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SCORPION-xisa [38]
3 years ago
13

A stainless steel ball (=8055 kg/m3, Cp= 480 J/kgK) of diameter D =15 cm is removed from theoven at a uniform temperature of 3

50oC. The ball is then subjectedto the flow of air at 1 atm pressure and 30oC with a velocityof 6 m/s. The surface temperature of the ball eventuallydrops to 250oC. Determine the average convection heat transfercoefficient during this cooling process and estimate howlong this process has taken.
Engineering
1 answer:
aleksandrvk [35]3 years ago
3 0

Answer:

i) 25.04 W/m^2 .k

ii) 23.82 minutes = 1429.2 secs

Explanation:

Given data:

Diameter of steel ball = 15 cm

uniform temperature = 350°C

p = 8055 kg/m^3

Cp = 480 J/kg.k

surface temp of ball drops to  250°C

average surface temperature = ( 350 + 250 ) / 2 = 300°C

<u>i) Determine the average convection heat transfer coefficient during the cooling process</u>

<em>Note : Obtain the properties of air at 1 atm at average film temp of 30°C from the table  " properties of air "  contained in your textbook</em>

average convection heat transfer coefficient = 25.04 W/m^2 .k

<u>ii) Determine how long this process has taken </u>

Time taken by the process = 23.82 minutes = 1429.2 seconds

Δt = Qtotal / Qavg = 683232 / 477.92 = 1429.59 secs

attached below is the detailed solution of the given question

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A heat pump receives heat from a lake that has an average wintertime temperature of 6o C and supplies heat into a house having a
Dafna1 [17]

Answer:

a) \dot W = 1.062\,kW

Explanation:

a) Let consider that heat pump is reversible, so that the Coefficient of Performance is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

COP_{HP} = \frac{298.15\,K}{298.15\,K-279.15\,K}

COP_{HP} = 15.692

The minimum heat received by the house must be equal to the heat lost to keep the average temperature constant. Hence:

\dot Q_{H} = 60000\,\frac{kJ}{h}

The minimum power supplied to the heat pump is:

\dot W = \frac{\dot Q_{H}}{COP}

\dot W = \frac{\left(60000\,\frac{kJ}{h}  \right)\cdot \left(\frac{1\,h}{3600\,s}  \right)}{15.692}

\dot W = 1.062\,kW

5 0
3 years ago
You are in charge of ordering the concrete for a basement wall concrete pour. The wall forms are all set up and ready. The wall
choli [55]

Answer:

189.15cy

Explanation:

To understand this problem we need to understand as well the form.

It is clear that there is four wall, two short and two long.

The two long are \rightarrow 120ft5in+2(10ft)

The two long are \rightarrow 122ft1in=122.08ft

The two shors are \rightarrow 86ft4.5in = 86.375ft

The height and the thickness are 14ft and 0.83ft respectively.

So we only calculate the Quantity of concrete,

Q_c = [(2*122.08)+(2*86-375)]*14*0.833\\Q_c=4864.02ft^3

That in cubic yards is equal to 180.15 (1cy=27ft^3)

Hence, we need order 5% plus that represent with the quantity

Q_{ordered}=1.05*180.15=189.15cy

8 0
3 years ago
A rectangular open box, 25 ft by 10 ft in plan and 12 ft deep weighs 40 tons. Sufficient amount of stones is placed in the box a
dimulka [17.4K]

Answer:

44.95 tonnes

Explanation:

According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces

It is given that empty weight of box = 40 tons

Let the mass of the stones to be placed be = M tonnes

Thus the combined mass of box and stones = (40+M) tonnes..........(i)

Since the box will displace water equal to it's volume V we have volume of box = 25ft*10ft*12ft= 3000ft^{3}

Volume= 84.95m^{3}

Since 1ft^{3} =0.028m^{3}

Now the weight of water displaced = Weight =\rho \times Volumewhererho is density of water = 1000kg/m^{3}

Thus weight of liquid displaced = \frac{84.95X1000}{1000}tonnes=84.95 tonnes..................(ii)

Equating i and ii we get

40 + M = 84.95

thus Mass of stones = 44.95 tonnes

3 0
3 years ago
Radioactive wastes are temporarily stored in a spherical container, the center of which is buried a distance of 10 m below the e
a_sh-v [17]

Answer:

Outside temperature =88.03°C

Explanation:

Conductivity of air-soil from standard table

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To find temperature we need to balance energy

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Now find the value

We know that for sphere

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Given that q=500 W

so

500=\dfrac{2\pi 2\times .6}{1-\dfrac{2}{4\times 10}}(T_1-25)

By solving that equation we get

T_2=88.03°C

So outside temperature =88.03°C

6 0
3 years ago
Which of the following explains the main reason to cut a piece of wood on the outside of the measurement mark?
maks197457 [2]
I think it’s D ?? I’m not completely sure tho
4 0
3 years ago
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