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2 years ago

Discuss the differences between conduction and convective heat transfer.

Engineering

1 answer:

FrozenT [24]2 years ago

8 0

Answer:

Basically there are two principal differences between the convection and conduction heat transfer

Explanation:

The conduction heat transfer is referred to the transfer between two solids due a temperature difference, while for, the convective heat transfer is referred to the transfer between a fluid (liquid or gas) and a solid. Also, they used different coefficients for its calculation.

We can include on the explanation that conduction thermal transfer is due to temperature difference, while convection thermal transfer is due to density difference.

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A steady stream (1000 kg/hr) of air flows through a compressor, entering at (300 K, 0.1 MPa) and leaving at (425 K, 1 MPa). The

AleksandrR [38]

Answer:

The work furnished by the compressor is $69.77kJ/s$

The minimum work required for the state to change is $55.26kW$

Explanation:

The explanation to these solution is on the first, second , third and fourth uploaded image respectively

8 0

2 years ago

Water vapor at 5 bar, 320°C enters a turbine operating at steady state with a volumetric flow rate of 0.65 m3/s and expands adia

harina [27]

Answer:

Power = 371.28 kW

Explanation:

Initial pressure, P1 = 5 bar

Final pressure, P2 = 1 bar

Initial temperature, T1 = 320°C

Final temperature, T2 = 160°C

Volume flow rate, V = 0.65m³/s

From steam tables at state 1,

h1 = 3105.6 kJ/kg, s1 = 7.5308 kJ/kgK

v1 = 0.5416 m³/kg

Mass flow rate, m = V/v1

m = 1.2 kg/s

From steam tables, at state 2

h2 = 2796.2 kJ/kg, s2 = 7.6597 kJ/kgK

Power developed, P = m(h1 - h2)

P = 1.2(3105.6-2796.2)

P = 371.28 kW

8 0

3 years ago

1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

a) The total number of users that can be accomodated in the system is:

$n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )$

$n = 1000\,users$

b) The length of the side of each cell is:

$l = \sqrt{1\,km^{2}}$

$l = 1\,km$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

c) The total number of users that can be accomodated in the system is:

$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

The length of each side of the cell is:

$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

8 0

3 years ago

he circular stream of water from a faucet is observed to taper from a diameter of 21 mm to 12 mm in a distance of 52 cm. Determi

Gemiola [76]

Answer: $1.52$ × $10^{-3}$ $\frac{m^{3} }{s}$

Explanation:

Please kindly find the attached document for the answer.

5 0

2 years ago

A heat engine that rejects waste heat to a sink at 520 R has a thermal efficiency of 35 percent and a second- law efficiency of

xeze [42]

Answer:

The source temperature is 1248 R.

Explanation:

Second law efficiency of the engine is the ratio of actual efficiency to the maximum possible efficiency that is reversible efficiency.

Given:

Temperature of the heat sink is 520 R.

Second law efficiency is 60%.

Actual thermal efficiency is 35%.

Calculation:

Step1

Reversible efficiency is calculated as follows:

$\eta_{II}=\frac{\eta_{a}}{\eta_{rev}}$

$0.6=\frac{0.35}{\eta_{rev}}$

$\eta_{rev}=0.5834$

Step2

Source temperature is calculated as follows:

$\eta_{rev}=1-\frac{T_{L}}{T}$

$\eta_{rev}=1-\frac{520}{T}$

$0.5834=1-\frac{520}{T}$

T = 1248 R.

The heat engine is shown below:

Thus, the source temperature is 1248 R.

6 0

3 years ago