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3 years ago

True False. First angle projection type used in United states.

Engineering

1 answer:

Sloan [31]3 years ago

5 0

Answer:

FALSE.

Explanation:

the correct answer is FALSE.

Projection is the process of representing the 3 D object on the flat surface.

there are four ways of representing the projection

1) First angle projection

2) second angle projection

3) third angle projection

4) fourth angle projection.

Generally, people prefer First and third angle projection because there is no overlapping of the projection take place.

In USA people uses the third angle of projection.

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A 15 ft high vertical wall retains an overconsolidated soil where OCR-1.5, c'-: O, ф , --33°, and 1 1 5.0 lb/ft3.

Tamiku [17]

Answer:

magnitude of thrust uis 11061.65 lb/ft

location is 5 ft from bottom

Explanation:

Given data:

Height of vertical wall is 15 ft

OCR is 1.5

$\phi = 33^o$

saturated uit weight $\gamma_{sat} = 115.0 lb/ft^3$

coeeficent of earth pressure $K_o$

$K_o = 1 -sin \phi$

= 1 - sin 33 = 0.455

for over consolidate

$K_{con} = K_o \times OCR$

$= 0.455 \times 1.5 = 0.683$

Pressure at bottom of wall is

$P =K_{con} \times (\gamma_{sat} - \gamma_{w}) + \gamma_w \times H$

$= 0.683 \times (115 - 62.4) \times 15 + 62.4 \times 15$

P = 1474.88 lb/ft^3

Magnitude pf thrust is

$F= \frac{1}{2} PH$

$=\frac{1}{2} 1474.88\times 15 = 11061.65 lb/ft$

the location must H/3 from bottom so

$x = \frac{15}{3} = 5 ft$

5 0

4 years ago

A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor-compression refrigeration cycle except

malfutka [58]

Answer:

(a) 0.0064 kg/s

(b) 800 KPa

Explanation:

The ideal vapor compression cycle consists of following processes:

Process 1-2 Isentropic compression in a compressor

Process 2-3 Constant-pressure heat rejection in a condenser

Process 3-4 Throttling in an expansion device

Process 4-1 Constant-pressure heat absorption in an evaporator

For state 4 (while entering compressor):

x₄ = 34% = 0.34

P₄ = 120 KPa

from saturated table:

h₄ = hf + x hfg = 22.4 KJ/kg + (0.34)(214.52 KJ/kg)

h₄ = 95.34 KJ/kg

For State 1 (Entering Compressor):

h₁ = hg at 120 KPa

h₁ = 236.99 KJ/kg

s₁ = sg at 120 KPa = 0.94789 KJ/kg.k

For State 3 (Entering Expansion Valve)

Since 3 - 4 is an isenthalpic process.

Therefore,

h₃ = h₄ = 95.34 KJ/kg

Since this state lies at liquid side of saturation line, therefore, h₃ must be hf. Hence from saturation table we find the pressure by interpolation.

P₃ = 800 KPa

For State 2 (Leaving Compressor)

Since, process 2-3 is at constant pressure. Therefore,

P₂ = P₃ = 800 KPa

T₂ = 70°C (given)

Saturation temperature at 800 KPa is 31.31°C, which is less than T₂. Thus, this is super heated state. From super heated property table:

h₂ = 306.9 KJ/kg

(a)

Compressor Power = m(h₂ - h₁)

where,

m = mass flow rate of refrigerant.

m = Compressor Power/(h₂ - h₁)

m = (0.450 KJ/s)/(306.9 KJ/kg - 236.99 KJ/kg)

m = 0.0064 kg/s

(b)

Condenser Pressure = P₂ = P₃ = 800 KPa

(c)

The COP of ideal vapor compression cycle is given as:

COP = (h₁ - h₄)/(h₂ - h₁)

COP = (236.99 - 95.34)/(306.9 - 236.99)

COP = 2.03

The Ph diagram is attached

7 0

4 years ago

engineering controls are devices such as self sheathing needles and sharps containers to block or eliminate the sharp risk. true

Paladinen [302]

Answer:

yes

Explanation:

The 1991 standard states, "engineering and work practice controls shall be used to eliminate or minimize employee exposure." The revision defines Engineering Controls as "controls (e.g., sharps disposal containers, self-sheathing needles, safer medical devices, such as sharps with engineered sharps injury protections ...

6 0

3 years ago

liq [111]

Answer:

Explanation:

3 0

4 years ago

A Carnot refrigeration cycle absorbs heat at -12 °C and rejects it at 40 °C. a)-Calculate the coefficient of performance of this

tresset_1 [31]

Answer:

a)COP=5.01

b) $W_{in}=2.998$ KW