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
TiliK225 [7]
3 years ago
12

The heat transfer coefficient decreases with increasing x for both the laminar and turbulent regions a. True b. False

Engineering
1 answer:
REY [17]3 years ago
3 0

Answer:

A) True  

Explanation:

Yes this is true when length is creases the heat transfer coefficient decease with length.

The heat transfer(h) coefficient is varying with x by given expression

For Laminar flow

h \alpha \dfrac{1}{x^{\frac{1}{2}}}

For turbulent flow

h \alpha \dfrac{1}{x^{\frac{1}{5}}}

But when flow is in transitional state the heat heat transfer(h) coefficient is increases with x.But for laminar as well as turbulent flow h is decrease when x increases.

You might be interested in
A piston/cylinder contains 1.5 kg of water at 200 kPa, 150°C. It is now heated by a process in which pressure is linearly relate
Fofino [41]

Answer:

final volume V2 = 0.71136 m³

work done in process W = -291.24 kJ

heat transfer Q = 164 kJ

Explanation:

given data

mass = 1.5 kg

pressure p1 = 200 kPa

temperature t1 = 150°C

final pressure p2 = 600 kPa

final temperature t2 = 350°C

solution

we will use here superheated water table that is

for pressure 200 kPa and 150°C temperature

v1 = 0.95964 m³/kg

u1 = 2576.87 kJ/kg

and

for pressure 600 kPa and 350°C temperature

v2 = 0.47424 m³/kg

u2 = 2881.12 kJ/kg

so v1 is express as

V1 = v1 × m    ............................1

V1 = 0.95964 × 1.5

V1 = 1.43946 m³

and

V2 = v2 × m    ............................2

V2 = 0.47424 × 1.5

final volume V2 = 0.71136 m³

and

W = P(avg) × dV      .............................3

P(avg) = \frac{p1+p2}{2}    = \frac{200+600}{2} = 400 × 10³

put here value

W = 400 × 10³ × (0.71136 - 1.43946 )

work done in process W = -291.24 kJ

and

heat transfer is

Q = m × (u2 - u1)  + W       .............................4

Q = 1.5 × (2881.12 - 2576.87)  + 292.24

heat transfer Q = 164 kJ

7 0
3 years ago
1. Springs____________<br> energy when compressed<br> And _________energy when they rebound.
densk [106]

Answer:

Springs store energy when compressed and release energy when they rebound

Explanation:

6 0
3 years ago
Which explanation best summarizes what went wrong during Paul’s cost analysis?
Valentin [98]

Answer:

wut is it

Explanation:

4 0
2 years ago
A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a
bazaltina [42]

Answer:

Q_{cv}=-339.347kJ

Explanation:

First we calculate the mass of the aire inside the rigid tank in the initial and end moments.

P_iV_i=m_iRT_i (i could be 1 for initial and 2 for the end)

State1

1bar*|\frac{100kPa}{1}|*2=m_1*0.287*295

m_1=232kg

State2

8bar*|\frac{100kPa}{1bar}|*2=m_2*0.287*350

m_2=11.946

So, the total mass of the aire entered is

m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg

At this point we need to obtain the properties through the tables, so

For Specific Internal energy,

u_1=210.49kJ/kg

For Specific enthalpy

h_1=295.17kJ/kg

For the second state the Specific internal Energy (6bar, 350K)

u_2=250.02kJ/kg

At the end we make a Energy balance, so

U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e

No work done there is here, so clearing the equation for Q

Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)

Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)

Q_{cv}=-339.347kJ

The sign indicates that the tank transferred heat<em> to</em> the surroundings.

8 0
3 years ago
Water vapor at 5 bar, 320°C enters a turbine operating at steady state with a volumetric flow rate of 0.65 m3/s and expands adia
harina [27]

Answer:

Power = 371.28 kW

Explanation:

Initial pressure, P1 = 5 bar

Final pressure, P2 = 1 bar

Initial temperature, T1 = 320°C

Final temperature, T2 = 160°C

Volume flow rate, V = 0.65m³/s

From steam tables at state 1,

h1 = 3105.6 kJ/kg, s1 = 7.5308 kJ/kgK

v1 = 0.5416 m³/kg

Mass flow rate, m = V/v1

m = 1.2 kg/s

From steam tables, at state 2

h2 = 2796.2 kJ/kg, s2 = 7.6597 kJ/kgK

Power developed, P = m(h1 - h2)

P = 1.2(3105.6-2796.2)

P = 371.28 kW

8 0
4 years ago
Other questions:
  • An aluminum cylinder bar ( 70 GPa E m = ) is instrumented with strain gauges and is subject to a tensile force of 5 kN. The diam
    9·1 answer
  • A level loop began and closed on BM_A (elevation = 823.368 ft). The plus and minus sights were kept approximately equal. Reading
    11·1 answer
  • To remove a spark plug the technician would need a(n) ___socket​
    7·2 answers
  • Which of the following is part of the highway
    11·2 answers
  • When you do a vehicle check, what do you NOT need to keep an eye on?
    9·1 answer
  • PLS HURRY!!!
    10·2 answers
  • In a movie theater in winter, 510 people, each generation sensible heat at a rate of 80 W, are watching a movie. The heat losses
    11·1 answer
  • How do you build a house.
    15·1 answer
  • What is the moment that the wrench puts on the bolt?
    13·1 answer
  • 3. Determine the most unfavorable arrangement of the crane loads and
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!