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marin [14]
2 years ago
15

A ____ is either in the pressure reducer or in the downstream side of the system to ensure that the control air pressure does no

t exceed about 30 psig. Group of answer choices
Engineering
1 answer:
Alika [10]2 years ago
3 0

Answer:

A relief valve is either in the pressure reducer or in the downstream side of the system to ensure that the control air pressure does not exceed about 30 psig.

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Two sections of a pressure vessel are to be held together by 5/8 in-11 UNC grade 5 bolts. You are told that the length of the bo
DiKsa [7]

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

7 0
3 years ago
A thin metal disk of mass m=2.00 x 10^-3 kg and radius R=2.20cm is attached at its center to a long fiber. When the disk is turn
topjm [15]

Answer:

k = 1.91 × 10^-5 N m rad^-1

Workings and Solution to this question can be viewed in the screenshot below:

6 0
3 years ago
Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of "84" m3/min and exits at 12
Nina [5.8K]

Answer:

W = - 184.8 kW

Explanation:

Given data:

P_1 = 1.05 bar

T_1 = 300K

\dot V_1 = 84 m^3/min

P_2 = 12 bar

T_2 = 400 K

We know that work is done as

W = - [ Q + \dor m[h_2 - h_1]]

forP_1 = 1.05 bar,  T_1 = 300K

density of air is 1.22 kg/m^3 and h_1 = 300 kJ/kg

for P_2 = 12 bar, T_2 = 400 K

h_2 = 400 kj/kg

\dot m = \rho \times \dor v_1 = 1.22 \frac{84}{60} =1.708 kg/s

W = -[14 + 1.708[400-300]]

W = - 184.8 kW

8 0
3 years ago
An ideal gas turbine operates using air coming at 355C and 350 kPa at a flow rate of 2.0 kg/s. Find the rate work output
GuDViN [60]

Answer:

The rate of work output = -396.17 kJ/s

Explanation:

Here we have the given parameters

Initial temperature, T₁ = 355°C = 628.15 K

Initial pressure, P₁ = 350 kPa

h₁ = 763.088 kJ/kg

s₁ = 4.287 kJ/(kg·K)

Assuming an isentropic system, from tables, we look for the saturation temperature of saturated air at 4.287 kJ/(kg·K) which is approximately

h₂ = 79.572 kJ/kg

The saturation temperature at the given

T₂ = 79°C

The rate of work output \dot W = \dot m×c_p×(T₂ - T₁)

Where;

c_p = The specific heat of air at constant pressure = 0.7177 kJ/(kg·K)

\dot m =  The mass flow rate = 2.0 kg/s

Substituting the values, we have;

\dot W = 2.0 × 0.7177 × (79 - 355) = -396.17 kJ/s

\dot W = -396.17 kJ/s

7 0
4 years ago
a piping system has an internal air pressure of 1,500 kpa. In addition to being subject to the air pressure, the piping supports
Alik [6]

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem.

5 0
4 years ago
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