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9 days ago

13

Radioactive wastes are temporarily stored in a spherical container, the center of which is buried a distance of 10 m below the e

arth's surface. The outside diameter of the container is 2 m, and 500 W of heat are released as a result of radioactive decay. If the soil surface temperature is 20°C, what is the outside surface temperature of the container under steady-state conditions? The thermal conductivity of soil is k = 0.52 W/mK.

1 answer:

mote1985 [204]9 days ago

4 0

Given:

outer radius, R' = 10 m

inner diameter, d = 2 m

inner radius, R = $\frac{d}{2}$ = 1 m

surface temperature, T' = $20^{\circ}C$

Thermal conductivity of soil, K = 0.52 W/mK

Solution:

In order to determine the thermal temperature of the conductor, T, we know the amount of heat, Q is defined as:

Q = $\frac{T - T'}{\frac{R' - R}{4\pi KRR'}}$

500 = $(T - 20)\frac{4\pi \times0.52\times 1\times 10}{10 - 1}$

T = 68.86 + 20 = $88.865^{\circ}C$

Thus, the temperature at the outer surface can be stated as $88.865^{\circ}C$

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An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,

Mrrafil [253]

Answer:

Volume change percentage is 2.60%

Water level increase is 4.138 mm

Explanation:

Provided data

Water volume V = 500 L

Initial temperature T1 = 20°C

Final temperature T2 = 80°C

Diameter of the vat = 2 m

Objective

We aim to determine percentage change in volume and the rise in water level.

Solution

We will apply the bulk modulus equation, which relates the change in pressure to the change in volume.

It can similarly relate to density changes.

Thus,

E = $-\frac{dp}{dV/V}$ ................1

And $-\frac{dV}{V} = \frac{d\rho}{\rho}$ ............2

Here, ρ denotes density. The density at 20°C = 998 kg/m³.

The density at 80°C = 972 kg/m³.

Plugging in these values into equation 2 gives

$-\frac{dV}{V} = \frac{d\rho}{\rho}$

$-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}$

dV = 0.0130 m³

Therefore, the percentage change in volume will be

dV % = $-\frac{dV}{V}$ × 100

dV % = $-\frac{0.0130}{500*10^{-3} }$ × 100

dV % = 2.60 %

Hence, the percentage change in volume is 2.60%

Initial volume v1 = $\frac{\pi }{4} *d^2*l(i)$ ................3

Final volume v2 = $\frac{\pi }{4} *d^2*l(f)$ ................4

From equations 3 and 4, subtract v1 from v2.

v2 - v1 = $\frac{\pi }{4} *d^2*(l(f)-l(i))$

dV = $\frac{\pi }{4} *d^2*dl$

Substituting all values yields

0.0130 = $\frac{\pi }{4} *2^2*dl$

Thus, dl = 0.004138 m.

Consequently, the water level rises by 4.138 mm.

8 0

25 days ago

Write cout statements with stream manipulators that perform the following:

grin007 [219]

Answer:

A)cout<<setw(9)<<fixed<<setprecision(2)<<34.789;

B)cout<<setw(5)<<fixed<<setprecision(3)<<7.0;

C)cout<<fixed<<5.789E12;

D)cout<<left<<setw(7)<<67;

Explanation:

Stream Manipulators are special functions for use with the insertion (<<) and extraction (>>) operators on C++ stream objects, while the 'cout' statement outputs content to the standard output device in C++ programming.

setw: specifies the minimum width of the output field

setprecision: defines the number of decimal places for floating-point value formatting.

fixed: sets the format flag for floating-point streams.

left: left-aligns the output.

A) This statement shows the number 34.789 in a field that provides eight character spaces with two decimal precision. cout<<setw(9)<<fixed<<setprecision(2)<<34.789;

B) Here, the number 7.0 is displayed within six spaces with three decimal precision. cout<<setw(5)<<fixed<<setprecision(3)<<7.0;

C) This command prints 5.789e+12 in fixed-point format. cout<<fixed<<5.789E12;

D) This statement left-aligns the number 67 across a field of six spaces. cout<<left<<setw(7)<<67;

7 0

23 days ago

The molecular weight of a 10g rubber band

Viktor [230]

The response to this query is 1 * 10 g/mole = 10.

8 0

9 days ago

Helium gas is compressed from 90 kPa and 30oC to 450 kPa in a reversible, adiabatic process. Determine the final temperature and

choli [191]

Answer:

T2 ( final temperature ) = 576.9 K

a) 853.4 kJ/kg

b) 1422.3 kJ / kg

Explanation:

given data:

pressure ( P1 ) = 90 kPa

Temperature ( T1 ) = 30°c + 273 = 303 k

P2 = 450 kPa

To determine final temperature in an Isentropic process

$T2 = T1 (\frac{p2}{p1} )^{(k-1)/k}$ ----------- ( 1 )

T2 = 303 $( \frac{450}{90})^{(1.667- 1)/1.667}$ = 576.9K

The work performed in a piston-cylinder device is calculated using the subsequent formula

$w_{in} = c_{v} ( T2 - T1 )$ ------- ( 2 )

where: cv = 3.1156 kJ/kg.k for helium gas

T2 = 576.9K, T1 = 303 K

substituting values into equation 2

$w_{in}$ = 853.4 kJ/kg

the work done in a steady flow compressor is determined using this

$w_{in} = c_{p} ( T2 - T1 )$

where: cp ( constant pressure of helium gas ) = 5.1926 kJ/kg.K

T2 = 576.9 k, T1 = 303 K

plugging values back into equation 3

$w_{in}$ = 1422.3 kJ / kg

4 0

3 days ago

Compute the sum with carry-wraparound (sometimes called the one's complement sum) of the following two numbers. Give answer in 8

grin007 [219]

Response:

00100111

Explanation:

We have been given;

10010110

10010000

Add them following standard binary addition rules

10010110

10010000

-------------

(1)00100110

-------------

ignore the leading (1) because it is a carry.

Increase the result by 1 to achieve a 1's complement sum

00100110 + 1 = 00100111

Final Result: 00100111

3 0

21 day ago