 0.00385.

The formula for RTD is

Rt = Ro ( 1 + alpha x t)

Where

Rt is Resistance at temperature t°C,

Ro is the Resistance at 0°C

Alpha is a constant temperature coefficient of class A RTD.

Here, Rt = 1000ohms,

Ro is taking as Ra = 18Ohms

Therefore,

1000 = 18 ( 1 + 0.00385t)

Divide both sides by 18

1 + 0.00385t = 1000/18

0.00385t = 55.55 - 1

0.00385t = 54.55

t = 54.55/0.00385

t = 14170.27°C

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Using Von Karman momentum integral equation, find the boundary layer thickness, the displacement thickness, the momentum thickne
Alex_Xolod [135]

Answer:

Explanation:

We can solve Von Karman momentum integral equation as seen below using following in the attached file

3 0
3 years ago
A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate
kherson [118]

Explanation:

thermal expansion ∝L = (δL/δT)÷L ----(1)

δL = L∝L + δT ----(2)

we have δL = 12.5x10⁻⁶

length l = 200mm

δT = 115°c - 15°c = 100°c

putting these values into equation 1, we have

δL = 200*12.5X10⁻⁶x100

= 0.25 MM

L₂ = L + δ L

= 200 + 0.25

L₂ = 200.25mm

12.5X10⁻⁶ *115-15 * 20

= 0.025

20 +0.025

D₂ = 20.025

as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0

3 0
3 years ago
(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor
serg [7]

Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm =  0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

<u>A) would the 1040 steel be a possible candidate for this application</u>

<em>Yield strength of 1040 steel < stress  ( in order to be a possible candidate )</em>

stress = p / A0 = ( 18000 ) / ( \frac{\pi }{4} ) * 0.0075^2

                      = 18,000 / (4.418 * 10^-5 )   =  407.424 MPa

Yield strength of 1040 steel = 450 MPa

stress = 407.424 MPa

∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )  

Therefore 1040 steel is not a possible candidate for this application

<u>B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm</u>

Area1 = ( \frac{\pi }{4} ) ( 0.006 )^2 = 2.83 * 10^-5 m^2

therefore % of cold work done = ( A0 - A1 ) / A0  * 100 = 35.94%

6 0
3 years ago
1. On a 2001 Honda Civic, while replacing fuel injectors, what do you coat the new O-rings with?
scoundrel [369]

Answer:

Coat new O-rings (D) with silicone oil or polyalkyleneglycol (PAG) oil, and pull them on the injectors.

3 0
3 years ago
An inventor claims to have invented a heat engine that operates between the temperatures of 627°C and 27°C with a thermal effici
Oksana_A [137]

Answer Explanation:

the efficiency of the the engine is given by=1-\frac{T_2}{T_1}

where T₂= lower temperature

           T₁= Higher temperature

we have given efficiency =70%

lower temperature T₂=27°C=273+27=300K

higher temperature T₁=627°C=273+627=900K

efficiency=1-\frac{T_2}{T_1}

                =1-\frac{300}{900}

                 =1-0.3333

                 =0.6666

                 =66%

66% is less than 70% so so inventor claim is wrong

3 0
3 years ago
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