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Nata [24]
3 years ago
15

A material condition used to indicate that a specified geometric tolerance applies at each increment of size of a feature within

its limits of size is called ___.
Engineering
1 answer:
Stolb23 [73]3 years ago
7 0

Answer: Regardless of feature size (RFS)

Explanation:

It is a pre-selected condition for every geometric tolerance which is the second rule of GD&T and it requires no repairs, it is adopted when size feature has no effect on defined tolerance. RFS implies that GD&T callout is not dependent on size dimension. Least Material Condition (LMC) cancels out this rule.

•Reasons for using RFS.

1. It can not be easily determined.

2. It can be used anytime but can be ignored if it is specified.

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You are an electrician on the job. The electrical blueprint shows that eight 500-W lamps are to be installed on the same circuit
Alchen [17]

Answer:

I = 14.44A

Explanation:

calculating 80% of the circuit breaker current

I = (80/100)20A = 16A

eight 500W lamps are to be installed on the circuit. assume they are connected in parallel

total power = 500 x 8 = 4000W

power = voltage x current

current = power/voltage = 4000W/277V = 14.4A

The current obtained is less than 80% of the circuit breaker.

20A circuit breaker is large enough to carry the load.

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3 years ago
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Answer:

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Explanation:

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4 years ago
A power plant operates on a regenerative vapor power cycle with one open feedwater heater. Steam enters the first turbine stage
ivolga24 [154]

Answer:

a. 46.15%

b. 261.73 kg/s

c. 54.79 kW/K

Explanation:

a. State 1

The parameters given are;

T₁ = 560°C

P₁ = 12 MPa = 120 bar

Therefore;

h₁ = 3507.41 kJ/kg,  s₁ = 6.6864 kJ/(kg·K)

State 2

p₂ = 1 MPa = 10 bar

s₂ = s₁ = 6.6864 kJ/(kg·K)

h₂ = (6.6864 - 6.6426)÷(6.6955 - 6.6426)×(2828.27 - 2803.52) + 2803.52

= (0.0438 ÷ 0.0529) × 24.75 = 2824.01 kJ/kg

State 3

p₃ = 6 kPa = 0.06 bar

s₃ = s₁ = 6.6864 kJ/(kg·K)

sg = 8.3291 kJ/(kg·K)

sf = 0.52087 kJ/(kg·K)

x = s₃/sfg = (6.6864- 0.52087)/(8.3291  - 0.52087) = 0.7896

(h₃ - 151.494)/2415.17 = 0.7896

∴ h₃ = 2058.56 kJ/kg

State 4

Saturated liquid state

p₄ = 0.06 bar= 6000 Pa, h₄ = 151.494 kJ/kg, s₄ = 0.52087 kJ/(kg·K)

State 5

Open feed-water heater

p₅ = p₂ =  1 MPa = 10 bar = 1000000 Pa

s₄ = s₅ = 0.52087 kJ/(kg·K)

h₅ = h₄ + work done by the pump on the saturated liquid

∴ h₅ = h₄ + v₄ × (p₅ - p₄)

h₅ = 151.494 + 0.00100645 × (1000000 - 6000)/1000 = 152.4944113 kJ/kg

Step 6

Saturated liquid state

p₆ = 1 MPa = 10 bar

h₆ = 762.683 kJ/kg

s₆ = 2.1384 kJ/(kg·K)

v₆ = 0.00112723 m³/kg

Step 7

p₇ = p₁ = 12 MPa = 120 bar

s₇ = s₆ = 2.1384 kJ/(kg·K)

h₇ = h₆ + v₆ × (p₇ - p₆)

h₇ = 762.683  + 0.00112723 * (12 - 1) * 1000 = 775.08253 kJ/kg

The fraction of flow extracted at the second stage, y, is given as follows

y = \dfrac{762.683 - 152.4944113 }{2824.01 - 152.4944113 } = 0.2284

The turbine control volume is given as follows;

\dfrac{\dot{W_t}}{\dot{m_{1}}} = \left (h_{1} - h_{2}  \right ) + \left (1 - y  \right )\left (h_{2} - h_{3}  \right )

= (3507.41  - 2824.01) + (1 - 0.22840)*(2824.01 - 2058.56) = 1274.02122 kJ/kg

For the pumps, we have;

\dfrac{\dot{W_p}}{\dot{m_{1}}} = \left (h_{7} - h_{6}  \right ) + \left (1 - y  \right )\left (h_{5} - h_{4}  \right )

= (775.08253 - 762.683) + (1 - 0.22840)*(152.4944113 -  151.494)

= 13.17 kJ/kg

For the working fluid that flows through the steam generator, we have;

\dfrac{\dot{Q_{in}}}{\dot{m_{1}}} = \left (h_{1} - h_{7}  \right )

= 3507.41 - 775.08253 = 2732.32747 kJ/kg

The thermal efficiency, η, is given as follows;

\eta = \dfrac{\dfrac{\dot{W_t}}{\dot{m_{1}}} -\dfrac{\dot{W_p}}{\dot{m_{1}}}}{\dfrac{\dot{Q_{in}}}{\dot{m_{1}}}}

η = (1274.02122 - 13.17)/2732.32747 = 0.4615 which is 46.15%

(762.683 - 152.4944113)/(2824.01 - 152.4944113)

b. The mass flow rate, \dot{m_{1}}, into the first turbine stage is given as follows;

\dot{m_{1}} = \dfrac{\dot{W_{cycle}}}{\dfrac{\dot{W_t}}{\dot{m_{1}}} -\dfrac{\dot{W_p}}{\dot{m_{1}}}}

\dot{m_{1}} = 330 *1000/(1274.02122 - 13.17) = 261.73 kg/s

c. From the entropy rate balance of the steady state form, we have;

\dot{\sigma }_{cv} = \sum_{e}^{}\dot{m}_{e}s_{e} - \sum_{i}^{}\dot{m}_{i}s_{i} = \dot{m}_{6}s_{6} - \dot{m}_{2}s_{2} - \dot{m}_{5}s_{5}

\dot{\sigma }_{cv} = \dot{m}_{6} \left [s_{6} - ys_{2} - (1 - y)s_{5}  \right ]

= 261.73 * (2.1384 - 0.2284*6.6864 - (1 - 0.2284)*0.52087 = 54.79 kW/K

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3 years ago
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Answer:

It looks like... A machine that reads electric pulse and surge... Not sure though.

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