Ask question

Ask question

All categories

2 months ago

5

6. You are evaluating flow through an airway. The current flow rate is 10 liters per minute with a fixed driving pressure (P1) o

f 20 cm H2O and a fixed downstream pressure (P2) of 5 cm H2O. Identify whether each statement is correct or incorrect if we pinch the lumen in the middle of the tube. a. The flow would decrease b. P1 would increase to maintain the flow rate c. The resistance would increase

1 answer:

iogann1982 [368]2 months ago

6 0

Answer:

B) P1 would have to increase to sustain the flow rate (correct)

C) Resistance would rise (correct)

Explanation:

Flow rate is measured at 10 liters per minute

Driving pressure (P1) stands at 20 cm H2O

Fixed downstream pressure (P2) is 5 cm H2O

The accurate statements when the lumen is pinched in the center of the tube are: P1 will increase to maintain the flow rate, and resistance will rise. This occurs because pinching the lumen decreases its diameter, leading to higher resistance, which is linearly related to pressure, thus P1 will also increase.

The incorrect statement is: the flow would decrease.

You might be interested in

Consider a rectangular fin that is used to cool a motorcycle engine. The fin is 0.15m long and at a temperature of 250C, while t

Kisachek [356]

Answer:

q' = 5826 W/m

Explanation:

Given:-

- The length of the fin in question, L = 0.15 m

- The fin's surface temperature, Ts = 250°C

- The velocity of free stream air, U = 80 km/h

- The air temperature, Ta = 27°C

- The flow is parallel over both sides of the fin, assuming turbulent flow conditions throughout.

Find:-

What is the heat removal rate per unit width of the fin?

Solution:-

- Steady state conditions are assumed, along with negligible radiation and turbulent flow conditions.

- From Table A-4, we gather air properties (T = 412 K, P = 1 atm ):

Dynamic viscosity, v = 27.85 * 10^-6 m²/s

Thermal conductivity, k = 0.0346 W / m.K

Prandtl number Pr = 0.69

- Compute the Nusselt Number (Nu) corresponding to - turbulent conditions - using the relevant relationship as follows:

$Nu = 0.037*Re_L^\frac{4}{5} * Pr^\frac{1}{3}$

Where, Re_L: Average Reynolds number for the entire length of fin:

$Re_L = \frac{U*L}{v} \\\\Re_L = \frac{80*\frac{1000}{3600} * 0.15}{27.85*10^-^6} \\\\Re_L = 119688.80909$

Consequently,

$Nu = 0.037*(119688.80909)^\frac{4}{5} * 0.69^\frac{1}{3}\\\\Nu = 378$

- The convective heat transfer coefficient (h) can now be derived from:

$h = \frac{k*Nu}{L} \\\\h = \frac{0.0346*378}{0.15} \\\\h = 87 \frac{W}{m^2K}$

- The heat loss rate q' per unit width can be established using the convection heat transfer formula and should be multiplied by (x2) since the airflow is present on both sides of the fin:

$q' = 2*[h*L*(T_s - T_a)]\\\\q' = 2*[87*0.15*(250 - 27)]\\\\q' = 5826\frac{W}{m}$

- Ultimately, the heat loss per unit width from the rectangular fin is q' = 5826 W/m

- The thermal loss per unit width (q') attributed to radiation:

$q' = 2*a*T_s^4*L$

Where, a signifies the Stefan-Boltzmann constant = 5.67*10^-8

$q' = 2*5.67*10^-^8*(523)^4*0.15\\\\q' = 1273 \frac{W}{m}$

- It is observed that radiation losses are not insignificant, accounting for 20% of thermal loss by convection. As the emissivity (e) of the fin is unspecified, this value is dismissed from the calculations as it pertains to the provided information.

7 0

2 months ago

Let Deterministic Quicksort be the non-randomized Quicksort which takes the first element as a pivot, using the partition routin

Daniel [329]

For Deterministic Quicksort, which operates by selecting the first element as the pivot, consider a scenario where the pivot consistently divides the array into segments of 1/3 and 2/3 for all recursive calls. (a) The runtime recurrence for this case needs to be determined. (b) Use a recursion tree to justify that this recurrence resolves to Theta(n log n). (c) Provide distinct sequences of 4 and 13 numbers that prompt this behavior.

3 0

2 months ago

You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an

pantera1 [306]

Answer:

$t'_{1\2} = 6.6 sec$

Explanation:

The half-life for the specified RC circuit can be expressed as

$t_{1\2} =\tau ln2$

where [/tex]\tau = RC[/tex]

$t_{1\2} = RCln2$

Given $t_{1\2} = 3 sec$

The circuit has a resistance of 40 ohms, and by adding a new resistor of 48 ohms, the total resistance becomes 40 + 48 = 88 ohms.

Thus, the new half-life is

$t'_{1\2} =R'Cln2$

Now, divide equation 2 by 1

$\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}$

$t'_{1\2} = t'_{1\2}\frac{R'}{R}$

After substituting all values, we can calculate the revised half-life

$t'_{1\2} = 3 * \frac{88}{40} = 6.6 sec$

$t'_{1\2} = 6.6 sec$

7 0

2 months ago

A hydrogen-filled balloon to be used in high altitude atmosphere studies will eventually be 100 ft in diameter. At 150,000 ft, t

mote1985 [299]

Answer:

The calculated result is 11.7 ft

Explanation:

You can apply the combined gas law, which incorporates Boyle's law, Charles's law, and Gay-Lussac's Law, because hydrogen demonstrates ideal gas behavior under these specific conditions.

$\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$

where the subscripts indicate "p" for pressure, "V" for volume, and "T" for temperature (in Kelvin) at varying moments. Let's denote $t_1$ as the balloon at 150,000 ft so

$p_1 = 0.14 \ lb/in^2$

$V_1 = \frac{4}{3} \pi R_1^3 = 523598.77 \ ft^3$

and $T_1 = -67^\circ F = 218.15\ K$ .

Then $t_2$ represents the point at which the balloon is on the ground.

$p_2 = 14.7 \ lb/in^2$ and $T_2 = 68^\circ F = 293.15\ K$ .

Based on the first equation

$V_2 = \frac{p_1 V_1 T_2}{T_1 p_2}$ , we find

$V_2 = 6701.07 ft^3$ and consequently the radius turns out to be

$R_2 = \sqrt[3]{\frac{3 V_2}{4 \pi}} = 11.7 \ ft$ .

5 0

2 months ago

A liquid food with 12% total solids is being heated by steam injection using steam at a pressure of 232.1 kPa (Fig. E3.3). The p

iogann1982 [368]

Answer:

$m_{s}=20kg/min$

$H_{s}=1914kJ/kg$

Explanation:

A liquid food containing 12% total solids is heated via steam injection at a pressure of 232.1 kPa (see Fig. E3.3). The product starts at a temperature of 50°C and has a flow rate of 100 kg/min, being elevated to a temperature of 120°C. The specific heat of the product varies with its composition as follows:

$c_{p}=c_{pw}(mass fraction H_{2}0)+c_{ps}(mass fraction solid)$ and the

specific heat of the product at 12% total solids is 3.936 kJ/(kg°C). The goal is to calculate the quantity and minimum quality of steam required to ensure that the leaving product has 10% total solids.

Given

Product total solids in ( $X_{A}$ ) = 0.12

Product mass flow rate ( $m_{A}$ ) = 100 kg/min

Product total solids out ( $X_{B}$ ) = 0.1

Product temperature in ( $T_{A}$ ) = 50°C

Product temperature out ( $T_{B}$ ) = 120°C

Steam pressure = 232.1 kPa at ( $T_{S}$ ) = 125°C

Product specific heat in ( $C_{PA}$ ) = 3.936 kJ/(kg°C)

The mass equation is:

$m_{A}X_{A}=m_{B}X_{B}$

$100(0.12)=m_{B}(0.1)\\m_{B}=\frac{100(0.12)}{0.1} =120$

Also $m_{a}+m_{s}=m_{b}\\$

Therefore: $100}+m_{s}=120\\\\m_{s}=120-100=20$

The energy balance equation is:

$m_{A}C_{PA}(T_{A}-0)+m_{s}H_{s}=m_{B}C_{PB}(T_{B}-0)$

$3.936 = (4.178)(0.88) +C_{PS}(0.12)\\C_{PS}=\frac{3.936-3.677}{0.12} =2.161$

$C_{PB}= 4.232*0.9+0.1C_{PS}= 4.232*0.9+0.1*2.161=4.025$ kJ/(kg°C)

By substituting values into the energy equation:

$100(3.936)(50-0)+20H_{s}=120(4.025)}(120-0)$

$19680+20H_{s}=57960\\20H_{s}=57960-19680 \\20H_{s}=38280\\H_{s}=\frac{38280}{20} =1914$

$H_{s}=1914kJ/kg$

From the properties of saturated steam at 232.1 kPa,

$H_{c}$ = 524.99 kJ/kg

$H_{v}$ = 2713.5 kJ/kg

% quality = $\frac{1914-524.99}{2713.5-524.99} =63.5%$

Any steam quality above 63.5% will result in higher total solids in the heated product.

3 0

2 months ago