Ask question

Ask question

All categories

bearhunter [10]

4 years ago

13

A composite wall is composed of 20 cm of concrete block with k = 0.5 W/m-K and 5 cm of foam insulation with k = 0.03 W/m-K. The

temperature of on the foam side is 25oC while the temperature on the block side is 0oC. What is the temperature at the interface between the block and the foam?

1 answer:

wariber [46]4 years ago

3 0

Answer:

4.8°C

Explanation:

The rate of heat transfer through the wall is given by:

$q=\frac{Ak}{L}dT$

$\frac{q}{A}=\frac{k}{L}dT$

Assumptions:

1) the system is at equilibrium

2) the heat transfer from foam side to interface and interface to block side is equal. There is no heat retention at any point

3) the external surface of the wall (concrete block side) is large enough that all heat is dissipated and there is no increase in temperature of the air on that side

${k_{fi}= 0.03 W/m.K$

${L_{fi}= 5 cm = 0.05 m$

${T_{fi}= 25 \°C$

${k_{cb} = 0.5 W/m.K$

${L_{cb}= 20 cm = 0.20 m$

${T_{cb}= 0 \°C$

${T_{m}= ? \°C =$ temperature at the interface

Solving for ${T_{m}$ will give the temperature at the interface:

$\frac{q}{A}=\frac{k_{fi} }{L_{fi} }(T_{fi} -T_{m})=\frac{k_{cb} }{L_{cb} }(T_{m} -T_{cb})$

$\frac{0.03}{0.05 }(25 -T_{m})=\frac{0.5}{0.2}(T_{m} -0})$

$15 -0.6T_{m}=2.5T_{m}$

$3.1T_{m}=15$

$T_{m}=4.8$

You might be interested in

What are the available motor sizes for 2023 ariya ac synchronous drive motor systems in kw?.

Anika [276]

The available motor sizes for 2023 Ariya AC synchronous drive motor systems are:

40 kW.

62 kW.

160 kW.

<h3>What is a synchronous motor?</h3>

A synchronous motor refers to an alternating current (AC) electric motor in which the rotational speed of the shaft is directly proportional (equal) to the frequency of the supply current, especially at a steady state.

In Engineering, the available motor sizes for 2023 Nissan Ariya AC synchronous drive motor systems include the following:

40 kW.

62 kW.

160 kW.

Read more on synchronous motor here:

brainly.com/question/12975042

#SPJ1

5 0

2 years ago

A large increase in elevation can cause a carbureted engine to run ________ if not properly adjusted for the altitude. a Rich b

mash [69]

Answer:

B - Poor

Explanation:

As you get higher up, There is less oxygen which causes the engine to create less power.

3 0

3 years ago

Q4. (20 points) For a bronze alloy, the stress at which plastic deformation begins is 271 MPa and the modulus of elasticity is 1

babunello [35]

Answer:

a) P = 86720 N

b) L = 131.2983 mm

Explanation:

σ = 271 MPa = 271*10⁶ Pa

E = 119 GPa = 119*10⁹ Pa

A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²

a) P = ?

We can apply the equation

σ = P / A ⇒ P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N

b) L₀ = 131 mm = 0.131 m

We can get ΔL applying the following formula (Hooke's Law):

ΔL = (P*L₀) / (A*E) ⇒ ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)

⇒ ΔL = 2.9832*10⁻⁴ m = 0.2983 mm

Finally we obtain

L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm

3 0

4 years ago

PELASEE HELPPP WITH MARK BRAINLIST!!!! You are stopped at a red light, and a long string of cars is crossing in front of you. Wh

choli [55]

Answer:

1st one.

Explanation:

I think that because they the one who is going to get a ticket and also you will not gonna get in a car accident.

7 0

3 years ago

Read 2 more answers

An ideal Otto cycle has a compression ratio of 9.2 and uses air as the working fluid. At the beginning of the compression proces

Allushta [10]

Answer:

(a) The amount of heat transferred to the air, $q_{out}$ is 215.5077 kJ/kg

(b) The net work output, $W_{net}$ , is 308.07 kJ/kg

(c) The thermal efficiency is 58.8%

(d) The Mean Effective Pressure, MEP, is 393.209 kPa

Explanation:

(a) The assumptions made are;

$c_p$ = 1.005 kJ/(kg·K), $c_v$ = 0.718 kJ/(kg·K), R = 0.287 kJ/(kg·K),

Process 1 to 2 is isentropic compression, therefore;

$T_{2}= T_{1}\left (\dfrac{v_{1}}{v_{2}} \right )^{k-1} = 300.15\times 9.2^{0.4} = 729.21 \, K$

From;

$\dfrac{p_{1}\times v_{1}}{T_{1}} = \dfrac{p_{2}\times v_{2}}{T_{2} }$

We have;

$p_{2} = \dfrac{p_{1}\times v_{1}\times T_{2}}{T_{1} \times v_{2}} = \dfrac{98\times 9.2\times 729.21}{300.15 } = 2190.43 \, kPa$

Process 2 to 3 is reversible constant volume heating, therefore;

$\dfrac{p_3}{T_3} =\dfrac{p_2}{T_2}$

p₃ = 2 × p₂ = 2 × 2190.43 = 4380.86 kPa

$T_3 = \dfrac{p_3 \times T_2}{p_2} =\dfrac{4380.86 \times 729.21}{2190.43} = 1458.42 \, K$

Process 3 to 4 is isentropic expansion, therefore;

$T_{3}= T_{4}\left (\dfrac{v_{4}}{v_{3}} \right )^{k-1}$

$1458.42= T_{4} \times \left (9.2 \right )^{0.4}$

$T_4 = \dfrac{1458.42}{(9.2)^{0.4}} = 600.3 \, K$

$q_{out} = m \times c_v \times (T_4 - T_1) = 0.718 \times (600.3 - 300.15) = 215.5077 \, kJ/kg$

The amount of heat transferred to the air, $q_{out}$ = 215.5077 kJ/kg

(b) The net work output, $W_{net}$ , is found as follows;

$W_{net} = q_{in} - q_{out}$

$q_{in} = m \times c_v \times (T_3 - T_2) = 0.718 \times (1458.42 - 729.21) = 523.574 \, kJ/kg$

$\therefore W_{net} = 523.574 - 215.5077 = 308.07 \, kJ/kg$

(c) The thermal efficiency is given by the relation;

$\eta_{th} = \dfrac{W_{net}}{q_{in}} \times 100= \dfrac{308.07}{523.574} \times 100= 58.8\%$

(d) From the general gas equation, we have;

$V_{1} = \dfrac{m\times R\times T_{1}}{p_{1}} = \dfrac{1\times 0.287\times 300.15}{98} =0.897\, m^{3}/kg$

The Mean Effective Pressure, MEP, is given as follows;

$MEP =\dfrac{W_{net}}{V_1 - V_2} = \dfrac{W_{net}}{V_1 \times (1- 1/r)}= \dfrac{308.07}{0.897\times (1- 1/9.2)} = 393.209 \, kPa$

The Mean Effective Pressure, MEP = 393.209 kPa.

3 0

3 years ago