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3 years ago

What is the saturation pressure, Vf, and Vg for saturated water at a temperature of 287 °C?

Engineering

1 answer:

Dvinal [7]3 years ago

6 0

Answer:

Vf = specific volume of saturated liquid = 0.0217158 ft^3/lb

Vg = specific volume of saturated steam = 0.430129 ft^3/lb

Explanation:

Given data:

water temperature is given as 287-degree Celcius

we have to find Vf and Vg

Vf = specific volume of saturated liquid

Vg = specific volume of saturated steam

we know that from the saturated steam table we can find these value

therefore for temperature 287-degree Celcius

Vf = specific volume of saturated liquid = 0.0217158 ft^3/lb

Vg = specific volume of saturated steam = 0.430129 ft^3/lb

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Answer:

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3 years ago

The sliders A and B are connected by a light rigid bar of length l = 20 in. and move with negligible friction in the slots, both

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Answer:

Explanation:

Given:

- The Length of the rigid bar L = 20 in

- The position of slider a, x_a = 16 in

- The position of slider b, y_b

- The velocity of slider a, v_a = 3 ft /s

- The velocity of slider b, v_b

- The acceleration of slider a, a_a

- The acceleration of slider b, a_b

Find:

-Determine the acceleration of each slider and the force in the bar at this instant.

Solution:

- The relationship between the length L of the rod and the positions x_a and x_b of sliders A & B is as follows:

L^2 = x_a^2 + y_b^2 ....... 1

y_b = sqrt( 20^2 - 16^2 )

y_b = 12

- The velocity expression can derived by taking a derivation of Eq 1 with respect to time t:

0 = 2*x_a*v_a + 2*y_b*v_b

0 = x_a*v_a + y_b*v_b ..... 2

0 = 16*36 + 12*v_b

v_b = - 48 in /s = -4 ft/s

- Similarly, the acceleration expression can be derived by taking a derivative of Eq 2 with respect to time t:

0 = v_a^2 + x_a*a_a + v_b^2 + y_b*a_b

0 = 9 + 4*a_a/3 + 16 + a_b

4*a_a/3 + a_b = -25

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- Use dynamics on each slider. For Slider A, Apply Newton's second law of motion in x direction:

F_x = m_a*a_a

P - R_r*16/20 = m_a*a_a

- For Slider B, Apply Newton's second law of motion in y direction:

F_y = m_b*a_b

- R_r*12/20 = m_b*a_b

- Combine the two dynamic equations:

P - 4*m_b*a_b / 3 = m_a*a_a

3P = 3*m_a*a_a + 4*m_b*a_b ... 4

- Where, P = Is the force acting on slider A

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3 years ago

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3 years ago

A tubular reactor has been sized to obtain 98% conversion and to process 0.03 m^3/s. The reaction is a first-order irreversible

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Answer:

The conversion in the real reactor is = 88%

Explanation:

conversion = 98% = 0.98

process rate = 0.03 m^3/s

length of reactor = 3 m

cross sectional area of reactor = 25 dm^2

pulse tracer test results on the reactor :

mean residence time ( tm) = 10 s and variance (∝2) = 65 s^2

note: space time (t) =

t = $\frac{A*L}{Vo}$ Vo = flow metric flow rate , L = length of reactor , A = cross sectional area of the reactor

therefore (t) = $\frac{25*3*10^{-2} }{0.03}$ = 25 s

since the reaction is in first order

X = 1 - $e^{-kt}$

$e^{-kt}$ = 1 - X

kt = In $\frac{1}{1-X}$

k = In $\frac{1}{1-X}$ / t

X = 98% = 0.98 (conversion in PFR ) insert the value into the above equation then

K = 0.156 $s^{-1}$

Calculating Da for a closed vessel

; Da = tk

= 25 * 0.156 = 3.9

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0.65 = $\frac{2}{Per} - \frac{2}{Per^2} ( 1 - e^{-per})$

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3 years ago