1. True/False The Pressure Relief valve maintains the minimum pressure in the hydraulic circuit

A 1 turn coil carries has a radius of 9.8 cm and a magnetic moment of 6.2 X 10 -2 Am 2. What is the current through the coil?

Alexus [3.1K]

Answer:

The current through the coil is 2.05 A

Explanation:

Given;

number of turns of the coil, N = 1

radius of the coil, r = 9.8 cm = 0.098 m

magnetic moment of the coil, P = 6.2 x 10⁻² A m²

The magnetic moment is given by;

P = IA

Where;

I is the current through the coil

A is area of the coil = πr² = π(0.098)² = 0.03018 m²

The current through the coil is given by;

I = P / A

I = (6.2 x 10⁻² ) / (0.03018)

I = 2.05 A

Therefore, the current through the coil is 2.05 A

6 0

4 years ago

The outer surface of an engine is situated in a place where oil leakage can occur. When leaked oil comes in contact with ahot su

VladimirAG [237]

Answer:

TBC thickness of 4 mm is insufficient to prevent fire hazard

Explanation:

Given:

- Temperature of hot-fluid inner surface T_i = 333°C

- The convection coefficient hot-fluid h_i = 7 W/m^2K

- The thermal conductivity of engine cover k_1 = 14 W/mK

- The thickness of engine cover L_1 = 0.01 m

- The thermal conductivity of TBC layer k_2 = 1.1 W/mK ... (Typing error)

- The thickness of TBC layer L_2 = 0.004 m

- Temperature of ambient air outer surface T_o = 69°C

- The convection coefficient ambient air h_o = 7 W/m^2K

Find:

Would a TBC layer of 4 mm thickness be sufficient to keep the engine cover surface below autoignition temperature of 200°C to prevent fire hazard?

Solution:

- We will use thermal circuit analogy for the 1-D problem and steady state conduction with no heat generation in the cover or TBC layer.

The temperature at each medium interface and the Thermal resistance for each medium is given in the attachment schematic and circuit analogy.

- We will calculate the total heat flux for the entire system q:

q = ( T_i - T_o ) / R_total

- R_total is the equivalent thermal resistance of the entire circuit. Since all resistances are in series we have:

R_total = 1 / h_i + L_1 / k_1 + L_2 / k_2 + 1 / h_o

- Plug in the values and compute:

R_total = 1 / 7 + 0.01 / 14 + 0.004 / 1.1 + 1 / 7

R_total = 0.2900649351 T-m^2 / W

- Calculate the Total heat flux q:

q = ( 333 - 69 ) / 0.2900649351

q = 910.141 W / m^2

- Just like the total current in a circuit remains same, the total heat flux remains same. We will use the total heat flux q to calculate the temperature of outer engine surface T_2 as follows:

q = ( T_i - T_2 ) / R_i2

Where,

R_i2 = 1 / h_i + L_1 / k_1

R_i2 = 1 / 7 + 0.01 / 14 = 0.14357 T-m^2 / W

Hence,

( T_i - T_2 ) = q*R_i2

T_2 = T_i - q*R_i2

Plug the values in:

T_2 = 333 - 910.141*0.14357

T_2 = 202.33°C

- The outer surface of the engine cover has a temperature above T_ignition = 200°C. Hence, the TBC thickness of 4 mm is insufficient to prevent fire hazard

3 0

3 years ago

**Please Help, ASAP**

Novay_Z [31]

Explanation:

We need to rearrange the following formula for the values given in parenthesis.

(1) x+xy = y, (x)

taking x common in LHS,

x(1+y)=y

$x=\dfrac{y}{1+y}$

(2) x+y = xy, (x)

Subrtacting both sides by xy.

x+y-xy = xy-xy

x+y-xy = 0

x-xy=-y

x(1-y)=-y

$x=\dfrac{-y}{1-y}$

(3) x = y+xy, (x)

Subrating both sides by xy

x-xy = y+xy-xy

x(1-y)=y

$x=\dfrac{y}{1-y}$

(4) E = (1/2)mv^2-(1/2)mu^2, (u)

Subtracting both sides by (1/2)mv^2

E-(1/2)mv^2 = (1/2)mv^2-(1/2)mu^2-(1/2)mv^2

E-(1/2)mv^2 =-(1/2)mu^2

So,

$2(E-\dfrac{1}{2}mv^2)=-mu^2\\\\u^2=\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)\\\\u=\sqrt{\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)}\\\\u=\sqrt{\dfrac{2}{m}(\dfrac{1}{2}mv^2-E)}$

(5) (x^2/a^2)-(y^2/b^2) = 1, (y)

$\dfrac{x^2}{a^2}-1=\dfrac{y^2}{b^2}\\\\y^2=b^2(\dfrac{x^2}{a^2}-1)\\\\y=b\sqrt{\dfrac{x^2}{a^2}-1}$

(6) ay^2 = x^3, (y)

$y^2=\dfrac{x^3}{a}\\\\y=\sqrt{\dfrac{x^3}{a}}$

Hence, this is the required solution.

3 0

3 years ago

You have been assigned the task of reviewing the relief scenarios for a specific chemical reactor in your plant. You are current

postnew [5]

Answer:

$D=0.016m$

Explanation:

From the question we are told that:

Discharge Rate $F_r=0.5kgls$

Pressure $P=15Kpa$

Temperature $T=25=>298K$

Ambient pressure is 1 atm.

Generally the equation for Density is mathematically given by

$\rho=\frac{PM}{RT}$

$\rho=\frac{15*10^5*28.0134*10^{-3}}{8.314*298}$

$\rho=16.958kg/m^2$

Generally the equation for Flow rate is mathematically given by

$F_r=\mu A\sqrt{Q \rho P(\frac{2}{Q+1})^{\frac{Q+1}{Q-1}}}$

Where

$Q=Heat coefficient\ ratio\ of\ Nitrogen$

$Q=1.4$

$\mu= Discharge\ coefficient$

$\mu=0.68$

Therefore

$0.5=0.68 A\sqrt{1.4 16.958 15*10^{5}(\frac{2}{1.4+1})^{\frac{1.4+1}{1.4-1}}}$

$A=2.129*10^{-4}$

Where

$A=\frac{\pi}{4}D^2$

$\frac{\pi}{4}D^2=2.129*10^{-4}$

$D=0.016m$

8 0

3 years ago

What should Kimi have done to ensure that she was acting legally and ethically?

Mila [183]

Answer:

Kimi have done to ensure that she was acting legally and ethically by following the measure that are discussed below in detail.

Explanation:

Legal measures are those measures that are outlined in regulatory laws. Ethical measures are based on the human postulates of morality and wrongdoing. The distinctions between them are these: Legal measures are based on transcribed law, while ethical measures are based on human rights and violations.

8 0

3 years ago

1. True/False The Pressure Relief valve maintains the minimum pressure in the hydraulic circuit​

1. True/False The Pressure Relief valve maintains the minimum pressure in the hydraulic circuit