REVVIVE ME MY MOM WALKED IN MY ROOM AND SCARED THE BAJESUS OUTTA ME

A venturi meter is to be installed in a 63 mm bore section of a piping system to measure the flow rate of water in it. From spac

Alex Ar [27]

Answer:

Throat diameter $d_2$ =28.60 mm

Explanation:

Bore diameter $d_1=63mm$ ⇒ $A_1=3.09\times 10^{-3} m^2$

Manometric deflection x=235 mm

Flow rate Q=240 Lt/min⇒ Q=.004 $\frac{m^3}{s}$

Coefficient of discharge $C_d$ =0.8

We know that discharge through venturi meter

$Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}$

$h=x(\dfrac{S_m}{S_w}-1)$

$S_m$ =13.6 for Hg and $S_w$ =1 for water.

$h=0.235(\dfrac{13.6}{1}-1)$

h=2.961 m

Now by putting the all value in

$Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}$

$0.004=0.8\times \dfrac{3.09\times 10^{-3} A_2\sqrt{2\times 9.81\times 2.961}}{\sqrt{(3.09\times 10^{-3})^2-A_2^2}}$

$A_2=6.42\times 10^{-4} m^2$

⇒ $d_2$ =28.60 mm

So throat diameter $d_2$ =28.60 mm

4 0

3 years ago

typedef struct bitNode { int data; struct bitNode *left; struct bstNode *right; } bstNode; int solve(bstNode* root) { if (root =

Sergio039 [100]

Answer:

The ten numbers to be filled in the blanks are: 18, 7, 7, 11, 18, 36, 3, 8, 13, 50.

Explanation:

keeps on going to the left node until node->left == NULL;

now at Node 18;

left = right = 0; hence condition is not satisfied

18 is printed first.

the value 18 is returned .

Then we reach at 4;

from there we move to 7;

just like 18, similar things happen with 7 and 7 is printed, the value 7 is returned.

Now coming to Node 4,

left = 0, right = 7 ; hence the condition is satisfied & res = 7; 7 is printed.

For Node 16, left = 7 ; right = 11(but for this we visit 11 first and 11 is printed)

for 16; condition is satisfied; res = 7 + 11 = 18 ; 18 is printed

Now for 5; left = right = 18; the condition is satisfied; so res = 18 + 18 = 36; 36 is printed

Next we visit Node 3; 3 is printed & 3 is returned

Then Node 8 ; 8 is printed & 8 is returned

for Node 13; left = 3, right = 8 ; condition is not satisfied, 13 is printed.

For Node 50; left = 36 right = 13 ; condition is not satisfied hence 50 is printed.

So the order of printing is 18 7 7 11 18 36 3 8 13 50.

4 0

3 years ago

Two mass streams of the same ideal gas are mixed in a steady-flow chamber while receiving energy by heat transfer from the surro

loris [4]

Answer:

(a)The final temperature of mixture is T₃ =m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp

(b) The final volume is V₃ =V₁ + V₂ + RQin/P₃Cp

(c) The volume flow rate at exit is V₃ =V₁ + V₂

Explanation:

Solution

Now

The system comprises of two inlets and on exit.

Mass flow rate enthalpy of fluid from inlet -1 be m₁ and h₁

Mass flow rate enthalpy of fluid from inlet -2 be m₂ and h₂

Mass flow rate enthalpy of fluid from exit be m₃ and h₃

Mixing chambers do not include any kind of work (w = 0)

So, both the kinetic and potential energies of the fluid streams are usually negligible (ke =0, pe =0)

(a) Applying the mass balance of mixing chamber, min = mout

Applying the energy balance of mixing chamber,

Ein = Eout

min hin =mout hout

miCpT₁ + m₂CpT₂ +Qin =m₃CpT₃

T₃ = miCpT₁/m₃CpT₃ + m₂CpT₂/m₃CpT₃ + Qin/m₃CpT₃ +

T₃ =m₁T₁/m₃+ m₂T/m₃ + Qin/m₃Cp

The final temperature of mixture is T₃ =m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp

(b) From the ideal gas equation,

v =RT/PT

v₃ = RT₃/P₃

The volume flow rate at the exit, V₃ =m₃v₃

V₃ = m₃ RT₃/P₃

Substituting the value of T₃, we have

V₃=m₃ R/P₃ (=m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp)

V₃ = R/P₃ (m₁T₁+ m₂T₂ + Qin/Cp)

Now

The mixing process occurs at constant pressure P₃=P₂=P₁.

Hence V₃ becomes:

V₃=m₁RT₁/P₁ +m₂RT₂/P₂ + RQin/P₃Cp

V₃ =V₁ + V₂ + RQin/P₃Cp

Therefore, the final volume is V₃ =V₁ + V₂ + RQin/P₃Cp

(c) Now for an adiabatic mixing, Qin =0

Hence V₃ becomes:

V₃ =V₁ + V₂ + r * 0/P₃Cp

V₃ =V₁ + V₂ + 0

V₃ =V₁ + V₂

Therefore the volume flow rate at exit is V₃ =V₁ + V₂

8 0

3 years ago

A satellite would have a mass of 270 kg on the surface of Mars. Determine the weight of the satellite in pounds if it is in orbi

koban [17]

Answer:

26 lbf

Explanation:

The mass of the satellite is the same regardless of where it is.

The weight however, depends on the acceleration of gravity.

The universal gravitation equation:

g = G * M / d^2

Where

G: universal gravitation constant (6.67*10^-11 m^3/(kg*s))

M: mass of the body causing the gravitational field (mass of Earth = 6*10^24 kg)

d: distance to that body

15000 miles = 24140 km

The distance is to the center of Earth.

Earth radius = 6371 km

Then:

d = 24140 + 6371 = 30511 km

g = 6.67*10^-11 * 6*10^24 / 30511000^2 = 0.43 m/s^2

Then we calculate the weight:

w = m * a

w = 270 * 0.43 = 116 N

116 N is 26 lbf

8 0

3 years ago

Just need someone to talk to pls dont just use me for points

Allushta [10]

Answer:

well what do you wanna talk about friend?

Explanation:

7 0

2 years ago