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Len [333]
3 years ago
6

Refrigerant 134a enters a well-insulated nozzle at 200 lbf/in.2, 200°F, with a velocity of 120 ft/s and exits at 50 lbf/in.2 wit

h a velocity of 1500 ft/s. For steady-state operation, and neglecting potential energy effects, determine the temperature, in °F, and the quality of the refrigerant at the exit.
Physics
1 answer:
riadik2000 [5.3K]3 years ago
5 0

Answer:

x = 0.75801 = 75.801%

T_2 = 72..78 degree F

Explanation:

From superheated R 134 a properties table

At 200 lb/in^2 and 200 degree F

h_1 = 138.99 Btu/lbm

steady flow energy equation is givena s

h_1 + \frac{v_1^2}{2}  = h_2 + \frac{v_2^2}{2}

138.99 + \frac{120^2}{2\times 25037} = h_2 + \frac{1500^2}{2 \times 25037}

h_2 = 94.344 Btu/lbm

At 90 lb/in2 Tsat = 72.78 degree F

h_f = 35.715 Btu/lbm

hfg  = 77.345 Btu/lbm

h = hf + x hfg

94.344 = 35.715+ x \times 77.345

solving for x we get

x = 0.75801 = 75.801%

T_2 = 72..78 degree F

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A uranium ion and an iron ion are separated by a distance of =61.10 nm. The uranium atom is singly ionized; the iron atom is dou
scoray [572]

Answer:

Explanation:

Charge on uranium ion = charge of a single electron

= 1.6 x 10⁻¹⁹ C

charge on doubly ionised iron atom = charge of 2 electron

= 2 x 1.6 x 10⁻¹⁹ C = 3.2 x 10⁻¹⁹ C

Let the required distance from uranium ion be d .

force on electron at distance d from uranium ion

= 9 x 10⁹ x 1.6 x 10⁻¹⁹ / r²

force on electron at distance 61.10 x 10⁻⁹ - r from iron  ion

= 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

For equilibrium ,

9 x 10⁹ x 1.6 x 10⁻¹⁹ / r² = 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

2 d² = (61.10 x 10⁻⁹ - r )²

1.414 r = 61.10 x 10⁻⁹ - r

2.414 r = 61.10 x 10⁻⁹

r = 25.31 nm .

4 0
3 years ago
Question 10 of 10
Alinara [238K]

it allows only a reduced number of electrons to flow through it.

4 0
3 years ago
Read 2 more answers
a baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45 with respect to the ground. (g=32 feet/
LiRa [457]

Answer:

hmax=81ft

Explanation:

Maximum height of the object is the highest vertical position along its trajectory.

The vertical velocity is equal to 0 (Vy = 0)

0=V_{y}-g*t=v_{0}*sin(\alpha)-g*th\\

we isolate th (needed to reach the maximum height hmax)

th = \frac{v_{0}*sin(\alpha)}{g}

The formula describing vertical distance is:

y = Vy * t-g* t^{2} / 2

So, given y = hmax and t = th, we can join those two equations together:

hmax = Vy* th-g*th^{2}/2

hmax =Vo^{2}*sin(\alpha )^{2}/(2*g)

if we launch a projectile from some initial height h all you need to do is add this initial elevation

hmax =h+Vo^{2}*sin(\alpha)^{2}/(2*g)

hmax =3+100^{2}*sin(45)^{2}/(2 * 32)=81 ft

6 0
3 years ago
A) An automobile light has a 1.0-A current when it is connected to a 12-V battery. Determine the resistance of the light.
kirill [66]

Answer:

The resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

Explanation:

Given:

(A)

Current I = 1 A

Voltage V = 12 V

For finding the resistance,

  V = IR

  R = \frac{V}{I}

  R = \frac{12}{1}

  R = 12Ω

(B)

For finding power delivered,

  P = I^{2} R

  P = (1) ^{2} \times 12

  P = 12 Watt

(C)

For finding the potential difference,

   V = IR

   V = 5 \times 10^{-3} \times 2

   V = 10 \times 10^{-3}

   V = 0.01 V

Therefore, the resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

4 0
3 years ago
a train traveling at 30m/s comes to a complete stop in 20 sec.how far did the train go during the deceleration
elixir [45]

Answer:

600m

Explanation:

v=30m/s

t=20s

s=VT

20×30=600m

3 0
3 years ago
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