The efficiency of otto cycle for petrol engine is 50% and the adiabatic constant is 1.4. calculate the compression ratio.

The efficiency of carnot engine is 50% and temperature of sink is 500K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of the sink will be : -

If the efficiency of an engine is 50% and its work output is 500 J then find input.

The efficiency of the cycle shown below in the figure ( consisting of one isobar , one adiabatic and one isotherm ( is 50 % the ratio ,x, between the highest and lowest temperature attained in this cycle obeys ( the working substance is an ideal gas )

The efficiency of a heat engine is defined as the ratio of the mechanical work done by the engine in one cycle to the heat absorbed from the high temperature source . eta = (W)/(Q_(1)) = (Q_(1) - Q_(2))/(Q_(1)) Cornot devised an ideal engine which is based on a reversible cycle of four operations in succession: isothermal expansion , adiabatic expansion. isothermal compression and adiabatic compression. For carnot cycle (Q_(1))/(T_(1)) = (Q_(2))/(T_(2)) . Thus eta = (Q_(1) - Q_(2))/(Q_(1)) = (T_(1) - T_(2))/(T_(1)) According to carnot theorem "No irreversible engine can have efficiency greater than carnot reversible engine working between same hot and cold reservoirs". A carnot engine whose low temperature reservoir is at 7^(@)C has an efficiency of 50% . It is desired to increase the efficiency to 70% . By how many degrees should the temperature of the high temperature reservoir be increased?

The efficiency of a heat engine is defined as the ratio of the mechanical work done by the engine in one cycle to the heat absorbed from the high temperature source . eta = (W)/(Q_(1)) = (Q_(1) - Q_(2))/(Q_(1)) Cornot devised an ideal engine which is based on a reversible cycle of four operations in succession: isothermal expansion , adiabatic expansion. isothermal compression and adiabatic compression. For carnot cycle (Q_(1))/(T_(1)) = (Q_(2))/(T_(2)) . Thus eta = (Q_(1) - Q_(2))/(Q_(1)) = (T_(1) - T_(2))/(T_(1)) According to carnot theorem "No irreversible engine can have efficiency greater than carnot reversible engine working between same hot and cold reservoirs". An inventor claims to have developed an engine working between 600K and 300K capable of having an efficiency of 52% , then -

The efficiency of a heat engine is defined as the ratio of the mechanical work done by the engine in one cycle to the heat absorbed from the high temperature source . eta = (W)/(Q_(1)) = (Q_(1) - Q_(2))/(Q_(1)) Cornot devised an ideal engine which is based on a reversible cycle of four operations in succession: isothermal expansion , adiabatic expansion. isothermal compression and adiabatic compression. For carnot cycle (Q_(1))/(T_(1)) = (Q_(2))/(T_(2)) . Thus eta = (Q_(1) - Q_(2))/(Q_(1)) = (T_(1) - T_(2))/(T_(1)) According to carnot theorem "No irreversible engine can have efficiency greater than carnot reversible engine working between same hot and cold reservoirs". Efficiency of a carnot's cycle change from (1)/(6) to (1)/(3) when source temperature is raised by 100K . The temperature of the sink is-

Determine efficiency of carnot cycle if in adiabatic expansion volume 3 times of initial and r=1.5.