Question

Air at 13 psia and 65°F enters an adiabatic diffuser steadily with a velocity of 500 ft/s and leaves with a low velocity at a pressure of 14.5 psia. The exit area of the diffuser is 3 times the inlet area. Determine (a) the exit temperature and (b) the exit velocity of the air. $P_1 = 13 \text{ psia}$ $T_1 = 65^\circ F$ $V_1 = 500 \text{ ft/s}$ Air $P_2 = 14.5 \text{ psia}$ $V_2 << V_1$ $A_2 = 3A_1$

          Air at 13 psia and 65°F enters an adiabatic diffuser steadily with a velocity of 500 ft/s and leaves with a low velocity at a pressure of 14.5 psia. The exit area of the diffuser is 3 times the inlet area. Determine (a) the exit temperature and (b) the exit velocity of the air.
$P_1 = 13 \text{ psia}$
$T_1 = 65^\circ F$
$V_1 = 500 \text{ ft/s}$
Air
$P_2 = 14.5 \text{ psia}$
$V_2 << V_1$
$A_2 = 3A_1$

Air at 13 psia and 65°F enters an adiabatic diffuser steadily with a velocity of 500 ft/s and leaves with a low velocity at a pressure of 14.5 psia. The exit area of the diffuser is 3 times the inlet area. Determine (a) the exit temperature and (b) the exit velocity of the air.
P1 = 13 psia
T1 = 65^∘ F
V1 = 500 ft/s
Air
P2 = 14.5 psia
V2 << V1
A2 = 3A1

Added by Jessica R.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

09/01/2023

Step 1

Given: T = 65°F To convert from Fahrenheit to Rankine, we use the formula: Rankine = Fahrenheit + 459.67 So, Rankine = 65 + 459.67 = 524.67 Rankine Show more…

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Texts: Struggling to use table, can't find my value. Air at 13 psia and 65°F enters an adiabatic diffuser steadily with a velocity of 500 ft/s and leaves with a low velocity at a pressure of 14.5 psia. The exit area of the diffuser is 3 times the inlet area. Determine the exit temperature. Given: P = 13 psia T = 65°F V = 500 ft/s P2 = 14.5 psia V << V A = 3A Air