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goldenfox [79]
3 years ago
7

Consider the mixing of 0.8 kg/s of hot water at 348 K and 1 kg/s of cool water at 298 K that is generating warm water. Assume no

work is being done and the system is in steady state, but heat is lost at the rate of 30 kJ/s during this mixing. Find the temperature of the warm water flow stream? Assume Cp =4.18 kJkg
Engineering
1 answer:
Drupady [299]3 years ago
3 0

Answer:

T_warm = 47.22 C

Explanation:

Using energy balance for the system:

m_1*h_1 + m_2*h_2 = m_3*h3   ... Eq1

h_i = c_p. T_i   ... Eq 2

m_1 + m_2 = m_3   ... steady flow system (Eq 3)

Substitute Eq 2 and Eq3 in Eq1

m_3 = 0.8 + 1 = 1.8 kg/s

(0.8)*(4.18)*( 348-273) + (1)*(4.18)*( 298-273) = 1.8 * 4.18 *T_3

T_3 = 355.3 / (1.8*4.18) = 47.22 C

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6. Find the heat flow in 24 hours through a refrigerator door 30.0" x 58.0" insulated with cellulose fiber 2.0" thick. The tempe
Ilia_Sergeevich [38]

Answer:

The heat flow in 24 hours through the refrigerator door is approximately 1,608.57 BTU

Explanation:

The given parameters are;

The duration of the heat transfer, t = 24 hours = 86,400 seconds

The area of the refrigerator door, A = 30.0" × 58.0" = 1,740 in.² = 1.122578 m²

The material of the insulator in the door = Cellulose fiber

The thickness of the insulator in the door, d = 2.0" = 0.0508 m

The temperature inside the fridge = 38° F = 276.4833 K

The temperature of the room = 78°F = 298.7056 K

The thermal conductivity of cellulose fiber = 0.040 W/(m·K)

By Fourier's law, the heat flow through a by conduction material is given by the following formula;

\dfrac{Q}{t} = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d}

Q = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d} \times t

Therefore, we have;

Q = \dfrac{0.04 \times 1.122578 \times (298.7056 - 276.4833 ) }{0.0508} \times 86,400 =1,697,131.73522

The heat flow in 24 hours through the refrigerator door, Q = 1,697,131.73522 J = 1,608.5705140685 BTU

7 0
3 years ago
Write the heat equation for each of the following cases:
jok3333 [9.3K]

Answer:

Explanation:

a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:

\dfrac{\partial^2T}{\partial x^2}=  \ 0  \  ;  \ if \  T = f(x)  \\ \\ \dfrac{\partial^2T}{\partial y^2}=  \ 0  \  ;  \ if \  T = f(y)  \\ \\ \dfrac{\partial^2T}{\partial z^2}=  \ 0  \  ;  \ if \  T = f(z)

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}

where;

Q_g = heat generated per unit volume

\alpha = Thermal diffusivity

c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0

where;

The radial directional term = \dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) and the axial directional term is \dfrac{\partial^2 T}{\partial z^2 }

d) The heat equation for a wire going through a furnace is:

\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]

since;

the steady-state is zero, Then:

\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]'

e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:

\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}

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A cylindrical bar of metal having a diameter of 17.8 mm and a length of 196 mm is deformed elastically in tension with a force o
ExtremeBDS [4]

Answer:

A) ΔL = 0.544 mm

B) Δd = -0.0168 mm

Explanation:

We are given;

Elastic modulus; E = 67.1 GPa = 67.1 × 10^(9) Pa

Force; F = 46400 N

Diameter; d = 17.8 mm = 17.8 × 10^(-3) m

Radius; r = d/2 = 17.8/2 = 8.9 mm = 8.9 × 10^(-3) m

Length; L = 196 mm = 0.196 m

Poisson ratio; ν = 0.34

A) We know that formula for elastic modulus is;

E = σ/ε

Where;

σ = F/A

ε = ΔL/L

Thus;

E = (FL/ΔL•A)

ΔL is change in length. Making it the subject of the formula, we have;

ΔL = FL/AE

Now, A = πr²

A = π × (8.9 × 10^(-3))²

ΔL = [(0.196 × 46400)/(π × (8.9 × 10^(-3))² × 67.1 × 10^(9)]

ΔL = 0.544 × 10^(-3) m

ΔL = 0.544 mm

B) formula for Poisson ratio is given as;

ν = -ε_x/ε_z

Where;

ε_x is transverse strain = Δd/d

ε_z is longitudinal strain = ΔL/L

Thus;

ν = -Δd•L/d•ΔL

Making Δd the subject, we have;

Δd = -νdΔL/L

Δd = -(0.34 × 17.8 × 10^(-3) × 0.544 × 10^(-3))/0.196

Δd = -0.0168 × 10^(-3) m = -0.0168 mm

7 0
3 years ago
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