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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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A square power screw has a mean diameter of 30 mm and a pitch of 4 mm with single thread. The collar diameter can be assumed to
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Answer:

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Given:

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The torque is:

T=F\frac{d_{m} }{2} (\frac{\pi *u*d_{m}+L }{\pi *d_{m}-uL } )+(u_{c} F+\frac{d_{2} }{2} )=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03+0.004}{\pi *0.03-0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=15.85Nm

ii) The torque to lowering the load is:

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iii)

T=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\  0=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *F*\frac{d_{c}}{2}\\\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *\frac{d_{c}}{2}\\\\\frac{0.03}{2} (\frac{u*\pi *0.03-0.004}{\pi *0.03+u*0.004} )=-0.05*\frac{0.035}{2}

Clearing u:

u = -0.016

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