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Jlenok [28]
3 years ago
13

Water at 20oC, with a free-stream velocity of 1.5 m/s, flows over a circular pipe with diameter of 2.0 cm and surface temperatur

e of 80oC. Calculate the average heat transfer coefficient and the heat transfer rate per meter length of pipe.\
Engineering
1 answer:
Lisa [10]3 years ago
3 0

Answer:

Average heat transfer coefficient =  31 kw/m^2 k

Heat transfer rate per meter length of pipe =  116.808 KW

Explanation:

water temperature = 20⁰c,  

free-stream velocity = 1.5 m/s

circular pipe diameter = 2.0 cm = 0.02 m

surface temperature = 80⁰c

A) calculate average heat transfer coefficient

we apply the formula below :

m = αAv

A (area) = \pi /4 (d)^2

m = 10^3 * \pi  / 4 ( 0.02)^2 * 1.5

   = 10^3 * 0.7857( 0.0004) * 1.5

   = 0.4714 kg/s

Average heat transfer coefficient  

h = \frac{m(cp)}{A}  ,  A = \pi DL

L = 1 m , m = 0.4714 kgs , cp = 4.18

back to equation

h = \frac{0.4714*4.18}{\pi * 0.02 }   = 1.970 / 0.0628 = 31.369 ≈ 31 kw/m^2 k

B) Heat transfer rate per meter length of pipe

Q = ha( ΔT ),  a = \pi DL

   = 31 * 0.0628 * ( 80 - 20 )

  = 31 * 0.0628 * 60 = 116.808 KW

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The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator at a locati
Lynna [10]

Answer:

a) 0.76

b) 0.80

c) 1964 kW

Explanation:

GIVEN DATA:

\dot m = 5000 kg/s

Assume Mechanical energy at exist is negligible

A) Take lake bottom as reference, and then kinetic and potential energy  are taken as zero.

change in mechanical energy is givrn as

e_{in} - e_{out} = \frac{P}{\rho} - 0 = gh = 9.81 \times 50( \frac{1 kJ/kg}{1000 m^2/s^2}

                         = 0.491 kJ/kg

\Delta \dot E_{mec} = \dot m (e_{in} - e_{out}) = 5000 \times 0.491 = 2455 kW

\eta_{OVERALL}  = \frac{\dot W}{\Delta \dot E_{mec}} = \frac{1862}{2455} = 0.76

B) \eta -{gen} = \frac{\eta_{overall}}{\eta_{gen}} = \frac{0.76}{0.95} = 0.80

c) \dot W_{shaft} = \eta_{overall} \left | \Delta \dot E_{mec} \right | = 0.80(2455)

\dot W_{shaft} = 1964 kW

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
A thick oak wall initially at 25°C is suddenly exposed to gases for which T =800°C and h =20 W/m2.K. Answer the following questi
Schach [20]

Answer:

a) What is the surface temperature, in °C, after 400 s?

   T (0,400 sec) = 800°C

b) Yes, the surface temperature is greater than the ignition temperature of oak (400°C) after 400 s

c) What is the temperature, in °C, 1 mm from the surface after 400 s?

   T (1 mm, 400 sec) = 798.35°C

Explanation:

oak initial Temperature = 25°C = 298 K

oak exposed to gas of temp = 800°C = 1073 K

h = 20 W/m².K

From the book, Oak properties are e=545kg/m³   k=0.19w/m.k   Cp=2385J/kg.k

Assume: Volume = 1 m³, and from energy balance the heat transfer is an unsteady state.

From energy balance: \frac{T - T_{\infty}}{T_i - T_{\infty}} = Exp (\frac{-hA}{evCp})t

Initial temperature wall = T_i

Surface temperature = T

Gas exposed temperature = T_{\infty}

6 0
3 years ago
1. A glass window of width W = 1 m and height H = 2 m is 5 mm thick and has a thermal conductivity of kg = 1.4 W/m*K. If the inn
emmasim [6.3K]

Answer:

1. \dot Q=19600\ W

2. \dot Q=120\ W

Explanation:

1.

Given:

  • height of the window pane, h=2\ m
  • width of the window pane, w=1\ m
  • thickness of the pane, t=5\ mm= 0.005\ m
  • thermal conductivity of the glass pane, k_g=1.4\ W.m^{-1}.K^{-1}
  • temperature of the inner surface, T_i=15^{\circ}C
  • temperature of the outer surface, T_o=-20^{\circ}C

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

\dot Q=k_g.A.\frac{dT}{dx}

here:

A = area through which the heat transfer occurs = 2\times 1=2\ m^2

dT = temperature difference across the thickness of the surface = 35^{\circ}C

dx = t = thickness normal to the surface = 0.005\ m

\dot Q=1.4\times 2\times \frac{35}{0.005}

\dot Q=19600\ W

2.

  • air spacing between two glass panes, dx=0.01\ m
  • area of each glass pane, A=2\times 1=2\ m^2
  • thermal conductivity of air, k_a=0.024\ W.m^{-1}.K^{-1}
  • temperature difference between the surfaces, dT=25^{\circ}C

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>

\dot Q=k_a.A.\frac{dT}{dx}

\dot Q=0.024\times 2\times \frac{25}{0.01}

\dot Q=120\ W

5 0
3 years ago
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