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3 years ago

10

A bona fide established commercial marketing agency is a business which is specifically devoted to public relations, advertising

and promoting the services of a client. True or False

1 answer:

Katarina [22]3 years ago

5 0

Answer:

True

Explanation:

Bona Fide is a Latin term which means in good faith or without any intention to deceive. The business established on a bona fide basis means that there is an absence of fraud. The marketing agency has devoted its services to public relations, advertising and promoting the services of clients. There is no intention of fraud in the business.

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Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa, 450°C, and 80 m/s, and the exit

8090 [49]

Answer:

a) The change in Kinetic energy, KE = -1.95 kJ

b) Power output, W = 10221.72 kW

c) Turbine inlet area, $A_1 = 0.0044 m^2$

Explanation:

a) Change in Kinetic Energy

For an adiabatic steady state flow of steam:

$KE = \frac{V_2^2 - V_1^2}{2} \\$ .........(1)

Where Inlet velocity, V₁ = 80 m/s

Outlet velocity, V₂ = 50 m/s

Substitute these values into equation (1)

$KE = \frac{50^2 - 80^2}{2} \\$

KE = -1950 m²/s²

To convert this to kJ/kg, divide by 1000

KE = -1950/1000

KE = -1.95 kJ/kg

b) The power output, w

The equation below is used to represent a steady state flow.

$q - w = h_2 - h_1 + KE + g(z_2 - z_1)$

For an adiabatic process, the rate of heat transfer, q = 0

z₂ = z₁

The equation thus reduces to :

w = h₁ - h₂ - KE...........(2)

Where Power output, $W = \dot{m}w$ ..........(3)

Mass flow rate, $\dot{m} = 12 kg/s$

To get the specific enthalpy at the inlet, h₁

At P₁ = 10 MPa, T₁ = 450°C,

h₁ = 3242.4 kJ/kg,

Specific volume, v₁ = 0.029782 m³/kg

At P₂ = 10 kPa, $h_f = 191.81 kJ/kg, h_{fg} = 2392.1 kJ/kg$ , x₂ = 0.92

specific enthalpy at the outlet, h₂ = $h_1 + x_2 h_{fg}$

h₂ = 3242.4 + 0.92(2392.1)

h₂ = 2392.54 kJ/kg

Substitute these values into equation (2)

w = 3242.4 - 2392.54 - (-1.95)

w = 851.81 kJ/kg

To get the power output, put the value of w into equation (3)

W = 12 * 851.81

W = 10221.72 kW

c) The turbine inlet area

$A_1V_1 = \dot{m}v_1\\\\A_1 * 80 = 12 * 0.029782\\\\80A_1 = 0.357\\\\A_1 = 0.357/80\\\\A_1 = 0.0044 m^2$

3 0

3 years ago

A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

abruzzese [7]

Answer:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂ = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

$\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}} \right )^{K-1}$

Otto cycle T-S diagram

T₂ = 288.706* $6.25^{0.393}$ = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

$\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}} \right )^{K-1}$

T₄ = 2888.89 / $6.25^{0.393}$ = 1406.5 K

Work done, W = $c_v$ ×(T₃ - T₂) - $c_v$ ×(T₄ - T₁)

0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.

8 0

4 years ago

A 625 g basketball and a 58.5 g tennis ball are dropped from a height of d = 1.5 m onto the floor. The coefficient of restitutio

stich3 [128]

Answer:

Maximum height=7.3535 m

Explanation:

Solution of the problem is given in the attachments.

3 0

3 years ago

A continuous random variable, X, whose probability density function is given by f(x) = ( λe−λx , if x ≥ 0 0, otherwise is said t

Ganezh [65]

Answer:

a) $F(x) = \lambda \int_0^{\infty} e^{-\lambda x} dx= -e^{-\lambda x} \Big|_0^{\infty} = 1- e^{-\lambda x} \$

b) $P(10 < X$

Explanation:

Previous concepts

The cumulative distribution function (CDF) F(x),"describes the probability that a random variableX with a given probability distribution will be found at a value less than or equal to x".

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution".

Part a

Let X the random variable of interest. We know on this case that $X\sim Exp(\lambda)$

And we know the probability denisty function for x given by:

$f(x) = \lambda e^{-\lambda x} , x\geq 0$

In order to find the cdf we need to do the following integral:

$F(x) = \lambda \int_0^{\infty} e^{-\lambda x} dx= -e^{-\lambda x} \Big|_0^{\infty} = 1- e^{-\lambda x} \$

Part b

Assuming that $X \sim Exp(\lambda =0.1)$ , then the density function is given by:

$f(x) = 0.1 e^{-0.1 x} dx , x\geq 0$

And for this case we want this probability:

$P(10 < X$

And evaluating the integral we got:

$P(10 < X$

4 0

3 years ago

A heat pump with an ideal compressor operates between 0.2 MPa and 1 MPa. Refrigerant R134a flows through the system at a rate of

solmaris [256]

Answer:

The mass flow rate of refrigerant is 0.352 kg/s

Explanation:

Considering the cycle of an ideal heat pump, provided in the attachment, we first find enthalpy at state B and D. For that purpose, we use property tables of refrigerant R134a:

<u>At State A</u>:

From table, we see the enthalpy and entropy value of saturated vapor at 0.2 MPa. Therefore:

ha = 244.5 KJ/kg

Sa = 0.93788 KJ/kg.k

<u>At State B</u>:

Since, the process from state A to B is isentropic. Therefore,

Sb = Sa = 0.93788 KJ/Kg

From table, we see the enthalpy value of super heated vapor at 1 MPa and Sb. Therefore:

hb = 256.85 KJ/kg (By interpolation)

<u>At State C</u>:

From table, we see the enthalpy and entropy value of saturated liquid at 1 MPa. Therefore:

hc = 107.34 KJ/kg

Now, from the diagram it is very clear that:

Heat Loss = m(hb = hc)

m = (Heat Loss)/(hb - hc)

where,

m = mass flow rate = ?

Heat Loss = (180,000 Btu/hr)(1.05506 KJ/1 Btu)(1 hr/3600 sec)

Heat Loss = 52.753 KW

Therefore,

m = (52.753 KJ/s)/(256.85 KJ/kg - 107.34 KJ/kg)

<u>m = 0.352 kg/s</u>

5 0

3 years ago