Question

An air tank of volume 1.5 m3 is initially at 800 kPa and 208C. At t 5 0, it begins exhausting through a converging nozzle to sea-level conditions. The throat area is 0.75 cm2 . Estimate (a) the initial mass fl ow in kg/s, (b) the time required to blow down to 500 kPa, and (c) the time at which the nozzle ceases being choked.

Answer 1

Answer:

(a) m = 0.141 kg/s

(b) t = 47.343 s

(c) t = 143.745 s

Explanation:

Given that:

The volume of air in the tank V = 1.5 m³

The initial temperature in the tank is supposed to be 20° C and not 208 C;

So  $T_o = 20^0 C = ( 20 +273) K = 293K$

The initial pressure in the tank $P_o= 800 \ kPa$

The throat area $A_t$ = 0.75 cm²

To find the initial mass flow in kg/s.

Lets first recall that:

Provided that $\dfrac{P_{amb}}{P_{tank}}< 0.528$, then the flow is choked.

Then;

$\dfrac{P_{amb}}{P_{tank}}= \dfrac{101.35}{800}$

$\dfrac{P_{amb}}{P_{tank}}= 0.1266$

From what we see above, it is obvious that the ratio is lesser than 0.528, therefore, the flow is choked.

Now, for a choked nozzle, the initial mass flow rate is determined by using the formula:

$m = \rho \times A \times V$

where;

$\rho = \rho_o \bigg ( \dfrac{2}{k+1} \bigg) ^{\dfrac{1}{k-1}}$

$\rho =\dfrac{P_o}{RT_o} \bigg ( \dfrac{2}{k+1} \bigg) ^{\dfrac{1}{k-1}}$

$\rho =\dfrac{800 \times 10^3}{287 \times 293} \bigg ( \dfrac{2}{1.4+1} \bigg) ^{\dfrac{1}{1.4-1}}$

$\rho =9.51350( 0.8333 ) ^{2.5}$

$\rho =6.03 \ kg/m^3$

$T = T_o \bigg ( \dfrac{2}{k+1}\bigg)$

where;

$T_o = 293 \ K$

$T = 293 \bigg ( \dfrac{2}{1.4+1}\bigg)$

$T = 293 \bigg ( \dfrac{2}{2.4}\bigg)$

$T = 293 ( 0.8333)$

T = 244.16 K

During a critical condition when Mach No. is equal to one;

$V = a = \sqrt{kRT}$

$V = \sqrt{1.4 \times 287 \times 244.16}$

$V = \sqrt{98103.488}$

V = 313.214  m/s

Thus, the initial mass flow rate $m = \rho \times A \times V$

m = 6.03 × 0.75 × 10⁻⁴ × 313.214

m = 0.141 kg/s

(b)

The mass balance formula for the control volume surrounding the tank can be expressed as:

$\dfrac{d}{dt}(\rho_o V) = \dfrac{d}{dt} \bigg ( \dfrac{P_o}{RT_o} V\bigg)$

$= \dfrac{V}{RT_o}\dfrac{dP_o}{dt}= -m$

When the air mass flow rate is:

$m = 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}$

Thus; replacing $m = 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}$ in the previous equation; we have:

$\dfrac{V}{RT_o}\dfrac{dP_o}{dt}= - 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}$

$\dfrac{dP_o}{P_o}= -0.6847 \dfrac{A\sqrt{RT_o}}{V} \ dt$

Taking the differential of both sides  from 0 → t

$In(P_o)^t_o = -0.6847 \dfrac{A\sqrt{RT}}{V} \times t$

$In \bigg ( \dfrac{P(t)}{P(0)} \bigg) = -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t$

$\dfrac{P(t)}{P(0)} =exp \bigg ( -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t \bigg )$

So, when the pressure P = 500 kPa, the time required is:

$\dfrac{500}{800} =exp \bigg ( -0.6847 \dfrac{0.75 \times 10^{-4}\sqrt{287 \times 293}}{1.5}\times t \bigg )$

t = 47.343 s

(c)

Let us recall that:

The choking on the nozzle occurred when $\dfrac{P_{amb}}{P_{tank}} = 0.528$

$\dfrac{P_{amb}}{0.528} = P_{tank}$

$\dfrac{101.35}{0.528} = P_{tank}$

$P_{tank}= 191.95 \ kPa \\ \\ P_{tank} \simeq 192 \ kPa$

From $\dfrac{P(t)}{P(0)} =exp \bigg ( -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t \bigg )$; the time required for $P_{tank} \simeq 192 \ kPa$ is:

$\dfrac{192}{800} =exp \bigg ( -0.6847 \dfrac{0.75 \times 10^{-4}\sqrt{287 \times 293}}{1.5}\times t \bigg )$

By solving:

t = 143.745 s