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Ugo [173]
3 years ago
8

A long rod of 60-mm diameter and thermophysical properties rho=8000 kg/m^3, c=500J/kgK, and k=50 W/mK is initally at a uniform t

emperature in a forced convection furnace maintained at 750K. The convection coeffiecent is estimated to be 1000 W/m^2K.
What is the centerline temperature of the rod when the surface temperature is 550K?
Engineering
1 answer:
Monica [59]3 years ago
3 0

Answer:

Tc = 424.85 K

Explanation:

Given that,

D = 60 mm = 0.06 m

\rho = 8000 kg/m^3

k = 50 w/m . kc = 500 j/kg.k

h_{\infty} = 1000 w/m^2t_{\infity} = 750 kt_w = 500 K

surface area = As = \pi dL \\\frac{As}{L} = \pi D = \pi \timeS 0.06

HEAT FLOW Q  is

Q = h_{\infty} As (T_[\infty} - Tw)  = 1000 \pi\times 0.06 (750-500)

 = 47123.88 w per unit length of rod

volumetric heat rate

q = \frac{Q}{LAs}

= \frac{47123.88}{\frac{\pi}{4} D^2 \times 1}

q = 1.66\times 10^{7} w/m^3

Tc = \frac{- qR^2}{4K} + Tw

= \frac{ - 1.67\times 10^7 \times (\frac{0.06}{2})^2}{4\times 50} +  500

  = 424.85 K

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A 0.06-m3 rigid tank initially contains refrigerant- 134a at 0.8 MPa and 100 percent quality. The tank is connected by a valve t
lesantik [10]

Answer:

a) 0.50613

b) 22.639 kJ

Explanation:

From table A-11 , we will make use of the properties of Refrigerant R-13a at 24°C

first step : calculate the  volume of R-13a  ( values gotten from table A-11 )

V = m1 * v1 = 5 * 0.0008261 = 0.00413 m^3

next : calculate final specific volume ( v2 )

v2 = V / m2 = 0.00413 / 0.25 ≈ 0.01652 m^3/kg

<u>a) Calculate the mass of refrigerant that entered the tank </u>

v2 = Vf + x2 * Vfg

v2 = Vf +  [ x2 * ( Vg - Vf ) ] ----- ( 1 )

where:  Vf = 0.0008261 m^3/kg, V2 = 0.01652 m^3/kg , Vg = 0.031834 m^3/kg  ( insert values into equation 1 above )

x2 = ( 0.01652 - 0.0008261 ) / 0.031834

     = 0.50613 ( mass of refrigerant that entered tank )

<u>b) Calculate the amount of heat transfer </u>

Final specific internal energy = u2 = Uf + ( x2 + Ufg ) ----- ( 2 )

uf = 84.44 kj/kg , x2 = 0.50613 , Ufg = 158.65 Kj/kg

therefore U2 = 164.737 Kj/kg

The mass balance  ( me ) = m1 - m2 --- ( 3 )

energy balance( Qin ) = ( m2 * u2 ) - ( m1 * u1 ) + ( m1 - m2 ) * he

therefore Qin = 41.184 - 422.2 + 403.655  = 22.639 kJ

3 0
3 years ago
What steps would you take to design an improved toothpaste container?
algol [13]

Answer:

A. Identify the need, recognize limitations of current toothpaste containers, and then brainstorm ideas on how to improve the existing

Explanation:

To design an improved toothpaste container, we must identify the needs of the customer, one of the major need is to make the container attractive to the sight. This is the first thing that will prompt a customer to wanting to buy the product (The reflectance/appearance).

Then recognize the limitation of the current design, what needed change. This will help in determining what is needed to be included and what should be removed based on identified customers need.

The last step is to brainstorm ideas on how to improve the existing designs. Get ideas from other colleagues because there is a saying that two heads are better than one. This will help in coming to a reasonable conclusion on the new design after taking careful consideration of people's opinion.

7 0
3 years ago
A market structure in which ther a single buyer
faltersainse [42]

Answer:

Monopsony

Explanation:

Kinda like a monopoly, except instead of one producer it is one buyer.

8 0
2 years ago
1 kg of saturated steam at 1000 kPa is in a piston-cylinder and the massless cylinder is held in place by pins. The pins are rem
BARSIC [14]

Answer:

The final specific internal energy of the system is 1509.91 kJ/kg

Explanation:

The parameters given are;

Mass of steam = 1 kg

Initial pressure of saturated steam p₁ = 1000 kPa

Initial volume of steam, = V₁

Final volume of steam = 5 × V₁

Where condition of steam = saturated at 1000 kPa

Initial temperature, T₁  = 179.866 °C = 453.016 K

External pressure = Atmospheric = 60 kPa

Thermodynamic process = Adiabatic expansion

The specific heat ratio for steam = 1.33

Therefore, we have;

\dfrac{p_1}{p_2} = \left (\dfrac{V_2}{V_1} \right )^k = \left [\dfrac{T_1}{T_2}   \right ]^{\dfrac{k}{k-1}}

Adding the effect of the atmospheric pressure, we have;

p = 1000 + 60 = 1060

We therefore have;

\dfrac{1060}{p_2} = \left (\dfrac{5\cdot V_1}{V_1} \right )^{1.33}

P_2= \dfrac{1060}{5^{1.33}}  = 124.65 \ kPa

\left [\dfrac{V_2}{V_1} \right ]^k = \left [\dfrac{T_1}{T_2}   \right ]^{\dfrac{k}{k-1}}

\left [\dfrac{V_2}{V_1} \right ]^{k-1} = \left \dfrac{T_1}{T_2}   \right

5^{0.33} = \left \dfrac{T_1}{T_2}   \right

T₁/T₂ = 1.70083

T₁ = 1.70083·T₂

T₂ - T₁ = T₂ - 1.70083·T₂

Whereby the temperature of saturation T₁ = 179.866 °C = 453.016 K, we have;

T₂ = 453.016/1.70083 = 266.35 K

ΔU = 3×c_v×(T₂ - T₁)

c_v = cv for steam at 453.016 K = 1.926 + (453.016 -450)/(500-450)*(1.954-1.926) = 1.93 kJ/(kg·K)

cv for steam at 266.35 K = 1.86  kJ/(kg·K)

We use cv given by  (1.93 + 1.86)/2 = 1.895 kJ/(kg·K)

ΔU = 3×c_v×(T₂ - T₁) = 3*1.895 *(266.35 -453.016) = -1061.2 kJ/kg

The internal energy for steam = U_g = h_g -pV_g

h_g = 2777.12 kJ/kg

V_g = 0.194349 m³/kg

p = 1000 kPa

U_{g1} = 2777.12 - 0.194349 * 1060 = 2571.11 kJ/kg

The final specific internal energy of the system is therefore, U_{g1} + ΔU = 2571.11 - 1061.2 = 1509.91 kJ/kg.

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3 years ago
A team of engineers is working on a design to increase the power of a hydraulic lever. They have brainstormed several ideas. Whi
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Answer:

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