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ss7ja [257]
3 years ago
5

A spherical steel container 3 feet in diameter is buried in a land fill. The container is filled with a chemical that keeps the

outer surface of the container at 100°F, whereas the earth's surface is at 50°F. Determine the heat transfer from the container if it is buried under 3 feet of earth.
Engineering
1 answer:
zmey [24]3 years ago
5 0

Answer:

Q = 378.247 Bt/hr

Explanation:

given data:

diameter of container = 3 m

so r =  1.5 m

T1 = 50°C

T2 = 100°C

depth y = 3 ft

Heat transfer is given as Q

Q = SK\Delta T

Where

S =  Shape factor for the object

S = \frac{4\pi r}{1-\frac{r}{2y}}

S = \frac{4\pi *1.5}{1-\frac{1.5}{2*3}}

S = 25.132 ft

Q = SK\Delta T

Q = 25.132*0.301 *(100-50)

Q = 378.247 Bt/hr

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meriva

Answer:

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7 0
2 years ago
If a motorist moves with a speed of 30 km/hr, and covers the distance from place A to place B
Sergio039 [100]

Answer:

105 km

Explanation:

The motorist was going 30 km/hr, and it took 3 hours 30 minutes. That's 3.5 hours. 3.5×30=105

5 0
3 years ago
A two-bus power system is interconnected by one transmission line. Bus 1 is a generator bus with specified terminal voltage magn
mixas84 [53]

Answer:

a) | v2 | ,  β2   ( load, bus voltage at bus 2 )

  p1 ,  q1 ( slack, bus power at bus 1 )

b) q2 , β2  

  p1 and q1 ( slack, bus power at bus 1 )

Explanation:

Attached below is a schematic representation of the solution

<u>a) Identify the variables in the solution vector assume Bus 2 is a load bus</u>

The specified parameters are ; P2 and q2

while | v2 | and β2 are not specified

given that bus 2 is a load bus, bus 1 is a slack bus with ; | v1 |  and β1 been specified while p1 and q1 are not specified

<em>Hence the variables in the solution </em>

<em>= | v2 | ,  β2   ( load, bus voltage at bus 2 )</em>

<em>   p1 ,  q1 ( slack, bus power at bus 1 ) </em>

<u>b) Identify the variables in the solution vector ( assume Bus 2 is a PV bus )</u>

specified at Bus 2 are ; | p2 | , | v2 |

unspecified : q2 , β2

Bus 1 ( still a slack bus )

specified parameter : | v1 |  and β1

unspecified : p1 and q1

<em>Hence the variables in the solution </em>

<em>= q2 , β2  </em>

<em>   p1 and q1 ( slack, bus power at bus 1 ) </em>

7 0
3 years ago
The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod
Vsevolod [243]

Answer:

Explanation:

From the information given:

E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa  \\ \\ d = 25 \ mm \  \\ \\ D = 45 \ mm \ \\ \\ L   = 761 \ mm  \\ \\ P = -88 kN

The total load is distributed across both the rod and tube:

P = P_1+P_2 --- (1)

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

\delta_1=\delta_2

\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}

\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}

P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})

P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}

P_2 = 6.6212 \ P_1

Replace P_2 into equation (1)

P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\  -88 = 7.6212 \ P_1  \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\  P_1 = -11.547 \ kN

Finally, to determine the normal stress in aluminum rod:

\sigma _1 = \dfrac{P_1}{A_1} \\ \\  \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}

\sigma_1 = - 23.523 \ MPa}

Thus, the normal stress = 23.523 MPa in compression.

8 0
3 years ago
A gas turbine receives a mixture having the following molar analysis: 10% CO2, 19% H2O, 71% N2 at 720 K, 0.35 MPa and a volumetr
Sliva [168]

Answer:

2074.2 KW

Explanation:

<u>Determine power developed at steady state </u>

First step : Determine mass flow rate  ( m )

m / Mmax = ( AV )₁ P₁ / RT₁   -------------------- ( 1 )

<em> where : ( AV )₁ = 8.2 kg/s,  P₁ = 0.35 * 10^6 N/m^2,   R = 8.314 N.M / kmol , </em>

<em>  T₁  = 720 K . </em>

insert values into equation 1

m  = 0.1871  kmol/s  ( mix )

Next : calculate power developed at steady state ( using ideal gas tables to get the h values of the gases )

W( power developed at steady state )

W = m [ Yco2 ( h1 - h2 )co2

Attached below is the remaining  part of the detailed solution

4 0
3 years ago
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