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3 years ago

When a specific property is multiplied by the mass of the system the results is always an intensive property. a)True b) False

Engineering

1 answer:

german3 years ago

6 0

Answer:

b). False

Explanation:

Thermodynamic property are of two types

1. Intensive property

2.Extensive property

Intensive properties are those properties that do not depend on mass. For example, Specific volume, temperature, specific heat capacity,density etc

Extensive properties are those properties which depends on mass.

For example, volume, heat capacity, etc.

Now when a specific property (intensive property) is multiplied with mass of the system, we get extensive property .

We know that density is an extensive property having unit of kg per meter cube.

Now the reciprocal of density is specific volume which is also an extensive property having unit of meter cube per kg.

Now if we multiply the specific volume with the mass we get volume which is an extensive property and not an intensive property.

Thus, we get extensive property when we multiply mass to any specific or intensive property.

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Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping stati

snow_tiger [21]

Answer:

$D=0.41m$

Explanation:

From the question we are told that:

Discharge rate $V_r=0.35 m3/s$

Distance $d=4km$

Elevation of the pumping station $h_p= 140 m$

Elevation of the Exit point $h_e= 150 m$

Generally the Steady Flow Energy Equation SFEE is mathematically given by

$h_p=h_e+h$

With

$P_1-P_2$

And

$V_1=V-2$

Therefore

$h=140-150$

$h=10$

Generally h is give as

$h=\frac{0.5LV^2}{2gD}$

$h=\frac{8Q^2fL}{\pi^2 gD^5}$

Therefore

$10=\frac{8Q^2fL}{\pi^2 gD^5}$

$D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}$

$D=0.41m$

8 0

3 years ago

Risk Management in a Business ModelLearning Objectives and OutcomesCreate a report documenting various aspects of how risk mana

zubka84 [21]

Answer:

The question is explained in detailed way in explanation section and in attached files.

Explanation:

The HIPAA Security Rule is designed to be flexible and appropriate for our organization’s particular size, structure, and inherent risks to business and personal information. Risk analysis is meant to be an ongoing process, during which we regularly review our records to track access to business and personal systems and data. With this in mind, I recommend that we expand our information security strategy to include more than just what is required in HIPAA. Just as a reminder below is the HIPAA ecompliance and implementation strategy that we came up with last week as given in attached file 1.

There are several areas in IT security that the above is incomplete or insufficient in. We recommend implementing several more complete or alternative controls in order to protect our systems, patients, employees, contractors, vendors, and assets beyond the HIPAA minimum requirements. The below section describes what the Centers for Medicare and Medicaid Services (CMS) recommend as additional areas to focus on in the effort to increase an organization's security. (See the attached file # example of some of the areas that we should monitor beyond what HIPAA requires are given in attached file # 03.

3 0

3 years ago

A rigid 14-L vessel initially contains a mixture of liquid water and vapor at 100°C with 12.3 percent quality. The mixture is th

tigry1 [53]

Answer:

Q = 65.388 KJ

Explanation:

To calculate the heat required for the given process Q, we recall the energy balance equation.

Therefore, : Q = Δ U = m (u₂ - u₁) ..................equation (1)

We should note that there are no kinetic or potential energy change so the heat input in the system is converted only to internal energy.

Therefore, we will start the equation with the mass of the water (m) using given the initial percentage quality as x₁ = 0.123 and initial temperature t₁ = 100⁰c , we can them determine the initial specific volume v₁ of the mixture. For the calculation, we will also need the specific volume of liquid vₙ = 0.001043m³/kg and water vapour (vₐ) = 1.6720m³/kg

Therefore, u₁ = vₙ + x₁ . ( vₐ - vₙ)

u₁ = 0.001043m³/kg + 0.123 . ( 1.6720m³/kg - 0.001043m³/kg)

u₁ = 0.2066m³/kg

Moving forward, the mass of the vapor can then be calculated using the given volume of tank V = 14 L but before the calculation, we need to convert the volume to from liters to m³.

Therefore, V = 14L . 1m² / 1000L = 0.014 m³

Hence, m = V / u₁

0.014m³ / 0.2066 m³/kg

m = 0. 0677 kg

Also, the initial specific internal energy u₁ can be calculated using the given the initial given quality of x₁ , the specific internal energy of liquid water vₐ = 419.06 kj / kg and the specific internal energy of evaporation vₐₙ = 2087.0 kj/kg.

Therefore, u₁ = vₐ + x₁ . vₐₙ

u₁ = 419.06 kj / kg + 0.123 . 2087.0 kj/kg

u₁ = 675.76 kj/kg

For the final specific internal energy u₂, we first need to calculate the final quality of the mixture x₂ . The tank is rigid meaning the volume does not change and it is also closed meaning the mass does not change.from this, we can conclude the the specific volume also does not change during the process u₁ = u₂. This allows us to use the given final temperature T₂ = 180⁰c to determine the final quality x₂ of the mixture. for the calculation, we will also need the specific volume of liquid vₙ=0.001091m³/kg and vapor vₐ = 0.39248m³/kg

Hence, x₂ = u₂ - vₙ / uₐ

x₂ = 0.2066 m³/kg - 0.001091m³/kg / 0.39248m³/kg

x₂ = 0.524

Moving forward to calculate the final internal energy u₂, we have :

u₂ = vₙ + x₂ . vₙₐ

u₂ = 631.66 kj/kg + 0.524 . 1927.4 kj/kg

u₂ = 1641.62 kj/kg

We now return to equation (1) to plug in the values generated thus far

Q = m (u₂ - u₁)

0. 0677 kg ( 1641.62 kj/kg - 675.76 kj/kg)

Q = 65.388KJ

7 0

3 years ago

Read 2 more answers

a ten station assembly machine has ideal cycle time of 6 sec. the fraction defect rate at each station 0.005 and defect always j

MAXImum [283]

Answer:

$T_{P}=(2.6667)(10^{-3})h$

Explanation:

Let's write the equation of the production rate for the assembly machine :

$T_{P}=T_{C}+(n).(m).(p).(T_{D})$

Where $T_{P}$ is the production rate for the assembly machine.

Where $T_{C}$ is the ideal cycle time

Where n is the number of stations.

Where m is the number stations that get jam when the defect occurs.

Where p is the defect rate at each station.

And where $T_{D}$ is the average downtime per breakdown

We are looking for the hourly production rate ⇒

$1h=60min\\1min=60s$ ⇒

$1h=3600s$ ⇒

$6s=\frac{(6s)(1h)}{(3600s)}= \frac{1}{600}h$

$60min=1h$ ⇒

$1.2min=\frac{(1.2min)(1h)}{(60min)}=0.02h$

$T_{P}=\frac{1}{600}h+(10)(1.0)(0.005)(0.02h)=\frac{1}{375}h=(2.6667)(10^{-3})h$

m = 1.0 in the equation.

3 0

3 years ago

A refrigerated space is maintained at -15℃， and cooling water is available at 30℃， the refrigerant is ammonia. The refrigeration

Illusion [34]

Answer:

(1) 5.74

(2) 5.09

(3) 3.05×10⁻⁵ kg/s

(4) 0.00573 kW

Explanation:

The parameters given are;

Working temperature, $T_C$ = -15°C = 258.15 K

Temperature of the cooling water, $T_H$ = 30°C = 303.15 K

(1) The Carnot coefficient of performance is given as follows;

$\gamma_{Max} = \dfrac{T_C}{T_H - T_C} = \dfrac{258.15}{303.15 - 258.15} = 5.74$

(2) For ammonia refrigerant, we have;

$h_2 = h_g = 1466.3 \ kJ/kg$

$h_3 = h_f = 322.42 \ kJ/kg$

$h_4 = h_3 = h_f = 322.42 \ kJ/kg$

s₂ = s₁ = 4.9738 kJ/(kg·K)

0.4538 + x₁ × (5.5397 - 0.4538) = 4.9738

∴ x₁ = (4.9738 - 0.4538)/(5.5397 - 0.4538) = 0.89

$h_1 = h_{f1} + x_1 \times h_{gf}$

h₁ = 111.66 + 0.89 × (1424.6 - 111.66) = 1278.5 kJ/kg

$\gamma = \dfrac{h_1 - h_4}{h_2 - h_1}$

$\gamma = \dfrac{1278.5 - 322.42}{1466.3 - 1278.5} = 5.09$

(3) The circulation rate is given by the mass flow rate, $\dot m$ as follows

$\dot m = \dfrac{Refrigeration \ capacity}{Refrigeration \ effect \ per \ unit \ mass}$

The refrigeration capacity = 105 kJ/h

The refrigeration effect, Q = (h₁ - h₄) = (1278.5 - 322.42) = 956.08 kJ/kg

Therefore;

$\dot m = \dfrac{105}{956.08} = 0.1098 \ kg/h$

$\dot m$ = 0.1098 kg/h = 0.1098/(60*60) = 3.05×10⁻⁵ kg/s

(4) The work done, W = (h₂ - h₁) = (1466.3 - 1278.5) = 187.8 kJ/kg

The rating power = Work done per second = W× $\dot m$

∴ The rating power = 187.8 × 3.05×10⁻⁵ = 0.00573 kW.

6 0

3 years ago