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Efficiency

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Rural Technology

Efficiency is a measure of ‘how effectively the resources are used?’. Ratio of output to input of any machine is called efficiency. It is usually calculated in percentage therefore

Calculate efficiency of some of the daily applications

1) A pump set is consuming 5 hp energy while lifting water. It needs 1 lit of diesel in one hour. Calculate efficiency of the pump set. (1 lit diesel contains 42 MJ energy i.e. 42 ´ 106 J) Step I: Power input = Input to pump set is 1 lit. diesel = 42 ´ 10 6 Joules Step II: Output of pump set = 5 hp Lets convert hp to joule as follows: 1 hp = 746 w Therefore, 5 hp = 746 ´ 5 = 3730w Watt is joule / second This means energy output per second is = 3730 J/s The pump was running for 1 hour.

1 hour = 60 min. = 3600 second Energy output in one hour = 3730 ´ 3600 = 13428000 Joule = 13.42 ´ 106 J Step III: The efficiency of the pump = (Output / Input) ´ 100 This shows efficiency of pump set is 31.95% 2) One Lit. milk was heated by 700 C. The weight of milk after heating was 1025g. The 30g of kerosene was used to heat this milk. Find out the efficiency of the system and how much energy spent in heating the pot and surrounding air. {Data: Milk lactometer –30, Energy needed to convert 1gm of water into vapour =540 Cal, kerosene has energy ~ 10000 cal/gm} Step I: Power Input = Total 30g kerosene was used. It is given that Kerosene has energy = 10000 cal/gm. Energy input from kerosene = 30g ´ 10,000 = 300,000 cal. Step II: Calculation of Energy output Write down the energy, mass balance Milk has a lactometer reading of 30. This means density of milk is 1.030 or 1 lit of milk weighs 1.030kg. 1025g of hot milk is obtained. Therefore 5 g of water became vapour. 1lit (1.030 g) of milk à heated à 1025g of hot milk + 5 g of vapour The energy supplied by kerosene is used to increase temperature of milk and to convert water into water vapour. a) 1 cal of energy is needed for 10C raise in temperature. Energy needed to heat the milk = mass in g ´ rise in temperature Here temperature of 1.030Kg or 1030g milk is raised by 700C. Therefore calories needed to raise the temperature of milk = Mass ´ temperature of milk = 1030 ´ 70 = 72100 calories b) 540cal needed to convert 1gm of water into vapor. Therefore, to convert 5 g of water into vapour. = 5 ´ 540 = 2700 calories

Total energy gone heating pots and surrounding air = Energy supplied – energy utilized for heating milk and water vapour. = 300,000 – 74800 = 225200 cal Therefore, 225200 cal. of energy must have been lost in heating the pot, air etc. % Efficiency = (Output / Input) ´ 100 = = (Useful energy / Total energy supplied) ´100 = 74800 / 225200 ´ 100 = 0.33 ´ 100 % Efficiency = 0.33 ´ 100 = 33 % This means only 33% energy supplied got utilized while heating the milk. 7.6 What you have learnt In this lesson, We know the Mass & Energy and understand their relationship. We learned that mass contains tremendous energy but it cannot be easily obtained. The conversion of mass to energy can be taken place in nuclear reactors. You also learnt to write mass and energy balance. You also learn to calculate efficiency of daily appliances like diesel engine and stove. 7.7 Terminal Questions 1) A diesel engine does 15MJ of work in 1 hour. It consumes diesel 1 lit diesel. (1 lit diesel has 42MJ energy). Calculate efficiency of the pump. 2) To prepare a rice, 9000 calories are required. It consumes 40gm of kerosene. Find out the efficiency of the system. (1 gm of kerosene contains 10000 calories) 7.8 ANSWER TO INTEXT QUESTIONS

7.1 (i) 10g (ii) c2

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