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galina1969 [7]
3 years ago
13

Shows a closed tank holding air and oil to which is connected a U-tube mercury manometer and a pressure gage. Determine the read

ing of the pressure gage, in lbf/in^2. (gage). The densities of the oil and mercury are 55 and 845, respectively, each in lb/ft^3. Let g=32.2ft/s^2.
Engineering
1 answer:
damaskus [11]3 years ago
3 0

Answer:

P_2-P_1=27209h

Explanation:

For pressure gage we can determine this by saying:

The closed tank with oil and air has a pressure of P₁ and the pressure of oil at a certain height in the U-tube on mercury is p₁gh₁. The pressure of mercury on the air in pressure gauge is p₂gh₂. The pressure of the gage is P₂.

P_1+p_1gh_1=p_2_gh_2+P_2

We want to work out P₁-P₂: Heights aren't given so we can solve it in terms of height: assuming h₁=h₂=h

P_1-P_2=p_1gh_1-p_2gh_2=(55)\cdot{32.2}h-845\cdot{32.2}h

P_2-P_1=27209h

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Nancy wants to buy a cooking stove that’s electrically insulated and resistant to heat. What material should she choose for the
astra-53 [7]

Answer:

Ceramics

Explanation:

Ceramics are used to make cooktops if you look it up and according to my engineering and design class ceramics have high heat resistance and melting points, and make good electrical insulators.

3 0
3 years ago
A 3-phase , 1MVA, 13.8kV/4160V, 60 Hz, transformer with Y-Delta winding connection is supplying a3-phase, 0.75 p.u. load on the
Tanya [424]

Answer:

a) 23.89 < -25.84 Ω

b) 31.38 < 25.84 A

c) 0.9323 leading

Explanation:

A) Calculate the load Impedance

current on load side = 0.75 p.u

power factor angle = 25.84

I_{load} = 0.75 < 25.84°

attached below is the remaining part of the solution

<u>B) Find the input current on the primary side in real units </u>

load current in primary = 31.38 < 25.84 A

<u>C) find the input power factor </u>

power factor = 0.9323 leading

<em></em>

<em>attached below is the detailed solution </em>

8 0
3 years ago
Given: A graphite-moderated nuclear reactor. Heat is generated uniformly in uranium rods of 0.05m diameter at the rate of 7.5 x
sineoko [7]

Answer:

The maximum temperature at the center of the rod is found to be 517.24 °C

Explanation:

Assumptions:

1- Heat transfer is steady.

2- Heat transfer is in one dimension, due to axial symmetry.

3- Heat generation is uniform.

Now, we consider an inner imaginary cylinder of radius R inside the actual uranium rod of radius Ro. So, from steady state conditions, we know that, heat generated within the rod will be equal to the heat conducted at any point of the rod. So, from Fourier's Law, we write:

Heat Conduction Through Rod = Heat Generation

-kAdT/dr = qV

where,

k = thermal conductivity = 29.5 W/m.K

q = heat generation per unit volume = 7.5 x 10^7 W/m³

V = volume of rod = π r² l

A = area of rod = 2π r l

using these values, we get:

dT = - (q/2k)(r dr)

integrating from r = 0, where T(0) = To = Maximum center temperature, to r = Ro, where, T(Ro) = Ts = surface temperature = 120°C.

To -Ts = qr²/4k

To = Ts + qr²/4k

To = 120°C + (7.5 x 10^7 W/m³)(0.025 m)²/(4)(29.5 W/m.°C)

To = 120° C + 397.24° C

<u>To = 517.24° C</u>

5 0
3 years ago
a) find the state-space representation of the system. b) is the system controllable? why? c) is the system observable? why
mart [117]

If a controlled input can transfer (alter) the control system's initial states to some other desired states in a finite amount of time, the control system is said to be controllable.

Using Kalman's test, we can determine whether a control system is controllable. The evolution model for the state variables (time-varying unknowns) and the observation model, which connects the observations to the state variables, make up the state space representation of a dynamical system. The capacity to move a system about in its full configuration space using just specific permitted actions is generally referred to as controllability. The precise definition changes slightly depending on the model type or framework used.

Learn more about control here-

brainly.com/question/28540307

#SPJ4

5 0
1 year ago
Wright Company deposits all cash receipts on the day when they are received and it makes all cash payments by check. At the clos
alina1380 [7]

Answer:

                                              Wright Company

                                             Bank Reconciliation

                                                 May 31, 2013

Credit side                                                                                   Debit side

Bank statement $26200                 |                          Book balance $27900

<em>Add;                                                    </em>

Deposit on May 31 $6400

Bank error $420

Sub-total=$33020

Deductions;                                        |                       Deduct

ions

Outstanding checks $5800              |                 Bank service charge $120

Adjusted bank balance $27220       |                  NSF check $560

                                                                             Total deduction $680

                                                      Adjusted book balance $27220

8 0
3 years ago
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