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Genrish500 [490]
2 years ago
6

What’s the answer please help

Engineering
1 answer:
irina1246 [14]2 years ago
4 0

Answer:

It B

Explanation:

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Answer:

I don't know sorry

Explanation:

5 0
3 years ago
A converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a te
Artyom0805 [142]

Explanation:

a converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a temperature of 360 K. For isentropic flow of an ideal gas with k = 1.4 and the gas constant R = Ru/MW = 287 J/kg-K, determine the mass flow rate in kg/s and the exit Mach number for back pressures

100% (3 ratings)

A_2 = 0.001 m^2 P_1 = 1 MPa, T_1 = 360 k P_2 = 500 kpa p^gamma - 1/gamma proportional T (1000/500)^1.4 - 1/1.4 = (360/T_2) 2^4/14 = 360/T_2 T_2

3 0
2 years ago
The compressed-air tank has an inner radius r and uniform wall thickness t. The gage pressure inside the tank is p and the centr
Sedaia [141]

Answer:

Explanation:

Given that:

The Inside pressure (p) = 1402 kPa

= 1.402 × 10³ Pa

Force (F) = 13 kN

= 13 × 10³ N

Thickness (t) = 18 mm

= 18 × 10⁻³ m

Radius (r) = 306 mm

= 306 × 10⁻³ m

Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)

Then;

the state of the plane stress can be expressed as follows:

(\sigma_ x)  = \dfrac{Pd}{4t}+ \dfrac{F}{2 \pi rt}

Since d = 2r

Then:

(\sigma_ x)  = \dfrac{Pr}{2t}+ \dfrac{F}{2 \pi rt}

(\sigma_ x)  = \dfrac{1402 \times 306 \times 10^3}{2(18)}+ \dfrac{13 \times 10^3}{2 \pi \times 306\times 18 \times 10^{-3} \times 10^{-3}}

(\sigma_ x)  = \dfrac{429012000}{36}+ \dfrac{13000}{34607.78467}

(\sigma_ x)  = 11917000.38

(\sigma_ x)  = 11.917 \times 10^6 \ Pa

(\sigma_ x)  = 11.917 \ MPa

\sigma_y = \dfrac{pd}{2t} \\ \\ \sigma_y = \dfrac{pr}{t} \\ \\  \sigma _y = \dfrac{1402\times 10^3 \times 306}{18} \ N/m^2 \\ \\ \sigma _y = 23.834 \times 10^6 \ Pa \\ \\ \sigma_y = 23.834 \ MPa

When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.

Thus;

\tau _{xy} =0

3 0
3 years ago
Two transmission belts pass over a double-sheaved pulley that is attached to an axle supported by bearings at A and D. The radiu
Fynjy0 [20]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

6 0
3 years ago
Refrigerant 134a enters an air conditioner compressor at 4 bar, 208C, and is compressed at steady state to 12 bar, 808C. The vol
SCORPION-xisa [38]

Answer:

heat transfer rate is -15.71 kW

Explanation:

given data

Initial pressure  = 4 bar

Final pressure  = 12 bar

volumetric flow rate = 4 m³ / min

work input to the compressor = 60 kJ per kg

solution

we use here super hated table for 4 bar and 20 degree temperature and 12 bar and 80 degree is

h1 = 262.96 kJ/kg

v1 = 0.05397 m³/kg

h2 = 310.24 kJ/kg

and here mass balance equation will be

m1  = m2

and mass flow equation is express as

m1 = \frac{A1\times V1}{v1}       .......................1

m1 = \frac{4\times \frac{1}{60}}{0.05397}  

m1 = 1.2353 kg/s

and here energy balance equation is express as

0 = Qcv - Wcv + m × [ ( h1-h2) + \frac{v1^2-v2^2}{2} + g (z1-z2) ]      ....................2

so here Qcv will be

Qcv =  m × [  \frac{Wcv}{m} + (h2-h1)  ]    ......................3

put here value and we get

Qcv =  1.2353 × [ {-60}+ (310.24-262.96) ]

Qcv =  -15.7130 kW

so here heat transfer rate is -15.71 kW

6 0
3 years ago
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