Question

An ideal gas expands in an adiabatic turbine from 1200 K and 900 kPa to 800 K. Determine the turbine inlet volume flow rate of gas, in m3/s, required to produce turbine work output at the rate of 650 kW. The average values of the specific heats for this gas over the temperature range and the gas constant are cp = 1.13 kJ/kgK, cv = 0.83 kJ/kgK, and R = 0.30 kJ/kgK.

Answer 1

Answer:

$V = 0.5752 m^3/s$

Explanation:

given data:

$T_1 = 1200 K$

$p_1 = 900 kPa$

$T_2 = 800 K$

$W_{out} = 650 kW$

cp= 1.13 kJ/kg K

CV = 0.83 kJ/kg K

FROM ENERGY BALANCE EQUATION

$E_{IN} - E_{OUT} = \Delta E_{SYS} = 0$

$E_{IN} = E_{OUT}$

$\dot m h_1 = W_{out} + \dot m h_2$

$\dot m =\frac{W_{out}}{h_1 -h_2}$

$\dot m  = \frac{w_{out}}{cp(T_1 -T_2)}$

$\dot m = \frac{650}{1.13 \times (1200-800)}$

$\dot m = 1438 kg/s$

sir specific volume is given as

$v_1 = \frac[RT_1}{P_1}$

$v_1 = \frac{0.3 \times 1200}{900}$

$v_1 = 0.4 m^3/kg$

volume of flow rate

$V = 1.438 \times 0.4 = 0.5752 m^3/s$