Carnot engine working between ice point and steam point it desired to increase it efficiency by 20% keeping temperature of source constant what is the change in temp of sink
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Well, efficiency can easily be calculated for the carnot engine by formula
Efficiency= (input/output)* 100 which will be equal to ((Q2- Q1)*Q1)*100
First let suppose Q2 and Q1 have the relationship with respect to T2 and T1
So this mean Q2/Q1=T2//T1
That's also means Q1 is equal to T1
so the (Q2-Q1)/Q1 will be equal to (T2 - T1)/T1*100
Put the value of temperature as the T1 =373K (steam point ) T2= 273K (ice point)
hence the efficiency = (T1-T2)/T1 * 100 =(373 - 273)/373*100 =26.80% answer
Efficiency= (input/output)* 100 which will be equal to ((Q2- Q1)*Q1)*100
First let suppose Q2 and Q1 have the relationship with respect to T2 and T1
So this mean Q2/Q1=T2//T1
That's also means Q1 is equal to T1
so the (Q2-Q1)/Q1 will be equal to (T2 - T1)/T1*100
Put the value of temperature as the T1 =373K (steam point ) T2= 273K (ice point)
hence the efficiency = (T1-T2)/T1 * 100 =(373 - 273)/373*100 =26.80% answer
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