A truss is subjected to three loads. The truss is supported by a roller at A and by a pin joint at B. What is most nearly the reaction force at B

Answer:

the temperature of the aluminum at this time is 456.25° C

Explanation:

Given that:

width w of the aluminium slab = 0.05 m

the initial temperature [tex]T_1[/tex] = 25° C

[tex]T{\infty} =600^0C[/tex]

h = 100 W/m²

The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;

density ρ = 2702 kg/m³

thermal conductivity k = 231 W/m.K

Specific heat c = 1033 J/Kg.K

Let's first find the Biot Number Bi which can be expressed by the equation:

[tex]Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{h \dfrac{w}{2}}{k}[/tex]

[tex]Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{100 \times \dfrac{0.05}{2}}{231}[/tex]

[tex]Bi = \dfrac{2.5}{231}[/tex]

Bi = 0.0108

The time constant value [tex]\tau_t[/tex] is :

[tex]\tau_t = \dfrac{pL_cc}{h} \\ \\ \tau_t = \dfrac{p \dfrac{w}{2}c}{h}[/tex]

[tex]\tau_t = \dfrac{2702* \dfrac{0.05}{2}*1033}{100}[/tex]

[tex]\tau_t = \dfrac{2702* 0.025*1033}{100}[/tex]

[tex]\tau_t = 697.79[/tex]

Considering Lumped capacitance analysis since value for Bi is less than 1

Then;

[tex]Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}][/tex]

where;

[tex]Q = -\Delta E _{st}[/tex] which correlates with the change in the internal energy of the solid.

So;

[tex]Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]= -\Delta E _{st}[/tex]

The maximum value for the change in the internal energy of the solid  is :

[tex](pVc)\theta_1 = -\Delta E _{st}max[/tex]

By equating the two previous equation together ; we have:

[tex]\dfrac{-\Delta E _{st}}{\Delta E _{st}{max}}= \dfrac{ (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]} { (pVc)\theta_1}[/tex]

Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75

Thus;

[tex]0.75= [1-e^{\dfrac {-t}{ \tau_1}}]}[/tex]

So;

[tex]0.75= [1-e^{\dfrac {-t}{ 697.79}}]}[/tex]

[tex]1-0.75= [e^{\dfrac {-t}{ 697.79}}]}[/tex]

[tex]0.25 = e^{\dfrac {-t}{ 697.79}}[/tex]

[tex]In(0.25) = {\dfrac {-t}{ 697.79}}[/tex]

[tex]-1.386294361= \dfrac{-t}{697.79}[/tex]

t = 1.386294361 × 697.79

t = 967.34 s

Finally; the temperature of Aluminium is determined as follows;

[tex]\dfrac{T - T _{\infty}}{T_1-T_{\infty}}= e ^ {\dfrac{-t}{\tau_t}}[/tex]

[tex]\dfrac{T - 600}{25-600}= e ^ {\dfrac{-967.34}{697.79}[/tex]

[tex]\dfrac{T - 600}{25-600}= 0.25[/tex]

[tex]\dfrac{T - 600}{-575}= 0.25[/tex]

T - 600 = -575 × 0.25

T - 600 = -143.75

T = -143.75 + 600

T = 456.25° C

Hence; the temperature of the aluminum at this time is 456.25° C

Answer:

[tex]\mathbf{F_{AB} = 13785.06 N }[/tex]

[tex]\mathbf{F_{AC} = -5062.38 N }[/tex]

[tex]\mathbf{F_{AD} = -5062.38 N }[/tex]

Explanation:

From the given information:

Let calculate the position vector of AB, AC, and AD

To start with AB; in order to calculate the position vector of AB ; we have:

[tex]r_{AB}^{\to} = r _{OA}^{\to} - r_{OB}^{\to} \\ \\ r_{AB}^{\to} = (2.6 \ \hat i + 2.6 \ \hat j - 2.2 \ \hat k ) - ( 1.5 \ \hat i + \ 1. 5 \hat j ) \\ \\ r_{AB}^{\to} = ( 2.6 \ \hat i - 1.5 \ \hat i + 2.6 \ \hat j - 1.5 \ \hat j - 2.2 \ \hat k) \\ \\ r_{AB}^{\to} = (1.1 \ \hat i + 1.1 \ \hat j - 2.2 \ \hat k ) m[/tex]

To calculate the position vector of AC; we have:

[tex]r_{AC}^{\to} = r _{OA}^{\to} - r_{OC}^{\to} \\ \\ r_{AC}^{\to} = (2.6 \ \hat i + 2.6 \ \hat j - 2.2 \ \hat k ) - ( 2\ \hat i + \ \hat j - 1.2 \ \hat k) \\ \\ r_{AC}^{\to} = ( 2.6 \ \hat i - 2\ \hat i + 2.6 \ \hat j - \ \hat j - 2.2 \ \hat k + 1.2 \ \hat k) \\ \\ r_{AC}^{\to} = (0.6 \ \hat i + 1.6 \ \hat j - \ \hat k ) m[/tex]

To calculate the position vector of AD ; we have:

[tex]r_{AD}^{\to} = r _{OA}^{\to} - r_{OD}^{\to} \\ \\ r_{AC}^{\to} = (2.6 \ \hat i + 2.6 \ \hat j - 2.2 \ \hat k ) - ( \hat i + \ 2 \hat j - 1.2 \ \hat k) \\ \\ r_{AD}^{\to} = ( 2.6 \ \hat i - \hat i + 2.6 \ \hat j - 2 \ \hat j - 2.2 \ \hat k + 1.2 \ \hat k) \\ \\ r_{AD}^{\to} = (1.6 \ \hat i + 0.6 \ \hat j - \ \hat k ) m[/tex]

However; let's calculate the force in AB, AC and AD in their respective unit vector form;

To start with unit vector AB by using the following expression; we have:

[tex]F_{AB}^{\to} = F_{AB} \dfrac{ r _{AB}^{\to} }{|r_{AB}^{\to}} \\ \\ \\ F_{AB}^{\to} = F_{AB} \dfrac{(1.1 \ \hat i + 1.1 \ \hat j - 2.2 \ \hat k ) }{\sqrt{ (1.1)^2 + (1.1)^2 + (-2.2 )^2 }} \\ \\ \\ F_{AB}^{\to} = F_{AB} \dfrac{(1.1 \ \hat i + 1.1 \ \hat j - 2.2 \ \hat k ) }{ \sqrt{7.26}} \\ \\ \\ F_{AB}^{\to} = F_{AB} \dfrac{(1.1 \ \hat i + 1.1 \ \hat j - 2.2 \ \hat k ) }{ 2.6944} \\ \\ \\ F_{AB}^{\to} = F_{AB} (0.408 \ \hat i+ 0.408 \ \hat j - 0.8165 \ \hat k ) N\\[/tex]

The force AC in unit vector form is ;

[tex]F_{AC}^{\to} = F_{AC} \dfrac{ r _{AC}^{\to} }{|r_{AC}^{\to}} \\ \\ \\ F_{AC}^{\to} = F_{AC} \dfrac{(0.6 \ \hat i + 1.6 \ \hat j - \ \hat k ) }{\sqrt{ (0.6)^2 + (1.6)^2 + (-1 )^2 }} \\ \\ \\ F_{AC}^{\to} = F_{AC} \dfrac{(0.6 \ \hat i + 1.6 \ \hat j - \ \hat k ) }{ \sqrt{3.92}} \\ \\ \\ F_{AC}^{\to} = F_{AC} \dfrac{(0.6 \ \hat i + 1.6 \ \hat j - \ \hat k ) }{1.9798} \\ \\ \\ F_{AC}^{\to} = F_{AC} (0.303 \ \hat i+ 0.808 \ \hat j - 0.505 \ \hat k ) N\\[/tex]

The force AD in unit vector form is ;

[tex]F_{AD}^{\to} = F_{AD} \dfrac{ r _{AD}^{\to} }{|r_{AD}^{\to}|} \\ \\ \\ F_{AD}^{\to} = F_{AD} \dfrac{(1.6 \ \hat i + 0.6 \ \hat j - \ \hat k ) }{\sqrt{ (1.6)^2 + (0.6)^2 + (-1 )^2 }} \\ \\ \\ F_{AD}^{\to} = F_{AD} \dfrac{(1.6 \ \hat i + 0.6 \ \hat j - \ \hat k ) }{ \sqrt{3.92}} \\ \\ \\ F_{AD}^{\to} = F_{AD} \dfrac{(1.6 \ \hat i + 0.6 \ \hat j - \ \hat k ) }{1.9798} \\ \\ \\ F_{AD}^{\to} = F_{AD} (0.808 \ \hat i+ 0.303 \ \hat j - 0.505 \ \hat k ) N\\[/tex]

Similarly ; the weight of the lunar Module is:

W = mg

where;

mass = 13500 kg

acceleration due to gravity=  1.82 m/s²

W = 13500 × 1.82

W = 24,570 N

Also. we known that the load is shared by four landing gears; Thus, the vertical reaction force exerted by the ground on each landing gear can be expressed as:

[tex]R =\dfrac{W}{4}[/tex]

[tex]R =\dfrac{24,570}{4}[/tex]

R = 6142.5 N

Now; the reaction force at point A in unit vector form is :

[tex]R^{\to} = Rk^{\to} \\ \\ R^{\to} = (6142.5 \ k ^{\to}) \ N[/tex]

Using the force equilibrium at the meeting point of the coordinates at A.

[tex]\sum F^{\to} = 0[/tex]

[tex]F_{AB}^{\to} +F_{AC}^{\to} + F_{AD}^{\to} + R^{\to} =0[/tex]

[tex][F_{AB} (0.408 \ \hat i + 0.408 \ \hat j - 0.8165 \ \hat k ) N + F_{AC} (0.303 \ \hat i + 0.808 \ \hat j - 0.505 \ \hat k ) N + F_{AD} (0.808 \ \hat i + 0.303 \ \hat j - 0.505 \ \hat k) N + (6142.5 \ k^ \to ) ][/tex]

[tex]= [ ( 0.408 F_{AB} +0.303 F_{AC} + 0.808F_{AD}) \hat i + (0.408 F_{AB}+0.808F_{AC}+0.303F_{AD}) \hat j + (-0.8165 F_{AB} -0.505F_{AC} -0.505 F_{AD} +6142.5 ) k ^ \to ] = 0[/tex]

From above; we need to relate and equate each coefficients i.e i ,j, and [tex]k ^ \to[/tex] on both sides ; so, we can re-write that above as;

[tex]0.408 F_{AB} +0.303 F_{AC} + 0.808F_{AD}) =0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ --- (1) \\ \\ 0.408 F_{AB}+0.808F_{AC}+0.303F_{AD}) =0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ --- (2) \\ \\ -0.8165 F_{AB} -0.505F_{AC} -0.505 F_{AD} +6142.5 = 0 --- (3)[/tex]

Making rearrangement and solving by elimination method;

[tex]\mathbf{F_{AB} = 13785.06 N }[/tex]

[tex]\mathbf{F_{AC} = -5062.38 N }[/tex]

[tex]\mathbf{F_{AD} = -5062.38 N }[/tex]

The force vector of each member, depends on the magnitude of the

force and the unit vector of the member.

Responses:

The force supported by the members are;

Resolving the given members into unit vectors gives;

[tex]\hat u_{AB} = \mathbf{\dfrac{(2.6 - 1.5) \cdot \hat i + (2.6 - 1.5)\cdot \hat j + (-2.2)\cdot \hat k }{\sqrt{(2.6- 1.5)^2 + (2.6 - 1.5)^2 + (-2.2)^2}}}[/tex][tex]\dfrac{(2.6 - 1.5) \cdot \hat i + (2.6 - 1.5)\cdot \hat j + (-2.2)\cdot \hat k }{\sqrt{(2.6- 1.5)^2 + (2.6 - 1.5)^2 + (-2.2)^2}}= 0.40825 \cdot \hat i + 0.40825\cdot \hat j - 0.81625\cdot \hat k[/tex]

Similarly, we have;

[tex]\hat u_{AC} =\mathbf{ \dfrac{(2.6 - 2) \cdot \hat i + (2.6 - 1)\cdot \hat j + (-2.2+1.2)\cdot \hat k }{\sqrt{(2.6- 2)^2 + (2.6 - 1)^2 + (-2.2+1.2)^2}}}[/tex]

[tex]\dfrac{(2.6 - 2) \cdot \hat i + (2.6 - 1)\cdot \hat j + (-2.2+1.2)\cdot \hat k }{\sqrt{(2.6- 2)^2 + (2.6 - 1)^2 + (-2.2+1.2)^2}} =\dfrac{0.6\cdot \hat i +1.6\cdot \hat j -1\cdot \hat k }{\sqrt{0.6^2 + 1.6^2 + (-1.)^2}}[/tex]

[tex]\dfrac{0.6\cdot \hat i +1.6\cdot \hat j -1\cdot \hat k }{\sqrt{0.6^2 + 1.6^2 + (-1.)^2}}= 0.303046\cdot \hat i + 0.80812\cdot \hat j - 0.50508\cdot \hat k[/tex]

[tex]\hat u_{AD} =\mathbf{ \dfrac{(2.6 - 1) \cdot \hat i + (2.6 -2)\cdot \hat j + (-2.2 + 1.2)\cdot \hat k }{\sqrt{(2.6-1)^2 + (2.6 -2))^2 + (-2.2 + 1.2)^2}}}[/tex]

[tex]\hat u_{AD} =\mathbf{0.80812\cdot \hat i+ 0.303046\cdot \hat j - 0.50508\cdot \hat k}[/tex]

The forces are therefore;

[tex]\vec F_{AB} =\mathbf{ F_{AB} \cdot \left ( 0.40825 \cdot \hat i + 0.40825\cdot \hat j - 0.81625\cdot \hat k \right)}[/tex]

[tex]\vec F_{AC} =\mathbf{ F_{AC} \cdot \left (0.303046\cdot \hat i + 0.80812\cdot \hat j - 0.50508\cdot \hat k\right)}[/tex]

[tex]\vec F_{AD} = \mathbf{F_{AD} \cdot \left (0.80812\cdot \hat i+ 0.303046\cdot \hat j - 0.50508\cdot \hat k\right)}[/tex]

[tex]Weight \ on \ the \ assembly = \dfrac{13,500 \, kg \times 1.82 \, m/s^2}{4} = 6,142.5 \, \hat k N[/tex]

Which gives;

[tex]\mathbf{0.40825 \cdot \hat i \cdot F_{AB}}[/tex] + [tex]0.303046\cdot \hat i \cdot F_{AC}[/tex] + [tex]0.80812\cdot \hat i \cdot F_{AD}[/tex] = 0

[tex]0.40825 \cdot \hat j \cdot F_{AB}[/tex] + [tex]0.80812\cdot \hat j \cdot F_{AC}[/tex] + [tex]0.303046 \cdot \hat j \cdot F_{AD}\left[/tex] = 0

[tex]-0.81625\cdot \hat k \cdot F_{AB}[/tex] - [tex]0.50508\cdot \hat k \cdot F_{AC}[/tex] - [tex]0.50508\cdot \hat k \cdot F_{AD}[/tex] +  [tex]\mathbf{6,142.5 \, \hat k}[/tex] = 0

Which gives;

[tex]-0.81625\cdot \hat k \cdot F_{AB}[/tex] - [tex]0.50508\cdot \hat k \cdot F_{AC}[/tex] - [tex]0.50508\cdot \hat k \cdot F_{AD}[/tex] =  [tex]-6,142.5 \, \hat k[/tex]

Solving gives;

Learn more about unit vectors here:

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Answer:

According to the Principles of Project management, the three factors which dominate the lifecycle of any project are:

The relationship between the three is usually governed by trade-offs.

Explanation:

In simple term, in executing a project, one must deal with the factors mentioned above.

It is always desirous for a project to be finished within a stipulated time. If the time required is reduced inconsiderably, it will most likely incur more cost and even impact performance.

On the other hand, if the project is cost-sensitive and is executed to a very minimalistic budget, performance will be impacted and it may take a protracted amount of time.

In addition to the above, if the principal decides to change the original design of the project, the performance expected is altered. This will attract additional time as well as cost.

It is possible for any of the above factors to be renegotiated and readjusted at any time during the project. It usually is a trade-off.. that is one for the other.

Cheers!

Given Information:

Frequency = f = 60 Hz

Transformer output voltage = Vrms = 6.3 V

Diode voltage drop = Vd = 1 V

Ripper voltage = Vr = 0.25 V

Load resistance = R = 0.5 Ω

Required Information:

a) dc output voltage = V₀ = ?

b) Capacitane = C = ?

Answer:

a) dc output voltage = V₀ = 2.52 V

b) Capacitane = C = 0.336 F

Explanation:

a) The average or dc output voltage of a half-wave rectifier is given by

[tex]V_0 = V_p/\pi[/tex]

Where Vp is given by

[tex]V_p = (V_{rms} \times \sqrt{2}) - V_d \\\\V_p = (6.3 \times \sqrt{2}) - 1 \\\\V_p = 8.91 - 1 \\\\V_p = 7.91 \: V \\\\[/tex]

So, the dc output voltage is

[tex]V_0 = 7.91/\pi \\\\V_0 = 2.52 \: V[/tex]

b) The minimum value of C required to maintain the ripple voltage to less than 0.25 V is given by

[tex]$ C = \frac{I}{Vr \cdot f} $[/tex]

Where I is current, Vr is the ripple voltage and f is the frequency

[tex]$ I = \frac{V_0}{R} $[/tex]

[tex]$ I = \frac{2.52}{0.5} $[/tex]

[tex]I = 5.04 \: A[/tex]

[tex]$ C = \frac{5.04}{0.25 \cdot 60} $[/tex]

[tex]C = 0.336 \: F[/tex]

Therefore, 0.336 F is the minimum value of capacitance required to maintain the ripple voltage to less than 0.25 V

Answer:

critical clearing angle = 70.3°

Explanation:

Generator operating at = 50 Hz

power delivered = 1 pu

power transferable when there is a fault = 0.5 pu

power transferable before there is a fault = 2.0 pu

power transferable after fault clearance = 1.5 pu

using equal area criterion to determine the critical clearing angle

Attached is the power angle curve diagram and the remaining part of the solution.

The power angle curve is given as

= Pmax sinβ

therefore :  2sinβo = Pm

                   2sinβo = 1

                   sinβo = 0.5 pu

                   βo = [tex]sin^{-1} (0.5) = 30[/tex]⁰

also ;   1.5sinβ1 = 1

               sinβ1 = 1/1.5

               β1 = [tex]sin^{-1} (\frac{1}{1.5} )[/tex] = 41.81⁰

∴ βmax = 180 - 41.81  = 138.19⁰

attached is the remaining solution

The critical clearing angle = [tex]cos^{-1} 0.3372[/tex]  ≈ 70.3⁰

Answer:

No, the velocity profile does not change in the flow direction.

Explanation:

In a fluid flow in a circular pipe, the boundary layer thickness increases in the direction of flow, until it reaches the center of the pipe, and fill the whole pipe. If the density, and other properties of the fluid does not change either by heating or cooling of the pipe, then the velocity profile downstream becomes fully developed, and constant, and does not change in the direction of flow.

Answer:

The answer is B. Avoid making a large wake.

Explanation:

When passing a fishing boat it is important to maintain a minimal wake due to the dangers a large wake could pose to the fishing boat you are passing, it is part of maintaining safety on the water.

You can not pass on the sides with the fishing lines also, and you are supposed to communicate to the fishing boat before taking the appropriate action.

Answer:

2.47  ft

Explanation:

Given that:

The initial temperature of air = 80°F

Diameter of the pipe = 72 ft

average velocity [tex]v_{air}[/tex] of the air flow through the pipe =  34 ft/s

The objective is to determine the diameter of the  pipe to  be used to move water at:

At a temperature = 60°F   &

An average velocity [tex]v_{water}[/tex] of 71 ft/s

Assuming Reynolds number similarity is enforced;

where :

kinematic viscosity (V_air) of air at 80 °F  (V_air)  = 1.69 × 10⁻⁴ ft²/s

kinematic viscosity of water  at 60 °F (V_water) = 1.21 × 10⁻⁵ ft²/s

The diameter of the pipe can be calculated by using the expression:

[tex]D_{water} = \dfrac{V_{water}}{V_{air}}*\dfrac{v_{air}}{v_{water}}* D_{air}[/tex]

[tex]D_{water} = \dfrac{1.21*10^{-5} \ ft^2/s}{1.69*10^{-4} \ ft^2/s}*\dfrac{34 \ ft/s}{71 \ ft/s}* 72 \ ft[/tex]

[tex]D_{water} =[/tex] 2.4686  ft

[tex]D_{water} =[/tex] 2.47 ft   ( to two decimal places)

Thus; diameter pipe to be use to move water at the given temperature and average velocity is 2.47 ft

Answer:

2.47  ft

Explanation:

Given that:

The initial temperature of air = 80°F

Diameter of the pipe = 72 ft

average velocity  of the air flow through the pipe =  34 ft/s

The objective is to determine the diameter of the  pipe to  be used to move water at:

At a temperature = 60°F   &

An average velocity  of 71 ft/s

Assuming Reynolds number similarity is enforced;

where :

kinematic viscosity (V_air) of air at 80 °F  (V_air)  = 1.69 × 10⁻⁴ ft²/s

kinematic viscosity of water  at 60 °F (V_water) = 1.21 × 10⁻⁵ ft²/s

The diameter of the pipe can be calculated by using the expression:

2.4686  ft

2.47 ft   ( to two decimal places)

Thus; diameter pipe to be use to move water at the given temperature and average velocity is 2.47 ft

Answer:

100000000000000000000000

Answer:

A ball bearing has been selected with the bore size specified in the catalog as 35.000 mm to 35.020 mm. Specify.... ... has been selected with the bore size specified in the catalog as 35.000 mm to 35.020 mm. Specify appropriate minimum and maximum shaft diameters to provide a locational interference fit.

Explanation:

Answer:

A 3-phase, 50 Hz, 110 KV overhead line has conductors

Explanation:

hope it will helps you

Answer:

(A)0.122 kW (B) Actual cost = 1.056 $/day, Theoretical cost =  0.234 $/day

Explanation:

Solution

Given that:

The coefficient of performance is =3

Heat transfer = 6000kJ/hr

Temperature = 20°C

Cost of electricity = 8 cents per kW-hr

Now

The next step is to find the power input to the refrigerator and compare with the power input considered by a reversed refrigeration cycle operating between reservoirs at the two temperatures.

Thus

(A)The coefficient of performance is given below:

COP = Heat transfer from freezer/Power input

3 =6000/P

P =6000/3

P= 2000

P =  2000 kJ/hr = 2000/(60*60) kW

= 2000 (3600)kW

= 0.55 kW

Thus

The ideal coefficient of performance = T_low/(T_high - T_low)

= (0+273)/(20-0)

= 13.65

So,

P ideal = 6000/13.65 = 439.6 kJ/hr

= 439.6/(60*60) kW

= 0.122 kW

(B)For the actual cost we have the following:

Actual cost = 0.55 kW* 0.08 $/kW-hr = $ 0.044 per hour

= 0.044*24 $/day

= 1.056 $/day

For the theoretical cost we have the following:

Theoretical cost = 0.122 kW* 0.08 $/kW-hr = $ 0.00976 per hour

= 0.00976*24 $/day

= 0.234 $/day

Answer:

Explanation:

We are assuming that there is

a steady state two dimensional conduction

constant properties

uniform volumetric heat generation

From symmetry, we will be determining 6 unknown temperatures.

See attachment for calculation and and tabulation

With T(s) = 300 K, the set of equations were written directly into the IHT work space and solved for nodal temperatures.

The result is seen in the second attachment

Answer:

1.887 minutes

Explanation:

We are given k = 0.07 dm3 /mol*min at 300 K and E = 20 kcal/mol = 20000 cal/mol

To solve this, first of all let's calculate the rate constant(k);

For this question, The formula is;

K(t) = k(300K) × exp[(E/R)((1/300) - (1/T2))]

R is gas constant = 1.987 cal/mol.K

For temperature of 70°C which is = 70 + 273K = 343K, we have;

K(343) = 0.07 × exp[(20000/1.987)((1/300) - (1/343))]

K(343) = 4.7 dm³/mol.min

The design equation is;

dX/dt = -(rA/C_Ao) = K•(C_Ao)²•(1 - X)²/(C_Ao) = (KC_Ao)(1 - X)²

Since there is no change in volume by cause of the state at which the reaction is carried out, that is liquid. Thus, integrating and solving for time for a 90% conversion we obtain;

(0.9,0)∫dX/(1 - X)².dX = (KC_Ao)((t, 0)∫dt

So, we'll get;

0.9/(1 - 0.9) = 4.77 × 1 × t

t = 9/4.77

t = 1.887 minutes

Answer:

The answer is 2.25 kΩ

Explanation:

Solution

Given that:

The resistance reading on a DMM'S meter face = 22.5 ohms

The range selector switch = R * 100 range,

We now have to find the actual measure resistance of the circuit which is given below:

The actual measured resistance of the circuit is=R * 100

= 22.5 * 100

=2.25 kΩ

Hence the measured resistance of the circuit is 2.25 kΩ

Answer:

b

Explanation:

a) ADC is located on DAQ filter but not the reconstruction filter

b) to remove aliasing, the sampling rate must be greater than or equal ot twice the highest frequency component in the input signal. In other words, all frequencies in input sgnal are less than fs/2. Therefore, frequencies greater than fs/2 are removed by anti-aliasing filter

c) the DAC can have different sampling rate from ADC

Answer:

a.) 4.46 m/s

b.) 0.41 m/s

Explanation:

a) Given that the mass M of the bullet = 30g = 30/1000 = 0.03 kg

Velocity V = 450 m/s

From conservative of linear momentum,

Sum of momentum before impact = Sum of momentum after impact

0.03 × 450 = (0.03 + 3 ) × v₂

v₂ = 13.5/3.03 = 4.4554 m/s

Therefore the velocity of the bullet and B after the first impact = 4.46 m/s approximately

(b) To calculate the velocity of the carrier, you will consider the conservation of linear momentum again.

(m₁ + m₂)×v₂  = (m₁ + m₂ + m₃)×v₃

Where:

Mass of the carrier m₃ = 30 kg

Substitute all the parameters into the formula

3.03×4.4554 = (3.03 +30) × v₃

v₃ = 13.5 / 33.03 = 0.40872 m/s

Therefore the velocity of the carrier is 0.41 m/s approximately.

Answer:

5.6 mm

Explanation:

Given that:

A cylindrical tank is required to contain a:

Gage Pressure P = 560 kPa

Allowable normal stress [tex]\sigma[/tex] = 150 MPa = 150000 Kpa.

The inner diameter of the tank = 3 m

In a closed cylinder  there exist both the circumferential stress and the longitudinal stress.

Circumferential stress [tex]\sigma = \dfrac{pd}{2t}[/tex]

Making thickness t the subject; we have

[tex]t = \dfrac{pd}{2* \sigma}[/tex]

[tex]t = \dfrac{560000*3}{2*150000000}[/tex]

t = 0.0056 m

t = 5.6 mm

For longitudinal stress.

[tex]\sigma = \dfrac{pd}{4t}[/tex]

[tex]t= \dfrac{pd}{4*\sigma }[/tex]

[tex]t = \dfrac{560000*3}{4*150000000}[/tex]

t = 0.0028  mm

t = 2.8 mm

From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value  with the maximum thickness = 5.6 mm

Answer:

The answer is explained below

Explanation:

Given that:

1 ft = 0.3048 m, 1 in = 0.0254 m, 1 pound = 4.44822 newton

[tex]b_1=3.6ft=1.1\ m[/tex], [tex]b_2=2.2 ft=0.67\ m[/tex], [tex]d_2=7.5 in=0.19\ m[/tex], [tex]d_1=6.8in=0.17\ m[/tex]. P = 2200 lb = 9786 N

The area (A) is given as:

[tex]A=\frac{\pi}{4} (d_2^2-d_1^2)=\frac{\pi}{4}(0.19^2-0.17^2)=5.65*10^{-3}m^2[/tex]

The moment of area is given as:

[tex]l=\frac{\pi}{64} (d_2^4-d_1^4)=\frac{\pi}{64}(0.19^4-0.17^4)=2.3*10^{-5}m^4[/tex]

The maximum tensile stress is given as:

[tex]\sigma_1=-\frac{P}{A}+\frac{M(\frac{d_2}{2} )}{l} =-\frac{9786\ N }{5.65*10^{-3}m^2}+\frac{11kNm(0.19 \ m)/2}{2.3*10^{-5}m^4} =-1.73\ MPa+45.4\ MPa=43.67\ MPa\\\sigma_1=43.67\ MPa[/tex]

The maximum compressive stress is given as:

[tex]\sigma_c=-\frac{P}{A}-\frac{M(\frac{d_2}{2} )}{l} =-\frac{9786\ N }{5.65*10^{-3}m^2}-\frac{11kNm(0.19 \ m)/2}{2.3*10^{-5}m^4} =-1.73\ MPa-45.4\ MPa=47.13\ MPa\\\sigma_c=47.13\ MPa[/tex]

The maximum shear stress is given as:

[tex]\tau_{max}=|\frac{\sigma_c}{2} |=\frac{47.13\ MPa}{2}=23.57\ MPa[/tex]

Answer:

650°C  or 1,200°F

Explanation:

Data provided in the question

Steel alloy contains 93.8 wt% Fe, 6.0 wt% Ni, and 0.2 wt% C

Plus we also assume that there are no changes in the boundaries of postions who have other phases but there is an addition of Ni.

Based on the above information, the approximate eutectoid temperature of this alloy for 6.1 wt% is 650°C  or 1,200°F

Answer:

the net work per cycle [tex]\mathbf{W_{net} = 0.777593696}[/tex]  Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

diameter of the four-cylinder bore = 3.7 in

length of the stroke = 3.4 in

The clearance volume = 16% = 0.16

The cylindrical volume [tex]V_2 = 0.16 V_1[/tex]

the crankshaft N rotates at a speed of  2400 RPM.

At the beginning of the compression , temperature [tex]T_1[/tex] = 60 F = 519.67 R    

and;

Otto cycle with a pressure =  14.5 lbf/in² = (14.5 × 144 ) lb/ft²

= 2088 lb/ft²

The maximum temperature in the cycle is 5200 R

From the given information; the change in volume is:

[tex]V_1-V_2 = \dfrac{\pi}{4}D^2L[/tex]

[tex]V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)[/tex]

[tex]V_1-0.16V_1= 36.55714291[/tex]

[tex]0.84 V_1 =36.55714291[/tex]

[tex]V_1 =\dfrac{36.55714291}{0.84 }[/tex]

[tex]V_1 =43.52040823 \ in^3 \\ \\ V_1 = 43.52 \ in^3[/tex]

[tex]V_1 = 0.02518 \ ft^3[/tex]

the mass in air ( lb) can be determined by using the formula:

[tex]m = \dfrac{P_1V_1}{RT}[/tex]

where;

R = 53.3533 ft.lbf/lb.R°

[tex]m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}[/tex]

m = 0.0018962 lb

From the tables  of ideal gas properties at Temperature 519.67 R

[tex]v_{r1} =158.58[/tex]

[tex]u_1 = 88.62 Btu/lb[/tex]

At state of volume 2; the relative volume can be determined as:

[tex]v_{r2} = v_{r1} \times \dfrac{V_2}{V_1}[/tex]

[tex]v_{r2} = 158.58 \times 0.16[/tex]

[tex]v_{r2} = 25.3728[/tex]

The specific energy [tex]u_2[/tex] at [tex]v_{r2} = 25.3728[/tex] is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

[tex]v_{r3} = 0.1828[/tex]

[tex]u_3 = 1098 \ Btu/lb[/tex]

To determine the relative volume at state 4; we have:

[tex]v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}[/tex]

[tex]v_{r4} =0.1828 \times \dfrac{1}{0.16}[/tex]

[tex]v_{r4} =1.1425[/tex]

The specific energy [tex]u_4[/tex] at [tex]v_{r4} =1.1425[/tex] is 591.84 Btu/lb

Now; the net work per cycle can now be calculated as by using the following formula:

[tex]W_{net} = Heat \ supplied - Heat \ rejected[/tex]

[tex]W_{net} = m(u_3-u_2)-m(u_4 - u_1)[/tex]

[tex]W_{net} = m(u_3-u_2- u_4 + u_1)[/tex]

[tex]W_{net} = m(1098-184.7- 591.84 + 88.62)[/tex]

[tex]W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)[/tex]

[tex]W_{net} = 0.0018962 \times (410.08)[/tex]

[tex]\mathbf{W_{net} = 0.777593696}[/tex]  Btu per cycle

the power developed by the engine, in horsepower. can be calculated as follows;

In the  four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:

[tex]W = 4 \times N' \times W_{net[/tex]

where ;

[tex]N' = \dfrac{2400}{2}[/tex]

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec ×  0.777593696

W = 62.20749568 Btu/s

W = 88.0144746 hp

The net work per cycle and the power developed by this combustion engine are 0.7792 Btu and 88.20 hp.

Given the following data:

Note: The cylindrical volume is equal to [tex]0.16V_1[/tex]

First of all, we would determine the volume, mass and specific energy as follows:

[tex]V_1-V_2=\frac{\pi D^2L}{4} \\\\V_1-0.16V_1=\frac{3.142 \times 3.7^2 \times 3.4}{4}\\\\0.84V_1=36.56\\\\V_1=\frac{36.56}{0.84} \\\\V_1=43.52\;in^3 \;to \;ft^3 = 0.0252\;ft^3[/tex]

For the mass:

[tex]M=\frac{PV}{RT} \\\\M=\frac{2088 \times 0.0252}{53.3533 \times 519.67} \\\\M=\frac{52.6176}{27726.109411}[/tex]

M = 0.0019 lb.

At a temperature of 519.67 R, the relative volume and specific energy are:

[tex]v_{r1}=158.58\\\\u_1 = 88.62\;Btu/lb[/tex]

For the relative volume at the second state, we have:

[tex]v_{r2}=v_{r1}\times \frac{V_2}{V_1} \\\\v_{r2}=158.58\times 0.16\\\\v_{r2}=25.3728[/tex]

Note: At 25.3728, specific energy ([tex]u_2[/tex]) is 184.7 Btu/lb.

At a maximum temperature of 519.67 R, the relative volume and specific energy are:

[tex]v_{r3}=0.1828\\\\u_3 = 1098\;Btu/lb[/tex]

For the relative volume at state 4, we have:

[tex]v_{r4}=v_{r3}\times \frac{V_1}{V_3} \\\\v_{r4}=0.1828\times \frac{1}{0.16}\\\\v_{r4}=1.1425[/tex]

Note: At 1.1425, specific energy ([tex]u_4[/tex]) is 591.84 Btu/lb.

Now, we can calculate the net work per cycle by using this following formula:

[tex]W=Heat\;supplied -Heat\rejected\\\\W=m(u_3-u_2)-m(u_4-u_1)\\\\W=0.0019(1098-184.7)-0.0019(591.84-88.62)\\\\W=1.73527-0.956118[/tex]

W = 0.7792 Btu.

In a four-cylinder, four-stroke internal combustion engine, power is given by this formula:

[tex]W=4N'W_{net}[/tex]

But;

[tex]N'=\frac{N}{2 \times 60} \\\\N'=\frac{2400}{120} \\\\N'=20\;cycle/sec[/tex]

Substituting the given parameters into the formula, we have;

[tex]W=4 \times 20 \times 0.7792[/tex]

W = 62.336 Btu/sec.

In horsepower:

W = 88.20 hp.

Read more on net work here: https://brainly.com/question/10119215

Answer:

Following are the conversion to this question:

Explanation:

In point (a):

[tex]\to \frac{36459080}{8} = 4557385 + \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{4557385}{8} = 569673 + \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{569673}{8} = 71209+ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{71209}{8}=8901+\ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{8901}{8}=1112+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{5}{8}\\\\\to \frac{1112}{8}=139+ \ \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{139}{8}=17+ \ \ \ \ \ \ \ \ \ \ \frac{3}{8}\\\\\to \frac{17}{8}=2+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\[/tex]

[tex]\to \frac{2}{8}=0+ \ \ \ \ \ \ \ \ \ \frac{2}{8}\\\\ \bold{(36459080)_{10}=(213051110)_8}[/tex]

In point (b):

[tex]\to \frac{20960032010}{16} = 13100020+ \ \ \ \ \ \ \ \ \ \frac{0}{16}\\\\\to \frac{13100020}{16} = 818751+ \ \ \ \ \ \ \ \ \ \frac{4}{16}\\\\\to \frac{818751}{16} = 51171+ \ \ \ \ \ \ \ \ \ \frac{15}{16}\\\\\to \frac{51171}{16}=3198+\ \ \ \ \ \ \ \ \ \ \ \frac{3}{16}\\\\\to \frac{3198}{16}=199+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{14}{1}\\\\\to \frac{199}{16}=12+ \ \ \ \ \ \ \ \ \ \ \frac{7}{16}\\\\\to \frac{12}{16}=0+ \ \ \ \ \ \ \ \ \ \ \frac{12}{16}\\\\ \bold{(20960032010)_{10}=(C7E3F40)_{16}}[/tex]

In point (c):

[tex]\to (2423233303003040)_s=(88757078520)_{10}\\\\\to \frac{88757078520}{25}= 3550283140+ \ \ \ \ \ \ \ \ \ \frac{20}{25}\\\\ \to \frac{3550283140}{25}= 142011325+ \ \ \ \ \ \ \ \ \ \frac{15}{25}\\\\\to \frac{142011325}{25}= 5680453+ \ \ \ \ \ \ \ \ \ \frac{0}{25}\\\\\to \frac{5680453}{25}= 227218+ \ \ \ \ \ \ \ \ \ \frac{3}{25}\\\\\to \frac{227218}{25}= 9088+ \ \ \ \ \ \ \ \ \ \frac{18}{25}\\\\\to \frac{9088}{25}= 363+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\[/tex]

[tex]\to \frac{363}{25}= 14+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\\to \frac{14}{25}= 0+ \ \ \ \ \ \ \ \ \ \frac{14}{25}\\\\\bold{(2423233303003040)_s=(EDDI30FK)_{25}}[/tex]

Symbols of Base 25 are as follows:

[tex]0, 1, 2, 3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,I,J,K,L,M,N, \ and \ O[/tex]

Answer:

the angle of twist of B with respect to D is -1.15°

the angle of twist of C with respect to D is 1.15°

Explanation:

The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity  G = 11 × 10³ Ksi =  11 × 10⁶ lb/in²

The objective are :

1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft [tex]J\tau[/tex]

shaft [tex]J\tau[/tex] = [tex]\dfrac{\pi}{2}r^4[/tex]

where ;

r = 1 in /2

r = 0.5 in

shaft [tex]J \tau[/tex] = [tex]\dfrac{\pi}{2} \times 0.5^4[/tex]

shaft [tex]J\tau[/tex] = 0.098218

Now; the angle of twist at  B with respect to D  is calculated by using the expression

[tex]\phi_{B/D} = \sum \dfrac{TL}{JG}[/tex]

[tex]\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}[/tex]

where;

[tex]T_{CD} \ \ and \ \ L_{CD}[/tex] are the torques at segments CD and length at segments CD

[tex]{T_{BC} \ \ and \ \ L_{BC}}[/tex] are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

[tex]L_{BC}}[/tex] = 2.5  in

[tex]J\tau[/tex] = 0.098218

G =  11 × 10⁶ lb/in²

[tex]T_{BC[/tex] = -60 lb.ft

[tex]T_{CD[/tex] = 0 lb.ft

[tex]L_{CD[/tex] = 5.5 in

[tex]\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}[/tex]

[tex]\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}[/tex]

[tex]\phi_{B/D} = \dfrac{-21600}{1079980}[/tex]

[tex]\phi_{B/D} =[/tex] − 0.02 rad

To degree; we have

[tex]\phi_{B/D} = -0.02 \times \dfrac{180}{\pi}[/tex]

[tex]\mathbf{\phi_{B/D} = -1.15^0}[/tex]

Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction

Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

For  the angle of twist of C with respect to D; we have:

[tex]\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}[/tex]

[tex]\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}[/tex]

[tex]\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}[/tex]

[tex]\phi_{C/D} = \dfrac{21600}{1079980}[/tex]

[tex]\phi_{C/D} =[/tex] 0.02 rad

To degree; we have

[tex]\phi_{C/D} = 0.02 \times \dfrac{180}{\pi}[/tex]

[tex]\mathbf{\phi_{C/D} = 1.15^0}[/tex]

Answer:

Your question is lacking some information attached is the missing part and the solution

A) AB = AD = BD = 0, BC = LC

    AC = [tex]\frac{5L}{3}T, CD = \frac{4L}{3} C[/tex]

B) AB = AD = BC = BD = 0

   AC = [tex]\frac{5L}{3} T, CD = \frac{4L}{3} C[/tex]

Explanation:

A) Forces in all members due to the load L in position A

assuming that BD goes slack from an inspection of Joint B

AB = 0 and BC = LC from Joint D, AD = 0 and CD = 4L/3 C

B) steps to arrive to the answer is attached below

AB = AD = BC = BD = 0

AC = [tex]\frac{5L}{3} T, CD = \frac{4L}{3}C[/tex]

Answer:

This doesn't represent an equilibrium state of stress

Explanation:

∝ = 1 , β = 1 ,  y = 1

x = 0 , y = 0 , z = 0 ( body forces given as 0 )

Attached is the detailed solution is and also the conditions for equilibrium

for a stress state to be equilibrium all three conditions has to meet the equilibrum condition as explained in the attached solution

Answer:

[tex]W_s =[/tex] 283.181 hp

Explanation:

Given that:

Air enters a compressor operating at steady state at a pressure [tex]P_1[/tex] =  176.4 lbf/in.^2  and Temperature [tex]T_1[/tex] at 260°F

Volumetric flow rate V = 424 ft^3/min

Air exits at a pressure [tex]P_2[/tex]  = 15.4 lbf/in.^2 and Temperature [tex]T_2[/tex] at 80°F.

Heat transfer occurs at a rate of 6800 Btu/h from the compressor to its surroundings; since heat is released to the surrounding; then:

[tex]Q_{cv}[/tex] = -6800 Btu/h  = - 1.9924 kW

Using the steady  state  energy in the process;

[tex]h_2 - h_1 + g(z_2-z_1)+ \dfrac{1}{2}(v^2_2-v_1^2) = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}[/tex]

where;

[tex]g(z_2-z_1) =0[/tex]  and  [tex]\dfrac{1}{2}(v^2_2-v_1^2) = 0[/tex]

Then; we have :

[tex]h_2 - h_1 = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}[/tex]

[tex]h_2 - h_1 = \dfrac{Q_{cv} - W_s}{m}[/tex]

[tex]{m}(h_2 - h_1) ={Q_{cv} - W_s}[/tex]

[tex]W_s ={Q_{cv} + {m}(h_2 - h_1)[/tex] ----- (1)

Using the relation of Ideal gas equation;

P₁V₁ = mRT₁

Pressure [tex]P_1[/tex] =  176.4 lbf/in.^2   = ( 176.4 ×  6894.76 ) N/m² = 1216235.664 N/m²

Volumetric flow rate V = 424 ft^3/min = (424 ×  0.0004719) m³  /sec

= 0.2000856 m³  /sec

Temperature = 260°F = (260°F − 32) × 5/9 + 273.15 = 399.817 K

Gas constant R=287 J/kg K

Then;

1216235.664 N/m² × 0.2000856 m³  /sec = m × 287 J/kg K × 399.817 K

[tex]m = \dfrac { 1216235.664 N/m^2 \times 0.2000856 m^3 /sec } {287 J/kg K \times 399.817 K }[/tex]

m = 2.121 kg/sec

The change in enthalpy:

[tex]m(h_1-h_2) = m * C_p * \Delta T= m* C_p * ( T_1 -T_2)[/tex]

[tex]= 2.121* 1.005* ( 399.817 -299.817)[/tex]

= 213.1605 kW

From (1)

[tex]W_s ={Q_{cv} + {m}(h_2 - h_1)[/tex]

[tex]W_s =[/tex]  - 1.9924 kW + 213.1605 kW

[tex]W_s =[/tex] 211.1681  kW

[tex]W_s =[/tex] 283.181 hp

The power input is [tex]W_s =[/tex] 283.181 hp

Explanation:

The value of enthalpy and entropy at state 1 will be determined according to the given pressure and temperature as follows using interpolation from A-13 is as follows.

[tex]h_{1}[/tex] = 247.88 kJ/kg,    [tex]S_{1}[/tex] = 0.9579 kJ/kg K

At state 2, isentropic enthalpy will be determined from the condition [tex]S_{2} = S_{1}[/tex] and given pressure at 2 with data from A-13 using interpolation is:

    [tex]h_{2s}[/tex] = 279.45 kJ/kg

We will calculate actual enthalpy at state 2 using given pressure and temperature from A-13 as follows.

        [tex]h_{2}[/tex] = 286.71 kJ/kg

Hence, isentropic compressor efficiency will be calculated using standard relation as:

      [tex]\eta_{c} = \frac{h_{2s} - h_{1}}{h_{2} - h_{1}}[/tex]  

                 = [tex]\frac{279.45 - 247.88}{286.71 - 247.88}[/tex]

                 = 0.813

Now, at state 3 enthaply is determined by temperature at state 3, that is, [tex]26^{o}C[/tex] for given pressure as per saturated liquid approximation and data from A-11.

   [tex]h_{3}[/tex] = 87.83 kJ/Kg

Using energy balance in 2-3, the rate of heat supplied to the heated room is as follows.

      [tex]Q_{H} = m(h_{2} - h_{3})[/tex]

                 = 0.022 (286.71 - 87.83) kW

                 = 4.38 kW

Now, COP will be calculated using power that is expressed through energy balance in 1-2 as follows.

     COP = [tex]\frac{Q_{H}}{W}[/tex]

              = [tex]\frac{Q_{H}}{m(h_{2} - h_{1})}[/tex]

              = [tex]\frac{4.38}{0.022 (286.71 - 246.88)}[/tex]

              = 5.13

In an ideal vapour-compression cycle, the enthalpy and entropy at state 1 will be obtained from given pressure and state with data from A-12:

  [tex]h_{1}[/tex] = 244.5 kJ/kg

  [tex]S_{1}[/tex] = 0.93788 kJ/kg K

  [tex]h_{2}[/tex] = 273.71 kJ/kg

At state 3, enthalpy will be determined from given pressure and state with data from A-12 as follows.

  [tex]h_{3}[/tex] = 95.48 kJ/kg

Hence, using energy balance in 2-3 the rate of heat supplied will be calculated as follows.

   [tex]Q_{H} = m(h_{2} - h_{3})[/tex]

              = 0.022 (273.31 - 95.48) kW

              = 3.91 kW

The power input which is expressed through energy balance in 1-2 will be used to determine COP as follows.

    COP = [tex]\frac{Q_{H}}{W}[/tex]

             = [tex]\frac{Q_{H}}{m (h_{2} - h_{1})}[/tex]

             = [tex]\frac{3.91}{0.022(273.31 - 244.5)}[/tex]

             = 6.17

Answer:

A) V_peak ≈ 165 V

B) V_rms ≈ 24 V

C) V_peak ≈ 311 V

D) V_peak ≈ 311 KV

Explanation:

Formula for RMS value is given as;

V_rms = V_peak/√2

Formula for peak value is given as;

V_peak = V_rms x √2

A) At RMS value of 117 V, peak value would be;

V_peak = 117 x √2

V_peak = 165.46 V

V_peak ≈ 165 V

B) At peak value of 33.9 V, RMS value would be;

V_rms = 33.9/√2

V_rms = 23.97 V

V_rms ≈ 24 V

C) At RMS value of 220 V, peak value is;

V_peak = 220 × √2

V_peak = 311.13 V

V_peak ≈ 311 V

D) At RMS value of 220 KV, peak value is;

V_peak = 220 × √2

V_peak = 311.13 KV

V_peak ≈ 311 KV

Answer:

A) W' = 178.568 KW

B) ΔS = 2.6367 KW/k

C) η = 0.3

Explanation:

We are given;

Temperature at state 1;T1 = 360 °C

Temperature at state 2;T2 = 160 °C

Pressure at state 1;P1 = 10 bar

Pressure at State 2;P2 = 1 bar

Volumetric flow rate;V' = 0.8 m³/s

A) From table A-6 attached and by interpolation at temperature of 360°C and Pressure of 10 bar, we have;

Specific volume;v1 = 0.287322 m³/kg

Mass flow rate of water vapour at turbine is defined by the formula;

m' = V'/v1

So; m' = 0.8/0.287322

m' = 2.784 kg/s

Now, From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific enthalpy;h1 = 3179.46 KJ/kg

Now, From table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific enthalpy;h2 = 3115.32 KJ/kg

Now, since stray heat transfer is neglected at turbine, we have;

-W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2 - h1)

Plugging in relevant values, the work of the turbine is;

W' = -2.784(3115.32 - 3179.46)

W' = 178.568 KW

B) Still From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific entropy: s1 = 7.3357 KJ/Kg.k

Still from table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific entropy; s2 = 8.2828 KJ/kg.k

The amount of entropy produced is defined by;

ΔS = m'(s2 - s1)

ΔS = 2.784(8.2828 - 7.3357)

ΔS = 2.6367 KW/k

C) Still from table A-6 attached and by interpolation at state 2 with s2 = s2s = 8.2828 KJ/kg.k and Pressure of 1 bar, we have;

h2s = 2966.14 KJ/Kg

Energy equation for turbine at ideal process is defined as;

Q' - W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Again, Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2s - h1)

W' = -2.784(2966.14 - 3179.46)

W' = 593.88 KW

the isentropic turbine efficiency is defined as;

η = W_actual/W_ideal

η = 178.568/593.88 = 0.3