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aev
2 months ago
14

A pipe in a district heating network is transporting over-pressurized hot water (10 atm) at a mass flow of 0.5 kg/s. The pipe is

5 m long, has an inner radius of 50 cm and pipe wall thickness of 50 mm. The pipe has a thermal conductivity of 20 W/m-K, and the inner pipe surface is at a uniform temperature of 110 ºC. The convection heat transfer coefficient of the air surrounding the pipe is 100W/m2 -K. The temperature of the water at inlet of pipe is 130 ºC and the constant pressure specific heat of hot water is 4000 J/kg-ºC. If the temperature of the air surrounding the pipe is 20 ºC, determine the exit temperature of the water at the end of the pipe.
Engineering
1 answer:
alex41 [359]2 months ago
6 0

Answer:3.6 kg/s 14 w/m-k

Explanation:

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A rod is 2m long at temperature of 10oC. Find the expansion of the rod, when the temperature is raised to 80oC. If this expansio
Kisachek [356]

Answer:

ΔL = 1.68 mm

σ = 84 MPa

Explanation:

Thermal expansion can be calculated as:

ΔL = α ΔT L

And thermal stress is given by:

σ = α ΔT E

Provided values:

α = 1.2×10⁻⁵ /°C

E = 1.0×10⁵ MPa

ΔT = 80°C − 10°C = 70°C

L = 2 m

Thus, ΔL can be calculated as follows:

ΔL = (1.2×10⁻⁵ /°C) (70°C) (2 m)

ΔL = 0.00168 m

ΔL = 1.68 mm

For the thermal stress:

σ = (1.2×10⁻⁵ /°C) (70°C) (1.0×10⁵ MPa)

σ = 84 MPa

8 0
2 months ago
A test machine that kicks soccer balls has a 5-lb simulated foot attached to the end of a 6-ft long pendulum arm of negligible m
choli [298]

Answer:

a) v₂ = 26.6 ft/s

b) v₂ = 31.9 ft/s

Explanation:

a) Applying the principle of energy conservation for the soccer arm from the point of impact to its lowest point:

mgh=\frac{1}{2} mv^{2} \\v=\sqrt{2gh} =\sqrt{2*32.2*6} =19.66ft/s

The ball's velocity in the tangential direction is given by:

v₂*sinθ = v₁*sin30

With the values:

if v₁ = 0

θ = 0º

The restitution coefficient is:

e=\frac{v_{2}*cosO-v_{2}cos30 }{v*cos30-(-v_{1}*cos30 )} \\0.8=\frac{v_{2}cos0-v_{2}*cos30 }{19.66*cos30-(0*cos30)} \\v_{2} =\frac{v_{2}-13.6 }{cos30}

where O=θ

The aggregate momentum equation is:

mv-mv_{1} =mv_{2} +mv_{2} cos(O+30)\\, where O = θ

mv-mv_{1} =m(\frac{v_{2}-13.6 }{cos30} )+mv_{2} cos(O+30)

When we incorporate gravitational acceleration:

mgv-mgv_{1} =mg(\frac{v_{2}-13.6 }{cos30} )+mgv_{2} cos(O+30)\\5*19.66-1*0=5*(\frac{v_{2}-13.6 }{cos30} )+1*v_{2} cos(O+30)

Isolating v₂ results in:

v₂ = 26.6 ft/s

b) To find the ball's velocity, we utilize:

v₂ * sinθ = v₁ * sin30

if v₁ = 10 ft/s

then v₂ * sinθ = 10 * sin30

thus sinθ = 5/v₂

The restitution coefficient remains:

e=\frac{v_{2}*cosO-v_{2}cos30 }{v*cos30-(-v_{2}*cos30) } \\0.8=\frac{v_{2}cosO-v_{2}cos30 }{19.66*cos30-(-10*cos30)} \\v_{2} =\frac{v_{2}cos30-20.55}{cos30}

where O = θ

mgv-mgv_{1} =mgv_{2} +mgv_{2} cos(O+30)\\5*19.66-1*0=5v_{2} +v_{2} cos(O+30)\\98.3=5v_{2}+v_{2} cosOcos30-v_{2} sinOsin30

Solving for v₂ yields:

v₂ = 31.9 ft/s

6 0
2 months ago
Consider a rectangular fin that is used to cool a motorcycle engine. The fin is 0.15m long and at a temperature of 250C, while t
Kisachek [356]

Answer:

q' = 5826 W/m

Explanation:

Given:-

- The length of the fin in question, L = 0.15 m

- The fin's surface temperature, Ts = 250°C

- The velocity of free stream air, U = 80 km/h

- The air temperature, Ta = 27°C

- The flow is parallel over both sides of the fin, assuming turbulent flow conditions throughout.

Find:-

What is the heat removal rate per unit width of the fin?

Solution:-

- Steady state conditions are assumed, along with negligible radiation and turbulent flow conditions.

- From Table A-4, we gather air properties (T = 412 K, P = 1 atm ):

    Dynamic viscosity, v = 27.85 * 10^-6 m²/s  

    Thermal conductivity, k = 0.0346 W / m.K

    Prandtl number Pr = 0.69

- Compute the Nusselt Number (Nu) corresponding to - turbulent conditions - using the relevant relationship as follows:

                          Nu = 0.037*Re_L^\frac{4}{5} * Pr^\frac{1}{3}

Where,    Re_L: Average Reynolds number for the entire length of fin:

                          Re_L = \frac{U*L}{v} \\\\Re_L = \frac{80*\frac{1000}{3600} * 0.15}{27.85*10^-^6} \\\\Re_L = 119688.80909

Consequently,

                         

Nu = 0.037*(119688.80909)^\frac{4}{5} * 0.69^\frac{1}{3}\\\\Nu = 378

- The convective heat transfer coefficient (h) can now be derived from:

                          h = \frac{k*Nu}{L} \\\\h = \frac{0.0346*378}{0.15} \\\\h = 87 \frac{W}{m^2K}

- The heat loss rate q' per unit width can be established using the convection heat transfer formula and should be multiplied by (x2) since the airflow is present on both sides of the fin:

                          q' = 2*[h*L*(T_s - T_a)]\\\\q' = 2*[87*0.15*(250 - 27)]\\\\q' = 5826\frac{W}{m}

- Ultimately, the heat loss per unit width from the rectangular fin is q' = 5826 W/m

- The thermal loss per unit width (q') attributed to radiation:

                  q' = 2*a*T_s^4*L

Where, a signifies the Stefan-Boltzmann constant = 5.67*10^-8

                  q' = 2*5.67*10^-^8*(523)^4*0.15\\\\q' = 1273 \frac{W}{m}

- It is observed that radiation losses are not insignificant, accounting for 20% of thermal loss by convection. As the emissivity (e) of the fin is unspecified, this value is dismissed from the calculations as it pertains to the provided information.

7 0
2 months ago
Gold and silver rings can receive an arc and turn molten. True or False
grin007 [323]
That statement is incorrect!
The claim is untrue
8 0
3 months ago
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The first step to merging is entering the ramp and _____.
alex41 [359]

Answer:

D.informing your passengers of your destination

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