ENGINEERING COLLEGE

A mass of 5 kg of saturated liquid-vapor mixture of water is contained in a piston-cylinder device at 125 kPa. Initially, 2 kg of the water is in the liquid phase and the rest is in the vapor phase. Heat is now transferred to the water, and the piston, which is resting on a set of stops, starts moving when the pressure inside reaches 300 kPa. Heat transfer continues until the total volume increases by 20%. (a) Determine the initial temperature. (b) Determine the final temperature. (c) Determine the mass of liquid water when the piston first starts moving. (d) Determine the work done during this process in kJ.

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Answer 1

Answer:

Answer:

See attached pictures.

Explanation:

See attached pictures for detailed explanation.

The answer is "45.3 NM".

There at end of the movement, the forging force is given by

$\to F = Y * \pi * r^2 * [1 + ((2 \mu r)/(3h))]$

h is the final height.

$\to h = (100)/(2)= 50 \ mm$

The ultimate radius is determined by following a volume constancy law, which states that volumes before deformation measured amount following distortion.

$\to \pi * 75^2 * 2 * 100 = \pi * r^2 * 2 * 50\n\n\to 75^2 * 2 = r^2\n\n\to r^2 = 11250\n\n\to r = √(11250)\n\n\to r = 106 \ mm\n\n\to E = \In((100)/(50))\n\n\to E = 0.69\n\n$

You may deduce from the graph flow that $Y = 1000\ MPa$ , thus we use the formula.

$= 1000 * 3.14 * 0.106^2 * [1 + (( 2 * 0.2 * 0.106)/(3 * 0.05))]\n\n= 1000 * 3.14 * 0.011236 * [1 + (( 0.0424)/(0.15))]\n\n= 35.3 * 1.2826\n\n = 45.3 \ MN\n\n\n$

Therefore, the answer is "45.3 NM".

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Answer:

45.3 MN

Explanation:

The forging force at the end of the stroke is given by

F = Y.π.r².[1 + (2μr/3h)]

The final height, h is given as h = 100/2

h = 50 mm

Next, we find the final radius by applying the volume constancy law

volumes before deformation = volumes after deformation

π * 75² * 2 * 100 = π * r² * 2 * 50

75² * 2 = r²

r² = 11250

r = √11250

r = 106 mm

E = In(100/50)

E = 0.69

From the graph flow, we find that Y = 1000 MPa, and thus, we apply the formula

F = Y.π.r².[1 + (2μr/3h)]

F = 1000 * 3.142 * 0.106² * [1 + (2 * 0.2 * 0.106/ 3 * 0.05)]

F = 35.3 * [1 + 0.2826]

F = 35.3 * 1.2826

F = 45.3 MN

A furnace is shaped like a long equilateral triangular duct where the width of each side is 2 m. Heat is supplied from the base surface, whose emissivity is ε1 = 0.8, at a rate of 800 W/m2 while the side surfaces, whose emissivities are 0.5, are maintained at 500 K. Neglecting the end effects, determine the temperature of the base surface. Can you treat this geometry as a two-surface enclosure?

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Answer:

The geometry is treated as a two surface enclosure because the two surfaces have the same properties.

Let's take the base surface to be surface 1, while the side surfaces are surface 2.

Let's take the heat transfer expression:

$Q_1_2 = (\sigma[(T_1)^4 - (T_2)^4])/((1 - E_1)/(A_1 E_1) + (1)/(A_1 F_1_2) + (1-E_2)/(A_2 E_2))$

Where,

$\sigma = Boltmanz constant = 5.68*10^-^8$

$T_1$ = base temperature

$T_2$ = surface 2 temperature = 500K

$E_1$ = emissivity of surface 1 = 0.8

$E_2$ = emissivity of surface 2 = 0.5

$A$ = Area

$F_1_2$ = shape factor

Substituting figures in the equation, we have:

$800= (5.67*10^-^8[(T_1)^4 - (500)^4])/((1 - 0.8)/(2*0.8) + (1)/(2*1) + (1-0.5)/(4*0.5))$

$[(T_1)^4 - (500)^4] = (700)/(5.67*10^-^8)$

$(T_1)^4 = 1.234*10^1^0 + 6.25*10^1^0$

$T_1 = (7.484*10^1^0)^0^.^2^5$

$T_1 = 523.038 K$

The base temperature is 523.038 k

Air expands adiabatically through a nozzle from a negligible initial velocity to a final velocity of 300 m/s, what is the temperature drop of the air, if air is assumed to be an ideal gas for which CP = (7/2)R?

Answers

The temperature drop of air if air is assumed to be an ideal gas for which C_p = ⁷/₂R is; Δt = 1546 K

We are given;

Final velocity; v₂ = 300 m/s

C_p = ⁷/₂R

At constant pressure, the change in enthalpy is;

Δh = C_p × Δt

Now, from first law of thermodynamics;

h₂ + (v₂²/2) = h₁ + (v₁²/2)

We are told initial velocity is negligible and as such v₁ = 0 m/s

Thus;

h₂ + (v₂²/2) = h₁ + 0

(h₁ - h₂) = (v₂²/2)

Thus; Δh = v₂²/2

Finally;

C_p × Δt = v₂²/2

Δt = v₂²/2/(C_p)

Δt = (300²/2)/(⁷/₂R)

where R is ideal gas constant = 8.314 Kj/kg.mol

Thus;

Δt = (300²/2)/(⁷/₂ × 8.314)

Δt = 1546 K

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Answer:

ΔH+U²/2=0

and

ΔH= $C{p$ ×ΔT

∴to get the temperature drop of air, you make ΔT subject of the formula

ΔT=-U²/2Cp

=-300²/2× $(7)/(2)$ ×8.314

∴ΔT=-1546K

Explanation:

Consider the products you use and the activities you perform on a daily basis. Describe three examples that use both SI units and customary units for measurement.

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Three activities that I can do on a daily basis that involve both metric units (SI units) and customary units are: measuring the length of a door with a tape measure, which includes both SI units and customary units (like feet, inches, and centimeters); baking a cake that calls for one teaspoon (customary unit) of baking soda, which can also be converted to four grams (SI unit); and weighing myself on a weighing scale, which can be measured in pound and kilogram (metric unit).

Answer: Three examples of activities that I can perform on a daily basis that involves both metric units (SI units) and customary units include: measuring the length of a door using a tape measure, which includes both SI units and customary units (like feet, inches, and centimeters); baking a cake that requires one teaspoon (customary unit) of baking soda, which could also be converted into four grams (SI unit); weighing myself on a weighing scale, which can be measured by pounds (customary unit) or kilograms (metric unit).

Explanation:I big brain :) (Not Really I Just Wanted To Help) I hope this helped! ;)

Input resistance of a FET is very high due to A) forward-biased junctions have high impedance B) gate-source junction is reverse-biased C) drain-source junction is reverse-biased D) none of the above

Answers

Answer:

B) gate-source junction is reverse-biased

Explanation:

FET is described as an electric field that controls the specific current and is being applied to a "third electrode" which is generally known as "gate". However, only the electric field is responsible for controlling the "current flow" in a specific channel and then the particular device is being "voltage operated" that consists of high "input impedance".

In FET, the different "charge carriers" tend to enter a particular channel via "source" and exits through "drain".

A 26-tooth pinion rotating at a uniform 1800 rpm meshes with a 55-tooth gear in a spur gear reducer. Both pinion and gear are manufactured to a quality level of 10. The transmitted tangential load is 22 kN. Conditions are such that Km = 1.7. The teeth are standard 20-degree, full-depth. The module is 5 and the face width 62 mm. Determine the bending stress when the mesh is at the highest point of single tooth contact.

Answers

Answer:

The bending stress of the face tooth is $\sigma _(bg) = 502.82 MPa$

Explanation:

From the question we are told that

The number of tooth of the pinion is $N_t = 26 \ tooth$

The velocity of rotation is given as $\omega_p = 1800 rpm$

The number of tooth is of the gear is $N_g = 55 \ tooth$

The quality level is $Q_r = 10$

The transmitted tangential load is $F_T = 22\ kN$ = $22 KN * (1000N)/(1KN) = 22*10^3 N$

$k_m = 1.7$

The angle of the teeth is $\theta_t = 20^o$

The module is $M= 5$

The face width is $W_f = 62mm$

The diameter of the pinion is mathematically represented as

$d_p = M * N_t$

Substituting the values

$d_p = 5 *26$

$= 130 mm = (130)/(1000) = 0.130m$

The pitch line velocity is mathematically represented as

$V_t = (d_p )/(2) (2 \pi \omega_p)/(60)$

Substituting values

$= (0.130)/(2) * (2 * 3.142 * 1800 )/(60)$

$= 12.25\ m/s$

Generally the dynamic factor is mathematically represented as

$K_v = [(A)/(A +√(200V_t) ) ]^B$

Now B is a constant that is mathematically represented as

$B = ((12 -Q_r )^(2/3))/(4)$

substituting values

$= ((12- 10 )^(2/3))/(4)$

$=0.3968$

A is also a constant that is mathematically represented as

$A = 50 + 56(1 -B)$

Substituting values

$= 50 +56 (1- 0.3968)$

$= 83.779$

Substituting these value into the equation for dynamic factor we have

$K_v = [(83.779)/(83.779 + √(200 * 12.25) ) ]^(0.3968)$

$= 0.831$

The geometric bending factor for a 20° profile from table

"AGMA Bending Geometry Factor J for 20°, Full -Depth Teeth with HPSTC Loading , Table 2-9"

That corresponds to 55 tooth gear meshing with 26 pinion is

$J_g = 0.41$

the diameter pitch can be mathematically represented as

$p_d = (1)/(M)$

Substituting values

$p_d = (1)/(5)$

$=0.2mm^(-1)$

The mathematically representation for gear tooth bending stress in the teeth face is as follows

$\sigma_(bg) = (F_T \cdot p_d )/(W_f * J_g)(K_a K_(dt) )/(K_v) K_s K_B K_t ----(1)$

Where $W_t$ is the tangential load

$W_f$ is the face width

$K_a$ is the application factor this is obtained from table "Application Factors, Table 12-17 " and the value is $K_a$ = 1

$K_(dt)$ is the load distributed factor

$K_s$ is the size factor

$K_B$ is the rim thickness factor which is obtained for M which has a value 1

$K_t$ is the idler

Substituting values into equation 1

$\sigma_(bg) = (22*10^3 *0.2)/(62 * 0.41) * (1 * 1.7 )/(0.831) * 1 *1 *1.42$

$= 502.82 N/mm^2$

$= 502.82 * 1000 * (N)/(m^2)$

$= 502.82 MPa$

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The answer is "45.3 NM".

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