Question

Air at 30°C and 2 MPa flows at steady state in a horizontal pipeline with a velocity of 25 m/s. It passes through a throttle valve where the pressure is reduced to 0.3 MPa. The pipe is the same diameter upstream and downstream of the valve. What is the outlet temperature and velocity of the gas

Answer 1

Answer:

The outlet velocity is 159.9m/s

The outlet temperature = 290.6K

Explanation:

At initial condition

P=2 MPa

T=30°C

V=25 m/s

At final condition

P=0.3 MPa

Now from first law for open system

$h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}$

We know that for air

h= 1.010 x T  KJ/kg

$1.01\times 303+\dfrac{25^2}{2000}=1.01\times T+\dfrac{V_2^2}{2000} -----1$

Now from mass balance

$\rho=\dfrac{P}{RT}\dfrac{P_1}{RT_1}V_1=\dfrac{P_2}{RT_2}V_2\dfrac{2}{R\times 303}\times 25=\dfrac{0.3}{RT_2}V_2T_2=1.81V_2 ----------2$

Now from equation 1 and 2

$V_2^2+3673.749V-612916.49=0$

So we can say that

$V_2=159.9\ m/s$

The outlet velocity is 159.9m/s

Now by putting the values in equation 2

$T_2=1.81\times 159.9 \\T_2=290.6K$

The outlet temperature = 290.6K