Answer back to question for la ,lot points

A converging-diverging nozzle has an area ratio of 5.9. (1) Determine the (P0/Pt) values corresponding to the 1st, 2nd, and 3rd

nata0808 [166]

Answer:

Check the explanation

Explanation:

The Total pressure is the overall of fixed or static pressure p, the dynamic pressure q, as well as gravitational head. Total pressure can also be referred to as the measure of the overall energy of the airstream, and is the same to static pressure plus velocity pressure.

kindly check the step by step solution in the attached image below to Determine the (P0/Pt) values corresponding to the 1st, 2nd, and 3rd critical points.

5 0

3 years ago

Express 2/16 in thirty-seconds

mafiozo [28]

Answer:

$\frac{2}{16}$ = $\frac{4}{32}$ in thirty seconds.

Explanation:

one thirty second is one part out of 32 equal section . It is used to describe amounts accurately.

$\frac{2}{16}$ can be easily expressed as $\frac{4}{32}$

3 0

2 years ago

11) If the evaporating pressure was 76 psig for r-22and the compressor inlet temperature was 65f, what would be the total superh

Karolina [17]

Saturated Pressure Temperature chart for R-22 shows 45 degF at 76 psig
65-45= 20 degF superheat

7 0

2 years ago

Does anybody know what plane this is? i saw it the other day doing a low pass through my community

Arlecino [84]

Answer:

Airbus A340-313

Explanation:

it is what it is

3 0

2 years ago

As part of a heat treatment process, cylindrical, 304 stainless steel rods of 100-mm diameter are cooled from an initial tempera

saveliy_v [14]

Answer:

Explanation:

Given that:

diameter = 100 mm

initial temperature = 500 ° C

Conventional coefficient = 500 W/m^2 K

length = 1 m

We obtain the following data from the tables A-1;

For the stainless steel of the rod $\overline T = 548 \ K$

$\rho = 7900 \ kg/m^3$

$K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K$

$\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\ B_i = \dfrac{h(\rho/4)}{K} \\ \\ =0.657$

Here, we can't apply the lumped capacitance method, since Bi > 0.1

$\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\$

$0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\ 0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81$

$t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\ t_f= 2162.5 \\ \\ t_f = 36 mins$

However, on a single rod, the energy extracted is:

$\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) ) \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\ \theta = 1.54 \times 10^7 \ J$

Hence, for centerline temperature at 50 °C;

The surface temperature is:

$T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}$

5 0

2 years ago