1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
tatyana61 [14]
3 years ago
15

Air enters a horizontal, constant-diameter heating duct operating at steady state at 290 K, 1 bar, with a volumetric flow rate o

f 0.25 m3/s, and exits at 325 K, 0.95 bar. The flow area is 0.04 m2. Assuming the ideal gas model with k 5 1.4 for the air, determine:
(a) the mass flow rate, in kg/s,
(b) the velocity at the inlet and exit, each in m/s, and
(c) the rate of heat transfer, in kW.
Engineering
2 answers:
Taya2010 [7]3 years ago
6 0

Answer:

a) m = 0.3003 kg/s

b) Vel1 = 6.25 m/s, Vel2 = 7.3725 m/s

c) 12.845 KW

Explanation:

a)

using ideal gas law:

PV = nRT

since, n = no. of moles = m/M

therefore,

PV = (m/M)RT

P1 v1 = RT1/M

where,

P1 = inlet pressure = 1 bar = 100000 Pa

v1 = specific volume at inlet = ?

R = universal gas constant = 8.314 KJ/Kmol.k

T1 = inlet temperature = 290 K

M = Molecular mass of air = 28.9628 Kg/kmol

Therefore,

v1 = (8.314 KJ/Kmol.k)(290 k)/(28.9628 kg/kmol)(100 KPa)

v1 = 0.83246 m³/kg

Now, the mass flow rate can be given as:

Mass Flow Rate = (Volume Flow Rate)/(v1)

Mass Flow Rate = (0.25 m³/s)/(0.83246 m³/kg)

<u>Mass Flow Rate = 0.3003 kg/s</u>

b)

using general gas equation to find the specific volume at the exit first, we get:

P1v1/T1 = P2v2/T2

v2 = P1 v1 T2/T1 P2

where,

P1 = inlet pressure = 1 bar

P2 = exit pressure = 0.95 bar

T1 = inlet temperature = 290 k

T2 = exit temperature = 325 k

v1 = specific volume at inlet = 0.83246 m³/kg

v2 = specific volume at exit = ?

Therefore,

v2 = (1 bar)(0.83246 m³/kg)(325 k)/(290 k)(0.95 bar)

v2 = 0.98203 m³/kg

Now, for velocity, we use formula:

Vel = v/A

where,

A = Area = 0.04 m²

For inlet:

Vel1 = Inlet Volume flow Rate/A = (0.25 m³/s)/(0.04 m²)

<u>Vel1 = 6.25 m/s</u>

For exit:

Vel2 = Exit Volume flow Rate/A = (Mass Flow Rate)(v2)/(0.04 m²)

Vel2 = (0.98203 m³/kg)(0.3003 kg/s)/0.04 m²

<u>Vel2 = 7.3725 m/s</u>

c)

using first law of thermodynamics, with no work done, we can derive the formula given below:

Q = m(h2 - h1) + (m/2)(Vel2² - Vel1²)

where,

Q = rate of heat transfer

m = mass flow rate = 0.3003 kg/s

m(h2 - h1) = change in enthalpy = mCpΔT   , for ideal gas

Vel1 = 6.25 m/s

Vel2 = 7.3725 m/s

Also,

Cp in this case will be function of temperature and given as:

Cp = KR/(K-1)

where,

K = 1.4

R = Gas constant per molecular mass of air = 0.28699 KJ/kg.k

Therefore,

Cp = 1.0045 KJ/kg.k

Now, using the values in the expression of first law of thermodynamics, we get:

Q = (0.3003 kg/s)(1.0045 KJ/kg.k)(35 K) + [(0.3003 kg/s)/2][(7.3725 m/s)² - (6.25 m/s)²]

Q = 10.55 KW + 2.29 KW

<u>Q = 12.845 KW</u>

Sliva [168]3 years ago
5 0

Answer:

A) Mass flow rate = 0.3004 Kg/s

B) Velocity at Inlet = 6.25 m/s

Velocity at exit = 7.3725 m/s

C) Rate of heat transfer = 12.858 Kw

Explanation:

T1 = 290K ;P1 = 1 bar = 100 KPa

T2= 325K ; P2 = 0.95 bar = 95 KPa

A = 0.04 m² ; k = 1.4

Molar mass of air = 28.97 Kg/Kmol

R = 8.314 J/molK

A) Since we are dealing with a steady state mass flow rate through an open system, we will treat this as an ideal gas. Thus ;

PV = nRT

n=m/M where m =1

Thus; V1 = RT1/MP1 = (8.314 X 290)/(28.97 x 100) = 0.8323 m³/Kg

Now, mass flow rate is given by;

Mass flow rate(m') = Volumetric flow rate(V')/Volume(v)

Thus; Mass flow rate(m') = 0.25/0.8323 = 0.3004 Kg/s

From Gay lussacs law; P1V1/T1 = P2V2/T2

So to find the volume at the exit which is V2, let's make V2 the subject of the formula;

(P1V1T2)/(P2T1) = V2

So; V2 = (100 x 0.8323 x 325)/(95 x 290) = 0.9818 m³/kg

B) we know that velocity = volumetric flow rate/area

Thus;

At inlet; Velocity (Vi) = 0.25/0.04 = 6.25m/s

At exit; Velocity (Ve) = Volumetric flow rate at exit/ Area.

We don't know the volumetric flow at exit so let's look for it. from earlier, we saw that;

volumetric flow rate/Volume = mass flow rate

And rearranging, volumetric flow rate(V') = mass flow rate(m) x volume(v)

So V' = 0.3004 x 0.9818 = 0.2949 Kg/s

So, Ve = 0.2949/0.04 = 7.3725 m/s

C) The steady state equation when potential energy is neglected is given by the formula;

Q'= m'Cp(T2 - T1) + (m'/2){(Ve)² - (Vi)²}

Where Q' is the rate of heat transfer.

Cp is unknown. The formula to find Cp is given as ;

Cp = KR/(K-1)

Since we are dealing with change in enthalpy here, the gas constant R will be expressed per molecular mass of the air and so R = 0.287 KJ/kg.k

Cp = (1.4 x 0.287)/(1.4 - 1) = 1.0045 Kj/KgK

And so, Q' = (0.3004 x 1.0045)(325 - 290) + (0.3004/2){(7.3725)² - (6.25)²} = 10.5613 + 2.2967 = 12.858 Kw

You might be interested in
30POINTS
garri49 [273]
Concentrating solar power (CSP) plants use mirrors to concentrate the sun's energy to drive traditional steam turbines or engines that create electricity. The thermal energy concentrated in a CSP plant can be stored and used to produce electricity when it is needed, day or night. Today, roughly 1,815 megawatts (MWac) of CSP plants are in operation in the United States.

Parabolic Trough
Parabolic trough systems use curved mirrors to focus the sun’s energy onto a receiver tube that runs down the center of a trough. In the receiver tube, a high-temperature heat transfer fluid (such as a synthetic oil) absorbs the sun’s energy, reaching temperatures of 750°F or higher, and passes through a heat exchanger to heat water and produce steam. The steam drives a conventional steam turbine power system to generate electricity. A typical solar collector field contains hundreds of parallel rows of troughs connected as a series of loops, which are placed on a north-south axis so the troughs can track the sun from east to west. Individual collector modules are typically 15-20 feet tall and 300-450 feet long.

Compact Linear Fresnel Reflector
CLFR uses the principles of curved-mirror trough systems, but with long parallel rows of lower-cost flat mirrors. These modular reflectors focus the sun's energy onto elevated receivers, which consist of a system of tubes through which water flows. The concentrated sunlight boils the water, generating high-pressure steam for direct use in power generation and industrial steam applications.
3 0
3 years ago
Read 2 more answers
What is CQ Thread Ball Valves​
hodyreva [135]
That is a thread ball valves

8 0
3 years ago
Read 2 more answers
What is the composition, in atom percent, of an alloy that consists of 4.5 wt% Pb and 95.5 wt% Sn?
jeka57 [31]

Answer: Option A is correct -- 2.6 at% Pb and 97.4 at% Sn.

Explanation:

Option A is the only correct option -- 2.6 at% Pb and 97.4 at% Sn. While option B, which is 7.6 at% Pb and 92.4 at% Sn. and option C, which is 97.4 at% Pb and 2.6 at% Sn. and option D, which is 92.4 at% Pb and 7.6 at% Sn. are wrong.

6 0
3 years ago
Maggie discovered that a pipe in her basement has sprung a leak. She calls a plumber but in the meantime she grabs a roll of duc
dem82 [27]

Answer:

The answer is "Option a".

Explanation:

Myelination was the myelinization mechanism of a neuron axon. The endothelium is enveloped all around the axon and isolates the axon that inhibits the neuronal message from leaking with the other neuronal axons. Inside this example, therefore, its tubes tape worked similarly to those of myelin sheath, which stops brain transmission.

3 0
3 years ago
What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that durin
DochEvi [55]

Answer:

m_{LP}=0.45\,kg

Explanation:

Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:

Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]

Q_{water} = 3599.435\,kJ

The heat liberated by the LP gas is:

Q_{LP} = \frac{3599.435\,kJ}{0.16}

Q_{LP} = 22496.469\,kJ

A kilogram of LP gas has a minimum combustion power of 50028\,kJ. Then, the required mass is:  

m_{LP} = \frac{22496.469\,kJ}{50028\,\frac{kJ}{kg} }

m_{LP}=0.45\,kg

6 0
3 years ago
Other questions:
  • Identify each statement as referring to a series or parallel circuit.
    15·1 answer
  • Which of the following is not a primary or fundamental dimension? (a)-mass m (b)-length L (c)- timer t (d)-volume V
    5·1 answer
  • 1 2 3 4 5 6 7 8 9 10
    14·1 answer
  • How do you calculate the dynamic lift in an aeroplane?
    5·1 answer
  • When a conductor is moved in a magnetic field, a voltage will be induced on the conductor. However, current won't flow through t
    11·1 answer
  • Write a statement that increases numPeople by 5. Ex: If numPeople is initially 10, the output is: There are 15 people.
    11·1 answer
  • // This program accepts data about 100 books and// determines a price for each.// The price is 10 cents per page for the// first
    12·1 answer
  • A kite is an airfoil that uses the wind to produce a lift. Held in place by a string, a kite can remain aloft indefinitely. The
    9·1 answer
  • What is the difference between POP3 and IMAP?
    5·1 answer
  • At the coast on a summer day, the land is hotter than the ocean. Warm air over the land rises and is replaced by cooler air, cau
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!