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Paul [167]
3 years ago
11

Consider a very long, cylindrical fin. The temperature of the fin at the tip and base are 25 °C and 50 °C, respectively. The dia

meter of the fin is 3 cm. The thermal conductivity of the fin is 150 W/m·K. The heat transfer coefficient is 123 W/m2·K. Estimate the fin temperature in °C at a distance of 10 cm from the base
Engineering
1 answer:
nekit [7.7K]3 years ago
7 0

Answer:

98°C

Explanation:

Total surface area of cylindrical fin = πr² + 2πrl , r = 0.015m; l= 0.1m; π =22/7

22/7*(0.015)² + 22/7*0.015*0.1 = 7.07 X 10∧-4 + 47.1 X 10∧-4 = (54.17 X 10∧-4)m²

Temperature change, t = (50 - 25)°C = 25°C = 298K

Hence, Temperature =  150 X (54.17 X 10∧-4) X 298/123 = 242.14/124 = 2.00K =

∴ Temperature change = 2.00K

But temperature, T= (373 - 2)K = 371 K

In °C = (371 - 273)K = 98°C         

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Explanation:

Thermodynamics system :

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Generally thermodynamics systems are of three types

1.Closed system(control mass system)

  Only energy transfer take place ,no mass transfer take place.

2.Open system(control volume system)

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3.Isolated system

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At steady state ,property is did not changes with respect to time.

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3 years ago
Two containers initially contain dry air at 40 °C. Water is added to both containers so that the relative humidity is 30% in con
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Answer:

A) Wet bulb temperature of #1 is less than that of #2

Explanation:

This can be gotten from pinpointing the states of the two containers on a psychometric chart.

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3 years ago
What is 7/8 + 1/2 on a ruler
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3 years ago
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Which option demonstrates when most vehicles lose their efficiency?
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3 years ago
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An industrial plant consists of several 60 Hz single-phase motors with low power factor. The plant absorbs 600 kW with a power f
Gelneren [198K]

Answer:

(a) Q=332 kvar and C=5.66 uF

(b) pf=0.90 lagging

Explanation:

Given Data:

P=600kW

V=12.47kV

f=60Hz

pf_{old} =0.75

pf_{new} =0.95

(a) Find the required kVAR rating of a capacitor

\alpha _{old}=cos^{-1}(0.75) =41.41°

\alpha _{new}=cos^{-1}(0.95) =18.19°

The required compensation reactive power can be found by

Q=P(tan(\alpha_{old}) - tan(\alpha_{new}))

Q=600(tan(41.41) - tan(18.19))

Q=332 kvar

The corresponding capacitor value can be found by

C=Q/2\pi fV^{2}

C=332/2*\pi *60*12.47^{2}

C=5.66 uF

(b) calculate the resultant supply power factor

First convert the hp into kW

P_{mech} =250*746=186.5 kW

Find the electrical power (real power) of the motor

P_{elec} =P_{mech}/n

where n is the efficiency of the motor

P_{elec} =186.5/0.80=233.125 kW

The current in the motor is

I_{m} =(P/\*V*pf)

The pf of motor is 0.85 Leading

Note that represents the angle in complex notation (polar form)

I_{m} =(233.125/12.47*0.85)

I_{m}=18.694+11.586j A

Now find the Load current

pf of load is 0.75 lagging (notice the minus sign)

I_{load} =(600/12.47*0.75)

I_{load} =48.115-42.433j A

Now the supply current is the current flowing in the load plus the current flowing in the motor

I_{supply} =I_{m} + I_{load}

I_{supply}= (18.694+11.586)+(48.115-42.433)

I_{supply} =66.809-30.847j A

or in polar form

I_{supply} =73.58°

Which means that the supply current lags the supply voltage by 24.78

therefore, the supply power factor is

pf=cos(24.78)=0.90 lagging

Which makes sense because original power factor was 0.75 then we installed synchronous motor which resulted in improved power factor of 0.90

8 0
3 years ago
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