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3 years ago

Print reading for industry unit 9 review questions

Engineering

2 answers:

Luda [366]3 years ago

7 0

Why do we need to print it out?

RideAnS [48]3 years ago

4 0

Hhhhhhhhhhhhhhh. High

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Two fluids, A and B exchange heat in a counter – current heat exchanger. Fluid A enters at 4200C and has a mass flow rate of 1 k

Volgvan

Answer:

Your question has some missing information below is the missing information

Given that ( specific heat of fluid A = 1 kJ/kg K and specific heat of fluid B = 4 kJ/kg k )

answer : 300 kW , 95°c

Explanation:

Given data:

Fluid A ;

Temperature of Fluid ( Th1 ) = 420° C

mass flow rate (mh) = 1 kg/s

Fluid B :

Temperature ( Tc1) = 20° C

mass flow rate ( mc ) = 1 kg/s

effectiveness of heat exchanger = 75% = 0.75

<u>Determine the heat transfer rate and exit temperature of fluid</u> <u>B</u>

Cph = 1000 J/kgk

Cpc = 4000 J/Kgk

Given that the exit temperatures of both fluids are not given we will apply the NTU will be used to determine the heat transfer rate and exit temperature of fluid B

exit temp of fluid B = 95°C

heat transfer = 300 kW

attached below is a the detailed solution

5 0

3 years ago

Copper spheres of 20-mm diameter are quenched by being dropped into a tank of water that is maintained at 280 K . The spheres ma

Ivenika [448]

Answer:

The height of the water is 1.25 m

Explanation:

copper properties are:

Kc=385 W/mK

D=20x10^-3 m

gc=8960 kg/m^3

Cp=385 J/kg*K

R=10x10^-3 m

Water properties at 280 K

pw=1000 kg/m^3

Kw=0.582

v=0.1247x10^-6 m^2/s

The drag force is:

$F_{D} =\frac{1}{2} Co*p_{w} A*V^{2}$

The bouyancy force is:

$F_{B} =V*p_{w} *g$

The weight is:

$W=V*p_{c} *g$

Laminar flow:

$v_{T} =\frac{p_{c}-p_{w}*g*D^{2} }{18*u} =\frac{(8960-1000)*9.8*(20x10^{-3})^{2} }{18*0.00143} =1213.48 m/s$

Reynold number:

$Re=\frac{1000*1213.48*20x10^{-3} }{0.00143} \\Re>>1$

Not flow region

For Newton flow region:

$v_{T} =1.75\sqrt{(\frac{p_{c}-p_{w} }{p_{w} })gD }=1.75\sqrt{(\frac{8960-1000}{1000} )*9.8*20x10^{-3} } =2.186m/s$

$Re=\frac{1000*2.186*20x10^{-3} }{0.00143} =30573.4$

$Pr=\frac{\frac{u}{p} }{\frac{K}{pC_{p} } } =\frac{u*C_{p} }{k} =\frac{0.0014394198}{0.582} =10.31$

$Nu=2+(0.4Re^{1/2} +0.06Re^{2/3} )Pr^{2/5} (u/us)^{1/4} \\Nu=2+(0.4*30573.4^{1/2}+0.06*30573.4^{2/3} )*10.31^{2/5} *(0.00143/0.00032)^{1/4} \\Nu=476.99$

$Nu=\frac{h*d}{K_{w} } \\h=\frac{476.99*0.582}{20x10^{-3} } =13880.44W/m^{2} K$

$\frac{T-T_{c} }{T_{w}-T_{c} } =e^{-t/T} \\T=\frac{m_{c}C_{p} }{hA_{c} } =\frac{8960*10x10^{-3}*385 }{13880.44*3} =0.828 s$

$e^{-t/0.828} =\frac{320-280}{360-280} \\t=0.573\\heightofthewater=2.186*0.573=1.25m$

8 0

3 years ago

Bài 3: Cho cơ cấu culít (hình 3.5) với các kích thước động lAB = 0,5lAC = 0,1m. Khâu 3 chịu tác dụng của mô men M3 = 500 N. Cơ c

lesya [120]

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3 0

2 years ago

The temperature at the bottom of a reservoir is TL = 280 K and the surface temperature is TH = 295 K. This temperature differenc

Tanzania [10]

a) For the thermal efficiency we have

$\eta_{th} = \frac{Q_{out}}{Q_{in}} = \frac{|W|}{|Q_h|}\\\eta_{th} = \frac{|W|}{|W|+|Q_2|}$

With the previously values we know that

$W=8kW$ and $Q_L = 1440/6kW$ (convert the min to sec)

Replacing the values

$\eta_{th}=\frac{8}{8+1440/6}=\frac{1}{31}\\\eta_{th}\% = 3.225\%$

b) We use the formula of carnot efficiency

$\eta_{th}=1-\frac{T_l}{T_h}\\\eta_{th}\% =(1-\frac{280}{295})*100\\\eta_{th}\%=5.085\%$

**Note that apply the formula of carnot cycle we need to consider that there is no exchange of heat, there is no friction and the reservior are completely insulated

8 0

3 years ago

A tax on the amount of money a person earns in a year is a

irina [24]

Answer: income tax

Explanation:

7 0

2 years ago