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Lorico [155]
3 years ago
11

A 5cm diameter copper sphere (of density = 8954 kg/m3, specific heat capacity = 0.3831 kJ/kg K) is initially at a uniform temper

ature of 10oC. The sphere is placed inside an environment having a temperature of 200oC. The surface heat transfer coefficient is h = 10 W/(m2K). Under these conditions, we can assume uniform internal temperature for the sphere. What is the temperature inside the sphere after 10 minutes in this environment?

Engineering
1 answer:
Korolek [52]3 years ago
6 0

Answer:

Temperature inside sphere after 10 minutes = 19924.33K

Explanation:

Detailed explanation and calculation is shown in the image below

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Sort the following alphabets using MergeSort and give required steps. [2 Marks]
OlgaM077 [116]

Answer:

I'm afraid i can't visualise it to you but visit the site below to help you out <3

Explanation:

https://opendsa-server.cs.vt.edu/embed/mergesortAV

4 0
3 years ago
A cylindrical resistor element on a circuit board dissipates 0.6 W of power. The resistor is 1.5 cm long, and has a diameter of
Burka [1]

Answer:

a. 51.84Kj

b. 2808.99 W/m^2

c. 11.75%

Explanation:

Amount of heat this resistor dissipates during a 24-hour period

= amount of power dissipated * time

= 0.6 * 24 = 14.4 Watt hour

(Note 3.6Watt hour = 1Kj )

=14.4*3.6 = 51.84Kj

Heat flux = amount of power dissipated/ surface area

surface area = area of the two circular end  + area of the curve surface

=2*\frac{\pi D^{2} }{4} + \pi DL\\=2*\frac{\pi *(\frac{0.4}{100} )^{2} }{4} + \pi *\frac{0.4}{100} *\frac{1.5}{100}

= 2.136 *10^-4 m^{2}

Heat flux =\frac{0.6}{2.136 * 10^{-4} } = 2808.99 W/m^{2}

fraction of heat dissipated from the top and bottom surface

=\frac{\frac{2*\pi D^{2} }{4} }{\frac{2*\pi D^{2}}{4} + \pi DL } \\\\=\\\frac{\frac{2*\pi *(\frac{0.4}{100} )^{2} }{4} }{\frac{2*\pi *(\frac{0.4}{100}  )^{2} }{4} +\pi *\frac{0.4}{100} *\frac{1.5}{100} } \\\\=\frac{2.51*10^{-5} }{2.136*10^{-4} } \\\\\= 0.1175

=11.75%

8 0
3 years ago
Read 2 more answers
Why is logging done during drilling?
Solnce55 [7]

Answer:

Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible

8 0
2 years ago
Consider the mixing of 0.8 kg/s of hot water at 348 K and 1 kg/s of cool water at 298 K that is generating warm water. Assume no
Drupady [299]

Answer:

T_warm = 47.22 C

Explanation:

Using energy balance for the system:

m_1*h_1 + m_2*h_2 = m_3*h3   ... Eq1

h_i = c_p. T_i   ... Eq 2

m_1 + m_2 = m_3   ... steady flow system (Eq 3)

Substitute Eq 2 and Eq3 in Eq1

m_3 = 0.8 + 1 = 1.8 kg/s

(0.8)*(4.18)*( 348-273) + (1)*(4.18)*( 298-273) = 1.8 * 4.18 *T_3

T_3 = 355.3 / (1.8*4.18) = 47.22 C

3 0
3 years ago
A cannon ball is fired with an arching trajectory such that at the highest point of the trajectory the cannon ball is traveling
ELEN [110]

Answer:

The radius of curvature is 979 meter

Explanation:

We have given velocity of the canon ball v = 98 m/sec

Acceleration due to gravity g=9.81m/sec^2

We know that at highest point of trajectory angular acceleration is equal to acceleration due to gravity

Acceleration due to gravity is given by a_c=\frac{v^2}{r}, here v is velocity and r is radius of curvature

So \frac{98^2}{r}=9.81

r = 979 meter

So the radius of curvature is 979 meter

8 0
3 years ago
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