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Basics of Fluid Mechanics, 2014a

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11.6. ISOTHERMAL FLOW 423<br />

Rearranging equation (11.122) and using the identify for perfect gas M 2 = ρU 2 /kP<br />

yields:<br />

− dP P − 4 fdx ( ) kPM<br />

2<br />

= kPM2 dU<br />

(11.123)<br />

D H 2<br />

U<br />

The pressure, P as a function <strong>of</strong> the Mach number has to substitute along with velocity,<br />

U as<br />

Differentiation <strong>of</strong> equation (11.124) yields<br />

U 2 = kRTM 2 (11.124)<br />

d(U 2 )=kR ( M 2 dT + Td(M 2 ) ) (11.125)<br />

d(M 2 )<br />

M 2 = d(U 2 )<br />

U 2 − dT T<br />

It can be noticed that dT =0for isothermal process and therefore<br />

d(M 2 )<br />

M 2 = d(U 2 )<br />

U 2 = 2UdU<br />

U 2<br />

= 2dU<br />

U<br />

The dimensionalization <strong>of</strong> the mass conservation equation yields<br />

(11.126)<br />

(11.127)<br />

dρ<br />

ρ + dU U<br />

= dρ<br />

ρ + 2 UdU<br />

2U 2<br />

= dρ<br />

ρ + d(U 2 )<br />

=0 (11.128)<br />

2 U<br />

2<br />

Differentiation <strong>of</strong> the isotropic (stagnation) relationship <strong>of</strong> the pressure (11.27) yields<br />

dP 0<br />

P 0<br />

⎛<br />

= dP P + ⎜<br />

⎝<br />

kM 2<br />

2<br />

1+ k − 1<br />

2<br />

M 2 ⎞<br />

⎟<br />

⎠ dM 2<br />

M 2 (11.129)<br />

Differentiation <strong>of</strong> equation (11.25) yields:<br />

(<br />

dT 0 = dT 1+ k − 1 )<br />

M 2 + T k − 1 dM 2 (11.130)<br />

2<br />

2<br />

Notice that dT 0 ≠0in an isothermal flow. There is no change in the actual temperature<br />

<strong>of</strong> the flow but the stagnation temperature increases or decreases depending on the Mach<br />

number (supersonic flow <strong>of</strong> subsonic flow). Substituting T for equation (11.130) yields:<br />

k − 1<br />

T 0 dM 2<br />

dT 0 = ( 2<br />

1+ k − 1 ) M 2<br />

M<br />

2<br />

2 M 2 (11.131)

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