Who is responsible for conducting a hazard assessment?

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

White raven [17]

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

8 0

2 years ago

*6–24. The beam is used to support a dead load of 400 lb>ft, a live load of 2 k>ft, and a concentrated live load of 8 k. D

lisabon 2012 [21]

Answer:

(a) maximum positive reaction at A = 64.0 k

(b) maximum positive shear at A = 32.0 k

(c) maximum negative moment at C = -540 k·ft

Explanation:

Given;

dead load Gk = 400 lb/ft

live load Qk = 2 k/ft

concentrated live load Pk =8 k

(a) from the influence line for vertical reaction at A, the maximum positive reaction is

$A_{ymax}$ = 2*(8) +(1/2(20 - 0)* (2))*(2 + 0.4) = 64 k

See attachment for the calculations of (b) & (c) including the influence line

3 0

2 years ago

Introduce JTA and JT

Vinil7 [7]

JT2 oaoaosnforeneomdocdmsnsksmsmsodnfnfj

7 0

2 years ago

A binary geothermal power plant uses geothermal water at 160°C as the heat source. The cycle operates on the simple Rankine cycl

bogdanovich [222]

A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined

Assumptions :

1. Steady operating conditions exist.

2. Kinetic and potential energy changes are negligible.

Properties: The specific heat of geothermal water ( $c_{geo}$ [) is taken to be 4.18 kJ/kg.ºC.

Analysis (a) We need properties of isobutane, we can obtain the properties from EES.

a. Turbine

P $P_{3} = 3.25mPa = (3.25*1000) kPa\\= 3250kPa\\from the EES TABLE\\h_{3} = 761.54 kJ/kg\\s_{3} = 2.5457 kJ/kg\\P_{4} = 410kPa\\\\s_{4} = s_{3} \\h_{4s} = 470.40kJ/kg\\\\T_{4} = 179.5^{0} C\\\\h_{4} = 689.74 kJ/KG\\\\ The isentropic efficiency, n_{T} = \frac{h_{3}-h_{4} }{h_{3}- h_{4s} }$

= $=\frac{761.54-689.74}{761.54-670.40} \\=\frac{71.8}{91.14} \\=0.788$

b. Pump

$h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} = \frac{v_{1}(P_{2}-P_{1}) }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\ = 273.01+5.81\\ = 278.82 kJ/kg\\\\w_{T,out} = m^{.} (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\$

$W^{.} _ {P,in} = m^{.} (h_{2} -h_{1}) \\=m^{.} w_{p,in \\=305.6(5.81)\\\\=1,777kW\\W^{.} _{net} = W^{.} _{T, out} - W^{.} _{P,in} \\= 21,942-1,777\\=20,166 kW\\\\HEAT EXCHANGER\\\\Q_{in} = m^{.} _{geo} c_{geo} (T_{in-T_{out} } )\\=555.9(4.18)(160-90)\\=162.656kW\\$

c. $The thermal efficiency of the cycle n_{th} =\frac{W^{.} _{net} }{Q^{._{in} } } \\\\= \frac{20,166}{162,656} \\=0.124\\=12.4%$

7 0

3 years ago

Read 2 more answers

Type the correct answer in the box. Spell all words correctly. Mike is constructing a project in which he uses a motor. How can

tankabanditka [31]

If it is. DC, direct current reverse the polarity of power leads on the motor.

If it is a 3 phase ac alternating current, reverse any of the two of three leads.

Disconnect power before attempting.

6 0

2 years ago