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andrezito [222]
3 years ago
15

For laminar flow of air over a flat plate that has a uniform surface temperature, the curve that most closely describes the vari

ation of the local heat transfer coefficient with position along the plate is
Engineering
1 answer:
Aliun [14]3 years ago
3 0

This question is incomplete, the missing diagram is uploaded along this answer below;

Answer:

from the diagram, the curve that most closely describes the variation of the local heat transfer coefficient with position along the plate is Option D

Explanation:

Given the data in the question;

We write the expression for the local Nusselt number for Laminar flow over the flat plate;

Nu = C(Re_x)^{0.5 (Pr)^{1/3

Nu = C(\frac{Vx}{v})^{0.5} (Pr)^{1/3

\frac{h_xx}{k} = C(\frac{V}{v})^{0.5}  (Pr)^{1/3  (x)^{0.5

h_x = \frac{1}{x^{1/2}}

Next we write down the expression for the local heat flux from the plate with  uniform surface temperature;

q = h_xA( T_s - T∞ )

q ∝ h_x

∴

q ∝  \frac{1}{x^{1/2}}

The local heat flux decreases with the position as it is inversely proportional to the square root of the position from the leading edge and it will not be zero at the end of the plate.

Therefore, from the diagram, the curve that most closely describes the variation of the local heat transfer coefficient with position along the plate is Option D

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A rigid 10-L vessel initially contains a mixture of liquid and vapor water at 100 °C, with a quality factor of 0.123. The mixtur
masya89 [10]

Answer:

Q_{in} = 46.454\,kJ

Explanation:

The vessel is modelled after the First Law of Thermodynamics. Let suppose the inexistence of mass interaction at boundary between vessel and surroundings, changes in potential and kinectic energy are negligible and vessel is a rigid recipient.

Q_{in} = U_{2} - U_{1}

Properties of water at initial and final state are:

State 1 - (Liquid-Vapor Mixture)

P = 101.42\,kPa

T = 100\,^{\textdegree}C

\nu = 0.2066\,\frac{m^{3}}{kg}

u = 675.761\,\frac{kJ}{kg}

x = 0.123

State 2 - (Liquid-Vapor Mixture)

P = 476.16\,kPa

T = 150\,^{\textdegree}C

\nu = 0.2066\,\frac{m^{3}}{kg}

u = 1643.545\,\frac{kJ}{kg}

x = 0.525

The mass stored in the vessel is:

m = \frac{V}{\nu}

m = \frac{10\times 10^{-3}\,m^{3}}{0.2066\,\frac{m^{3}}{kg} }

m = 0.048\,kg

The heat transfer require to the process is:

Q_{in} = m\cdot (u_{2}-u_{1})

Q_{in} = (0.048\,kg)\cdot (1643.545\,\frac{kJ}{kg} - 675.761\,\frac{kJ}{kg} )

Q_{in} = 46.454\,kJ

3 0
3 years ago
The rolling process is governed by the frictional force between the rollers and the workpiece. The frictional force at the entra
adell [148]

Answer:

b)false

Explanation:

Rolling is a process in which work piece passes through rolls to produce desired out put of the work piece.Rolling  is a metal forming process.

We know that friction force is responsible for motion of work piece between rolls.If friction force is so small at the entrance side then work piece will not enter in the forming zone and forming process will not occurs.So the friction force should be high at the entrance side and low at the exit side.

So given statement is wrong.

3 0
3 years ago
Consider two different types of motors. Motor A has a characteristic life of 4100 hours (based on a MTTF of 4650 hours) and a sh
Daniel [21]

Answer:B

Explanation:

Given

For motor A

Characteristic life(r)=4100 hr

MTTF=4650 hrs

shape factor(B )=0.8

For motor B

Characteristic life(r)=336 hr

MTTF=300 hr

Shape Factor (B)=3

Reliability for 100 hours

R_a=e^{-\left ( \frac{T-r}{n}\right )B}

R_a=e^{-\left ( \frac{4650-4100}{100}\right )0.8}

R_a=e^{-4.4}=0.01227

For B

R_b=e^{-\left ( \frac{300-336}{100}\right )3}

R_b=e^{1.08}=2.944

B is better for 100 hours

(b)For 750 hours

R_a=e^{-0.5866}=0.55621

R_b=e^{0.144}=1.154

So here B is more Reliable.

3 0
3 years ago
Explain the concept of energy conversion as applied to the generation of electricity also known as electrical energy​
Allisa [31]

Answer:

Electrical energy is energy derived as a result of movement of electrons. When used loosely, electrical energy refers to energy that has been converted from electric potential energy. ... Once converted from potential energy, electrical energy can always be called another type of energy (heat, light, motion, etc.)

Explanation:

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5 0
2 years ago
Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb
guajiro [1.7K]

Answer:

\eta_{turbine} = 0.603 = 60.3\%

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in  vapor state:

h₂ = h_{g\ at\ 125KPa} = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than s_g and greater than s_f at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88

Now, we will find h_{2s}(enthalpy at the outlet for the isentropic process):

h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg

Now, the isentropic efficiency of the turbine can be given as follows:

\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%

3 0
3 years ago
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