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Electrical Engineering

Simple fraction form of 0.25

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by itszee asked Jul 18, 2021 0 views
Two persons of masses 55 kg and 65 kg respectively, are at the opposite ends of a boat. The length of the boat is 3.0 m and weighs 100 kg. The 55 kg man walks up to the 65 kg man and sits with him. If the boat is in still water the centre of mass of the system shifts by

Two persons of masses 55 kg and 65 kg respectively, are at the opposite ends of a boat. The length of the boat is 3.0 m and weighs 100 kg. The 55 kg man walks up to the 65 kg man and sits with him. If the boat is in still water the centre of mass of the system shifts by
Answer : (4) zero...

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by anonymous asked Jun 17, 2018 125 views
The characteristic equation of a simple servo system is s2+6s+25=0. Damping factor of the system is:

The characteristic equation of a simple servo system is s2+6s+25=0. Damping factor of the system is: A) 3.2 B) 2.4 C) 1.8 D) 2.9

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by anonymous asked Feb 11, 2021 12 views
Determine the amount of heat required to produce 1Kg of steam at a pressure of 7 bar, at a temperature 29 degree Celsius, under the following . (i) When the steam is wet having dryness fraction 0.87 (ii) When the steam is dry saturated Assume specific heat = 2.35KJ/KgK

Determine the amount of heat required to produce 1Kg of steam at a pressure of 7 bar, at a temperature 29 degree Celsius, under the following . (i) When the steam is wet having dryness fraction 0.87 (ii) When the steam is dry saturated Assume specific heat = 2.35KJ/KgK 
Answer : P 7bar T 29ᵒC x=0.9 m =1kg C p =2.35 KJ/KgK At pressure 7 bar saturation Temp T sat =165 ᵒC h f =697.2 KJ/Kg h fg= 2064.8 KJ/Kg i) Heat required when steam is wet ... ) When the stem is dry saturated hg=hf+hfg =2762 kj Heat actually required = 2762-121.8 = 2640.2 kj ...

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by anonymous asked Aug 3, 2020 407 views
A sample of 35 Kg of dry steam contains 0.7 Kg of water is in suspension, find its dryness fraction.

A sample of 35 Kg of dry steam contains 0.7 Kg of water is in suspension, find its dryness fraction. 
Answer : Mass of dry steam=35 kg Mass water suspension=0.7 kg Weight of wet steam=35+0.7=35.7 kg So,  Dryness fraction X=Actual mass of dry steam/ weight of wet steam  = 35 / (35+0.7)  =0.098039...

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