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3 years ago

11. Which of these is NOT true when dealing with refrigerants?

Engineering

1 answer:

Alexus [3.1K]3 years ago

4 0

Answer is an increase in pressure will cause an decrease in the pressure

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A cartridge electrical heater is shaped as a cylinder of length L=200mm and outer diameter D=20 mm. Under normal operating condi

lara31 [8.8K]

Answer:

T(water)=50.32℃

T(air)=3052.6℃

Explanation:

Hello!

To solve this problem we must use the equation that defines the transfer of heat by convection, which consists of the transport of heat through fluids in this case water and air.

The equation is as follows!

$Q=ha(Ts-T\alpha )$

Q = heat

h = heat transfer coefficient

Ts = surface temperature

T = fluid temperature

a = heat transfer area

The surface area of a cylinder is calculated as follows

$a=\pi D(\frac{D}{2} +L)$

Where

D=diameter=20mm=0.02m

L=leght=200mm)0.2m

solving

$a=\pi (0.02)(\frac{0.02}{2} +0.2)=0.01319m^2$

For water

Q=2Kw=2000W

h=5000W/m2K

a=0.01319m^2

Tα=20C

$Q=ha(Ts-T\alpha )$

solving for ts

$Ts=T\alpha +\frac{Q}{ha}$

$Ts=20+\frac{2000}{(0.01319)(5000)} =50.32C$

for air

Q=2Kw=2000W

h=50W/m2K

a=0.01319m^2

Tα=20C

$Ts=20+\frac{2000}{(0.01319)(50)}=3052.6C$

3 0

3 years ago

A cross beam in a highway bridge experiences a stress of 14 ksi due to the dead weight of the bridge structure. When a fully loa

zlopas [31]

Answer:

a) 2.452

b) 1.256

Explanation:

Stress due to dead weight. = 14 Ksi

Stress due to fully loaded tractor-trailer = 45Ksi

ultimate tensile strength of beam = 76 Ksi

yield strength = 50 Ksi

endurance limit = 38 Ksi

Determine the safety factor for an infinite fatigue life

a) If mean stress on fatigue strength is ignored

β = ( 45 - 14 ) / 2

= 15.5 Ksi

hence FOS ( factor of safety ) = endurance limit / β

= 38 / 15.5 = 2.452

b) When mean stress on fatigue strength is considered

β2 = 45 + 14 / 2

= 29.5 Ksi

Ratio = β / β2 = 15.5 / 29.5 = 0.5254

Next step: applying Goodman method

Sa = [ ( 0.5254 * 38 *76 ) / ( 0.5254*76 + 38 ) ]

= 19.47 Ksi

hence the FOS ( factor of safety ) = Sa / β

= 19.47 / 15.5 = 1.256

8 0

2 years ago

Design a PLC ladder logic program to control the operation of a conveyor-storage system using the following sequence: - 1. Progr

Phantasy [73]

Answer:

See explaination

Explanation:

Kindly check attachment for the step by step solution of the given problem.

8 0

3 years ago

A fair die is thrown, What is the probability gained if you are told that 4 will

Semmy [17]

Answer:

1/6

Explanation:

A dice has 6 sides, the probability of 4 appearing is 1/6.

7 0

2 years ago

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2

Svetllana [295]

$79 f t^{3}$ is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

$V_{1}=100\ \mathrm{ft}^{3}$

First has a natural water content of 25% = $\frac{25}{100}$ = 0.25

Shrinkage limit, $w_{1}=12 \%=\frac{12}{100}=0.12$

$G_{s}=2.70$

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

$V \propto[1+e]$

$\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}$ ------> eq 1

$e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}$

The above equation is at $S_{r}=1$ ,

$e_{1}=w_{1} \times G_{s}$

Applying the given values, we get

$e_{1}=0.25 \times 2.70=0.675$

Shrinkage limit is lowest water content

$e_{2}=w_{2} \times G_{s}$

Applying the given values, we get

$e_{2}=0.12 \times 2.70=0.324$

Applying the found values in eq 1, we get

$\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904$

$V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}$

7 0

3 years ago