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Xelga [282]
3 years ago
6

The exhaust steam from a power station turbine is condensed in a condenser operating at 0.0738 bar(abs). The surface of the heat

transfer surface is held at 20°C. What percentage change does the inclusion of the sensible heat correction term make to the estimated heat transfer condensing film coefficient?
Engineering
1 answer:
lozanna [386]3 years ago
3 0

Answer:

Percentage change 5.75 %.

Explanation:Given ;

Given

 Pressure of condenser =0.0738 bar

Surface temperature=20°C

Now from steam table

Properties of steam at 0.0738 bar  

Saturation temperature corresponding to saturation pressure =40°C      

 h_f= 167.5\frac{KJ}{Kg},h_g= 2573.5\frac{KJ}{Kg}

So Δh=2573.5-167.5=2406 KJ/kg

Enthalpy of condensation=2406 KJ/kg

So total heat=Sensible heat of liquid+Enthalpy of condensation

Total\ heat\ =C_p\Delta T+\Delta h

Total heat =4.2(40-20)+2406

Total heat=2,544 KJ/kg

Now film coefficient before inclusion of sensible heat

  h_1=\dfrac{\Delta h}{\Delta T}

  h_1=\dfrac{2406}{20}

h_1=120.3\frac{KJ}{kg-m^2K}

Now film coefficient after inclusion of sensible heat

 h_2=\dfrac{total\ heat}{\Delta T}

 h_2=\dfrac{2,544}{20}

h_2=127.2\frac{KJ}{kg-m^2K}

So\ Percentage\ change=\dfrac{h_2-h_1}{h_1}\times 100

             =\dfrac{127.2-120.3}{120.3}\times 100

                   =5.75 %

So Percentage change 5.75 %.

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Answer:

Answer: (a) = 3.8187m/s, (b) =24.0858m/s (c) = = 3220.071m/s

   

Explanation:

du/u² = dt = ∫du/2.3183 = ∫dt

0.4313 u = t + c

(a) t = 0, u= 15m/s, c = 0.647

u = t+c/0.4313 = t + 0.647/0.4313

(a) when t= 1   u = 1+ 0.647/0.4313 = 3.8187m/s

(b) when t= 10   u = 10 + 0.647/0.4313 = 24.0858m/s

(c)when t= 1000  u = 1000 + 0.647/0.4313 = 3220.071m/s

5 0
3 years ago
If the atomic radius of copper is 0.128 nm, calculate the volume of its unit cell in cubic meters.
Alex_Xolod [135]

Answer:

Volume of face centered cubic cell=4.74531*10^{-29} m^3

Explanation:

Consider the face centered cubic cell:

1 atom at each corner of cube.

1 atom at center of each face.

Consider the one face (ABCD) as shown in attachment for calculation:

Length of the all sides of face centered cubic cell is L.

Volume of face centered cubic cell= L^3

Now Consider the figure shown in attachment:

According to Pythagoras theorem on ΔADC.

L^{2}+L^2=(4a)^2     (a is the atomic radius)

L=\frac{4a}{\sqrt{2}} (Put in the formula of Volume)

Volume of face centered cubic cell= L^3

Volume of face centered cubic cell= (\frac{4a}{\sqrt{2}})^3

Volume of face centered cubic cell= (\frac{4(0.128*10^{-9}}{\sqrt{2}})^3

Volume of face centered cubic cell=4.74531*10^{-29} m^3

3 0
4 years ago
Steam flows steadily through a turbine at a rate of 420 kg/min. The enthalpy of the steam decreases by 600 kJ/kg as it flows ste
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Answer:

the rate of heat loss from the steam turbine  is Q = 200 kW

Explanation:

From the first law of thermodynamics applied to open systems

Q-W₀ = F*(ΔH + ΔK + ΔV)

where

Q= heat loss

W₀= power generated by the turbine

F= mass flow

ΔH = enthalpy change

ΔK = kinetic energy change

ΔV = potencial energy change

If we neglect the changes in potential and kinetic energy compared with the change in enthalpy , then

Q-W₀ = F*ΔH

Q =  F*ΔH+ W₀

replacing values

Q =  F*ΔH+ W₀ = 420 kg/min * (-600 kJ/kg) * 1 min/60 s * 1 MW/1000 kW + 4 MW = -0.2 MW = -200 kW (negative sign comes from outflow of energy)

4 0
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The resistivity of mercury drops suddenly to zero at a critical temperature, maki mercury a superconductor below that temperatur
seropon [69]

Resistance zero meaning superconductor, so True.

4 0
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A tension test is carried out on an Al alloy specimen which has an original diameter of 0.505 in and an original gauge length of
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Detailed solution is given in attached image

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