Respuesta :
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 [tex]T_o = 20^0 C = ( 20 +273) K = 293K[/tex]
The initial pressure in the tank [tex]P_o= 800 \ kPa[/tex]
The throat area [tex]A_t[/tex] = 0.75 cm²
To find the initial mass flow in kg/s.
Lets first recall that:
Provided that [tex]\dfrac{P_{amb}}{P_{tank}}< 0.528[/tex], then the flow is choked.
Then;
[tex]\dfrac{P_{amb}}{P_{tank}}= \dfrac{101.35}{800}[/tex]
[tex]\dfrac{P_{amb}}{P_{tank}}= 0.1266[/tex]
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:
[tex]m = \rho \times A \times V[/tex]
where;
[tex]\rho = \rho_o \bigg ( \dfrac{2}{k+1} \bigg) ^{\dfrac{1}{k-1}}[/tex]
[tex]\rho =\dfrac{P_o}{RT_o} \bigg ( \dfrac{2}{k+1} \bigg) ^{\dfrac{1}{k-1}}[/tex]
[tex]\rho =\dfrac{800 \times 10^3}{287 \times 293} \bigg ( \dfrac{2}{1.4+1} \bigg) ^{\dfrac{1}{1.4-1}}[/tex]
[tex]\rho =9.51350( 0.8333 ) ^{2.5}[/tex]
[tex]\rho =6.03 \ kg/m^3[/tex]
[tex]T = T_o \bigg ( \dfrac{2}{k+1}\bigg)[/tex]
where;
[tex]T_o = 293 \ K[/tex]
[tex]T = 293 \bigg ( \dfrac{2}{1.4+1}\bigg)[/tex]
[tex]T = 293 \bigg ( \dfrac{2}{2.4}\bigg)[/tex]
[tex]T = 293 ( 0.8333)[/tex]
T = 244.16 K
During a critical condition when Mach No. is equal to one;
[tex]V = a = \sqrt{kRT}[/tex]
[tex]V = \sqrt{1.4 \times 287 \times 244.16}[/tex]
[tex]V = \sqrt{98103.488}[/tex]
V = 313.214 m/s
Thus, the initial mass flow rate [tex]m = \rho \times A \times V[/tex]
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:
[tex]\dfrac{d}{dt}(\rho_o V) = \dfrac{d}{dt} \bigg ( \dfrac{P_o}{RT_o} V\bigg)[/tex]
[tex]= \dfrac{V}{RT_o}\dfrac{dP_o}{dt}= -m[/tex]
When the air mass flow rate is:
[tex]m = 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}[/tex]
Thus; replacing [tex]m = 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}[/tex] in the previous equation; we have:
[tex]\dfrac{V}{RT_o}\dfrac{dP_o}{dt}= - 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}[/tex]
[tex]\dfrac{dP_o}{P_o}= -0.6847 \dfrac{A\sqrt{RT_o}}{V} \ dt[/tex]
Taking the differential of both sides from 0 → t
[tex]In(P_o)^t_o = -0.6847 \dfrac{A\sqrt{RT}}{V} \times t[/tex]
[tex]In \bigg ( \dfrac{P(t)}{P(0)} \bigg) = -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t[/tex]
[tex]\dfrac{P(t)}{P(0)} =exp \bigg ( -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t \bigg )[/tex]
So, when the pressure P = 500 kPa, the time required is:
[tex]\dfrac{500}{800} =exp \bigg ( -0.6847 \dfrac{0.75 \times 10^{-4}\sqrt{287 \times 293}}{1.5}\times t \bigg )[/tex]
t = 47.343 s
(c)
Let us recall that:
The choking on the nozzle occurred when [tex]\dfrac{P_{amb}}{P_{tank}} = 0.528[/tex]
[tex]\dfrac{P_{amb}}{0.528} = P_{tank}[/tex]
[tex]\dfrac{101.35}{0.528} = P_{tank}[/tex]
[tex]P_{tank}= 191.95 \ kPa \\ \\ P_{tank} \simeq 192 \ kPa[/tex]
From [tex]\dfrac{P(t)}{P(0)} =exp \bigg ( -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t \bigg )[/tex]; the time required for [tex]P_{tank} \simeq 192 \ kPa[/tex] is:
[tex]\dfrac{192}{800} =exp \bigg ( -0.6847 \dfrac{0.75 \times 10^{-4}\sqrt{287 \times 293}}{1.5}\times t \bigg )[/tex]
By solving:
t = 143.745 s
