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Anna71 [15]
3 years ago
8

A displacement transducer has the following specifications: Linearity error ± 0.25% reading Drift ± 0.05%/○C reading Sensitivity

error ± 0.25% reading Excitation 10 – 25 V dc Output 0 – 5 V dc Range 0 – 5 cm The transducer output is to be indicated on a voltmeter having a stated accuracy of ± 0.1% reading with a resolution of 10 μV. The system is to be used at room temperature, which can vary by ±10○C. Estimate an uncertainty in a nominal displacement of 2 cm at the design stage. Assume 95% confidence. Assume sensitivity is K = 1 V/cm.
Engineering
1 answer:
White raven [17]3 years ago
8 0

Answer:

The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm

Explanation:

uncertainty in a nominal displacement

= (u^2 + v^2)^(1/2)

assume from specifications that k = 5v/5cm

                                                         = 1v/cm

u^2 = (0.0025*2)^(2) + (0.005*10*2)^2 + (0.0025*2)^2

      = 0.01225v

v = 2v * 0.001

  = 0.002v

uncertainty in a nominal displacement

= (u^2 + v^2)^(1/2)

= ((0.01225)^2 + (0.002)^2)^(1/2)

= 0.0124 cm

Therefore, The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm

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Air at 2.5 bar, 400 K is extracted from a main jet engine compressor for cabin cooling. The extracted air enters a heat exchange
Shkiper50 [21]

Answer:

a) Power developed by the turbine = 132.89 kW

b) magnitude of the rate of heat transfer from the air to the ambient, in kw = 251.25 kW

Explanation:

b) The process is a constant pressure process (Isobaric process)

The constant pressure specific heat of air, c_{p} = 1.005 kJ/kg -K

Specific heat ratio for air, \gamma = 1.4

The mass flow rate of air, \dot{m} = 2.5 kg/s

P₁ = 2.5 bar, T₁ = 400 K

P₂ = 2.5 bar, T₂ = 300 K

Using the steady flow energy equation:

Q_{1-2}  = \dot{m} c_{p} (T_{2} - T_{1} \\Q_{1-2}  = 2.5 * 1.005 * (300 - 400)\\Q_{1-2}  = -251.25 kW

Therefore, the magnitude of the rate of heat transfer from the air to the ambient, in kw, Q_{1-2} = 251.25 kW

a) For the isentropic process:

Power developed by the turbine is given by the relation \dot{W} = \dot{M}  c_{p} (T_{2} - T_{3})

Isentropic efficiency, \eta_{t} = 80%

P₂ = 2.5 bar, T₂ = 300 K

P₃ = 1 bar, T_{3s} = ? where T_{3s} is the isentropic temperature at 100% efficiency

The isentropic relation is given by:

\frac{T_{3s} }{T_{2} } = (\frac{P_{3} }{P_{2} }) ^{\frac{\gamma - 1}{\gamma} } \\\frac{T_{3s} }{300 } = (\frac{1 }{2.5 }) ^{\frac{1.4 - 1}{1.4 }

T_{3s} = 230.9 K

To get the temperature at 80% efficiency, we will use the relation:

\eta_{t} = \frac{T_{2} - T_{3}  }{T_{2} - T_{3s} } \\0.8= \frac{300 - T_{3}  }{300 - 230.9 }

T₃ = 244.72 K

Power developed by the turbine is given by the relation:

\dot{W} = \dot{M}  c_{p} (T_{2} - T_{3})\\ \dot{W} = 2.5 * 1.005* (300-244.72)\\ \dot{W} = 138.89 kW

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3 years ago
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vesna_86 [32]

Answer:

O False

Explanation:

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2 years ago
The European Space Agency launched a probe called Rosetta in March 2004. In August​ 2014, Rosetta reached its​ destination: a co
Art [367]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The of the lander would be 75.66 pounds

Explanation:

From the question we are given that the

Weight of the Philae on Earth  is = 220 pound-mass

gravity of Mars is = 3.71 m/s^{2}

Weight of the Philae on Mars is = ?

Now Mass is quantity of matter in an object so it is always constant for a particular object

Generally Weight = Mass × Gravity :

                      \frac{Weight Of Earth}{Gravity Of Earth} = \frac{Weight Of Mars}{Gravity Of Mars}

Hence  

         Weight of the Philae on Mars is = \frac{Weight Of Earth *Gravity Of Mars}{Gravity Of Earth}

                                                              =\frac{200 *3.71}{9.81}

                                                              = 75.66 pounds

6 0
3 years ago
A steam turbine receives 8 kg/s of steam at 9 MPa, 650 C and 60 m/s (pressure, temperature and velocity). It discharges liquid-v
adelina 88 [10]

Answer:

The power produced by the turbine is 23309.1856 kW

Explanation:

h₁ = 3755.39

s₁ = 7.0955

s₂ = sf + x₂sfg  =

Interpolating fot the pressure at 3.25 bar gives;

570.935 +(3.25 - 3.2)/(3.3 - 3.2)*(575.500 - 570.935) = 573.2175

2156.92 +(3.25 - 3.2)/(3.3 - 3.2)*(2153.77- 2156.92) = 2155.345

h₂ = 573.2175 + 0.94*2155.345 = 2599.2418 kJ/kg

Power output of the turbine formula =

Q - \dot{W } = \dot{m}\left [ \left (h_{2}-h_{1}  \right )+\dfrac{v_{2}^{2}- v_{1}^{2}}{2} + g(z_{2}-z_{1})\right ]

Which gives;

560 - \dot{W } = 8\left [ \left (2599.2418-3755.39  \right )+\dfrac{15^{2}- 60^{2}}{2} \right ]

= -8*((2599.2418 - 3755.39)+(15^2 - 60^2)/2 ) = -22749.1856

- \dot{W } = -22749.1856 - 560 = -23309.1856 kJ

\dot{W } = 23309.1856 kJ

Power produced by the turbine = Work done per second = 23309.1856 kW.

5 0
3 years ago
What is the modulus of resilience for a tensile test specimen with a nearly linear elastic region if the yield strength is 500MP
ella [17]

Answer:

The modulus of resilience is 166.67 MPa

Explanation:

Modulus of resilience is given by yield strength ÷ strain

Yield strength = 500 MPa

Strain = 0.003

Modulus of resilience = 500 MPa ÷ 0.003 = 166.67 MPa

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3 years ago
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