Thermodynamics

Consider a heat engine working with a hot reservoir of the furnace gases at 2100 °C, when the cooling water is available at -35.7 °C. then, choose the correct option(s).

Q: Consider a heat engine working with a hot reservoir of the furnace gases at 2100 °C , when the cooling water is available at -35.7 °C. then, choose the correct option(s).

(a) Engine efficiency can be made more than 90% without altering operating temperatures

(b) Engine efficiency always remains less than 90 %

(c) Engine efficiency is much lower practically due to irreversibility in the actual cycle

(d) Engine efficiency is 90% but such temperatures are not available practically

Click to See Answer :
Ans: (c)

Sol: Temperature of hot Reservoir , T1 = 2100 + 273 = 2373 k

Temperature of cold Reservoir , T2 = -35.7 + 273 = 237.3 K

Maximum Efficiency , $\displaystyle \eta_{max} = 1 – \frac{T_2}{T_1}$

$\displaystyle \eta_{max} = 1 – \frac{237.3}{2373}$

$\displaystyle \eta_{max} = 1 – \frac{1}{10} = \frac{9}{10}$

Percentage efficiency $\displaystyle = \frac{9}{10} \times 100$

= 90 %

 

During a process a system receives 30 kJ of heat from a reservoir and does 60 kJ of work. Then, choose the correct option(s).

Q: During a process a system receives 30 kJ of heat from a reservoir and does 60 kJ of work. Then, choose the correct option(s).

(a) It is possible to reach initial state by a quasistatic isothermal process

(b) It is possible to reach initial state using a constant volume process

(c) It is possible to reach initial state using a constant pressure process

(d) It is possible to reach initial state using an adiabatic process

Click to See Answer :
Ans: (d)
Sol: Here ΔQ = 30 kJ , ΔW = 60 kJ

For Process 1-2 ,

ΔQ1-2 = ΔU1-2 + ΔW1-2

30 = ΔU1-2 + 60

ΔU1-2 = – 30 k J

For Process 2-1 ,

ΔQ2-1 = ΔU2-1 + ΔW2-1

0 = U1 – U2 + ΔW2-1 (for Adiabatic Process , ΔQ = 0 )

0 = 30 + ΔW2-1

ΔW2-1 = – 30 kJ

Therefore , by doing a work of 30 kJ on the System , it can go back to Initial state

 

A heat engine receives heat at rate of 1500 kJ/min and gives an output of 8.2 kW. Rate of heat rejection by the engine is

Q: A heat engine receives heat at rate of 1500 kJ/min and gives an output of 8.2 kW. Rate of heat rejection by the engine is

(a) 1500 kJmin-1

(b) 8.2 kJs-1

(c) 16.8 kJs-1

(d) 8.2 kJmin-1

Click to See Answer :
Ans: (c)
Sol: Heat received per sec = 1500/60 = 25 kJ/sec

Work output per sec = 8.2 kW = 8.2 kJ/sec

Rate of heat rejection by engine = 25 – 8.2 = 16.8 kJ/sec

 

One mole of an ideal gas undergoes a process P = P_0/(1+(V_0/V)^2 ) . Here , P0 and V0 are constants….

Q: One mole of an ideal gas undergoes a process $\displaystyle P = \frac{P_0 }{1 + (\frac{V_0}{V})^2}$ . Here , P0 and V0 are constants. Change in temperature of the gas when volume is change from V = V0 and 2 V0 is

(a) $\displaystyle \frac{2 P_0 V_0}{5 R}$

(b) $\displaystyle \frac{11 P_0 V_0}{10 R}$

(c) $\displaystyle \frac{5 P_0 V_0}{4 R}$

(d) P0 V0

Click to See Answer :
Ans: (b)

 

A Carnot engine is designed to operate between 480 K and 300 K. if the engine actually produces 1.2 J of mechanical energy per cal….

Q: A Carnot engine is designed to operate between 480 K and 300 K . If the engine actually produces 1.2 J of mechanical energy per cal. of heat absorbed, then the ratio of actual efficiently to theoretical efficiency is

(a) 3/4

(b) 4/3

(c) 1/3

(d) 3/1

Click to See Answer :
Ans: (a)