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Makovka662 [10]
3 years ago
12

Annie has collected several items from around her house. She is using these objects to investigate which objects are attracted t

o magnets and which ones are not. Which item should Annie expect to be attracted to her magnet?
A) ice cube

B) metal screw

C) leather glove

D) notebook paper
Engineering
1 answer:
Margarita [4]3 years ago
5 0
The metal screw would be attracted to her magnet since its metal and the others don’t attract to magnets.
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Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value
masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} }  )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

q_{n} =uK(\frac{g(\rho_{L}-\rho _{v})     }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr  } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

8 0
3 years ago
You are to design two CONCEPTUALLY different synchronous state machines (Mealy and Moore) that perform the task described below.
allochka39001 [22]
Answer:








Explanation:









I hope this helps!
3 0
3 years ago
Why do I eat Takis please
sammy [17]

Answer:

because they taste good :)

7 0
10 months ago
Read 2 more answers
A pump operating at steady state receives liquid water at 20°C, 100 kPa with a mass flow rate of 53 kg/min. The pressure of the
VARVARA [1.3K]

Answer:

Input Power = 6.341 KW

Explanation:

First, we need to calculate enthalpy of the water at inlet and exit state.

At inlet, water is at 20° C and 100 KPa. Under these conditions from saturated water table:

Since the water is in compresses liquid state and the data is not available in compressed liquid chart. Therefore, we use approximation:

h₁ = hf at 20° C = 83.915 KJ/kg

s₁ = sf at 20° C = 0.2965 KJ/kg.k

At the exit state,

P₂ = 5 M Pa

s₂ = s₁ = 0.2965 K J / kg.k    (Isentropic Process)

Since Sg at 5 M Pa is greater than s₂. Therefore, water is in compresses liquid state. Therefore, from compressed liquid property table:

h₂ = 88.94 KJ/kg

Now, the total work done by the pump can be calculated as:

Pump Work = W = (Mass Flow Rate)(h₂ - h₁)

W = (53 kg/min)(1 min/60 sec)(88.94 KJ/kg - 83.915 KJ/kg)

W = 4.438 KW

The efficiency of pump is given as:

efficiency = η = Pump Work/Input Power

Input Power = W/η

Input Power = 4.438 KW/0.7

<u>Input Power = 6.341 KW</u>

5 0
3 years ago
In a reversing 2-high mill, a series of cold rolling process is used to reduce the thickness of a plate from 45mm down to 20mm.
podryga [215]

Answer:

Explanation:

Given

Initial Thickness=45 mm

Final thickness=20 mm

Roll diameter=600 mm

Radius(R)=300 mm

coefficient of friction between rolls and strip (\nu)=0.15

maximum draft(d_{max})=\nu ^2R

=0.15^2\times 300=6.75 mm

Minimum no of passes=\frac{45-20}{6.75}=3.70\approx 4

(b)draft per each pass

d=\frac{Initial\ Thickness-Final\ Thickness}{min.\ no.\ of\ passes}

d=\frac{45-20}{4}=6.25 mm

5 0
3 years ago
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