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](https://tex.z-dn.net/?f=T_o%20%3D%2020%5E0%20C%20%3D%20%28%2020%20%2B273%29%20K%20%3D%20293K)
The initial pressure in the tank ![P_o= 800 \ kPa](https://tex.z-dn.net/?f=P_o%3D%20800%20%5C%20kPa)
The throat area
= 0.75 cm²
To find the initial mass flow in kg/s.
Lets first recall that:
Provided that
, then the flow is choked.
Then;
![\dfrac{P_{amb}}{P_{tank}}= \dfrac{101.35}{800}](https://tex.z-dn.net/?f=%5Cdfrac%7BP_%7Bamb%7D%7D%7BP_%7Btank%7D%7D%3D%20%5Cdfrac%7B101.35%7D%7B800%7D)
![\dfrac{P_{amb}}{P_{tank}}= 0.1266](https://tex.z-dn.net/?f=%5Cdfrac%7BP_%7Bamb%7D%7D%7BP_%7Btank%7D%7D%3D%200.1266)
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](https://tex.z-dn.net/?f=m%20%3D%20%5Crho%20%5Ctimes%20A%20%5Ctimes%20V)
where;
![\rho = \rho_o \bigg ( \dfrac{2}{k+1} \bigg) ^{\dfrac{1}{k-1}}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Crho_o%20%5Cbigg%20%28%20%5Cdfrac%7B2%7D%7Bk%2B1%7D%20%5Cbigg%29%20%5E%7B%5Cdfrac%7B1%7D%7Bk-1%7D%7D)
![\rho =\dfrac{P_o}{RT_o} \bigg ( \dfrac{2}{k+1} \bigg) ^{\dfrac{1}{k-1}}](https://tex.z-dn.net/?f=%5Crho%20%3D%5Cdfrac%7BP_o%7D%7BRT_o%7D%20%5Cbigg%20%28%20%5Cdfrac%7B2%7D%7Bk%2B1%7D%20%5Cbigg%29%20%5E%7B%5Cdfrac%7B1%7D%7Bk-1%7D%7D)
![\rho =\dfrac{800 \times 10^3}{287 \times 293} \bigg ( \dfrac{2}{1.4+1} \bigg) ^{\dfrac{1}{1.4-1}}](https://tex.z-dn.net/?f=%5Crho%20%3D%5Cdfrac%7B800%20%5Ctimes%2010%5E3%7D%7B287%20%5Ctimes%20293%7D%20%5Cbigg%20%28%20%5Cdfrac%7B2%7D%7B1.4%2B1%7D%20%5Cbigg%29%20%5E%7B%5Cdfrac%7B1%7D%7B1.4-1%7D%7D)
![\rho =9.51350( 0.8333 ) ^{2.5}](https://tex.z-dn.net/?f=%5Crho%20%3D9.51350%28%200.8333%20%29%20%5E%7B2.5%7D)
![\rho =6.03 \ kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D6.03%20%5C%20kg%2Fm%5E3)
![T = T_o \bigg ( \dfrac{2}{k+1}\bigg)](https://tex.z-dn.net/?f=T%20%20%3D%20T_o%20%5Cbigg%20%28%20%5Cdfrac%7B2%7D%7Bk%2B1%7D%5Cbigg%29)
where;
![T_o = 293 \ K](https://tex.z-dn.net/?f=T_o%20%3D%20293%20%5C%20K)
![T = 293 \bigg ( \dfrac{2}{1.4+1}\bigg)](https://tex.z-dn.net/?f=T%20%20%3D%20293%20%5Cbigg%20%28%20%5Cdfrac%7B2%7D%7B1.4%2B1%7D%5Cbigg%29)
![T = 293 \bigg ( \dfrac{2}{2.4}\bigg)](https://tex.z-dn.net/?f=T%20%20%3D%20293%20%5Cbigg%20%28%20%5Cdfrac%7B2%7D%7B2.4%7D%5Cbigg%29)
![T = 293 ( 0.8333)](https://tex.z-dn.net/?f=T%20%20%3D%20293%20%20%28%200.8333%29)
T = 244.16 K
During a critical condition when Mach No. is equal to one;
![V = a = \sqrt{kRT}](https://tex.z-dn.net/?f=V%20%3D%20a%20%3D%20%5Csqrt%7BkRT%7D)
![V = \sqrt{1.4 \times 287 \times 244.16}](https://tex.z-dn.net/?f=V%20%3D%20%5Csqrt%7B1.4%20%5Ctimes%20287%20%5Ctimes%20244.16%7D)
![V = \sqrt{98103.488}](https://tex.z-dn.net/?f=V%20%3D%20%5Csqrt%7B98103.488%7D)
V = 313.214 m/s
Thus, the initial mass flow rate ![m = \rho \times A \times V](https://tex.z-dn.net/?f=m%20%3D%20%5Crho%20%5Ctimes%20A%20%5Ctimes%20V)
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)](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%7D%7Bdt%7D%28%5Crho_o%20V%29%20%3D%20%5Cdfrac%7Bd%7D%7Bdt%7D%20%5Cbigg%20%28%20%5Cdfrac%7BP_o%7D%7BRT_o%7D%20V%5Cbigg%29)
![= \dfrac{V}{RT_o}\dfrac{dP_o}{dt}= -m](https://tex.z-dn.net/?f=%3D%20%5Cdfrac%7BV%7D%7BRT_o%7D%5Cdfrac%7BdP_o%7D%7Bdt%7D%3D%20-m)
When the air mass flow rate is:
![m = 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}](https://tex.z-dn.net/?f=m%20%3D%200.6847%20%5Cdfrac%7BP_oA%7D%7B%5Csqrt%7BRT_o%7D%7D)
Thus; replacing
in the previous equation; we have:
![\dfrac{V}{RT_o}\dfrac{dP_o}{dt}= - 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}](https://tex.z-dn.net/?f=%5Cdfrac%7BV%7D%7BRT_o%7D%5Cdfrac%7BdP_o%7D%7Bdt%7D%3D%20-%200.6847%20%5Cdfrac%7BP_oA%7D%7B%5Csqrt%7BRT_o%7D%7D)
![\dfrac{dP_o}{P_o}= -0.6847 \dfrac{A\sqrt{RT_o}}{V} \ dt](https://tex.z-dn.net/?f=%5Cdfrac%7BdP_o%7D%7BP_o%7D%3D%20-0.6847%20%5Cdfrac%7BA%5Csqrt%7BRT_o%7D%7D%7BV%7D%20%5C%20dt)
Taking the differential of both sides from 0 → t
![In(P_o)^t_o = -0.6847 \dfrac{A\sqrt{RT}}{V} \times t](https://tex.z-dn.net/?f=In%28P_o%29%5Et_o%20%3D%20-0.6847%20%5Cdfrac%7BA%5Csqrt%7BRT%7D%7D%7BV%7D%20%5Ctimes%20t)
![In \bigg ( \dfrac{P(t)}{P(0)} \bigg) = -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t](https://tex.z-dn.net/?f=In%20%5Cbigg%20%28%20%5Cdfrac%7BP%28t%29%7D%7BP%280%29%7D%20%5Cbigg%29%20%3D%20-0.6847%20%5Cdfrac%7BA%5Csqrt%7BRT_o%7D%7D%7BV%7D%5Ctimes%20t)
![\dfrac{P(t)}{P(0)} =exp \bigg ( -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t \bigg )](https://tex.z-dn.net/?f=%5Cdfrac%7BP%28t%29%7D%7BP%280%29%7D%20%20%3Dexp%20%5Cbigg%20%28%20-0.6847%20%5Cdfrac%7BA%5Csqrt%7BRT_o%7D%7D%7BV%7D%5Ctimes%20t%20%5Cbigg%20%29)
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 )](https://tex.z-dn.net/?f=%5Cdfrac%7B500%7D%7B800%7D%20%20%3Dexp%20%5Cbigg%20%28%20-0.6847%20%5Cdfrac%7B0.75%20%5Ctimes%2010%5E%7B-4%7D%5Csqrt%7B287%20%5Ctimes%20293%7D%7D%7B1.5%7D%5Ctimes%20t%20%5Cbigg%20%29)
t = 47.343 s
(c)
Let us recall that:
The choking on the nozzle occurred when ![\dfrac{P_{amb}}{P_{tank}} = 0.528](https://tex.z-dn.net/?f=%5Cdfrac%7BP_%7Bamb%7D%7D%7BP_%7Btank%7D%7D%20%3D%200.528)
![\dfrac{P_{amb}}{0.528} = P_{tank}](https://tex.z-dn.net/?f=%5Cdfrac%7BP_%7Bamb%7D%7D%7B0.528%7D%20%3D%20P_%7Btank%7D)
![\dfrac{101.35}{0.528} = P_{tank}](https://tex.z-dn.net/?f=%5Cdfrac%7B101.35%7D%7B0.528%7D%20%3D%20P_%7Btank%7D)
![P_{tank}= 191.95 \ kPa \\ \\ P_{tank} \simeq 192 \ kPa](https://tex.z-dn.net/?f=P_%7Btank%7D%3D%20191.95%20%5C%20kPa%20%5C%5C%20%5C%5C%20P_%7Btank%7D%20%5Csimeq%20192%20%5C%20kPa)
From
; the time required for
is:
![\dfrac{192}{800} =exp \bigg ( -0.6847 \dfrac{0.75 \times 10^{-4}\sqrt{287 \times 293}}{1.5}\times t \bigg )](https://tex.z-dn.net/?f=%5Cdfrac%7B192%7D%7B800%7D%20%20%3Dexp%20%5Cbigg%20%28%20-0.6847%20%5Cdfrac%7B0.75%20%5Ctimes%2010%5E%7B-4%7D%5Csqrt%7B287%20%5Ctimes%20293%7D%7D%7B1.5%7D%5Ctimes%20t%20%5Cbigg%20%29)
By solving:
t = 143.745 s