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SCORPION-xisa [38]
4 years ago
7

A composite wall is to be used to insulate a freezer chamber at -350C. Two insulating materials are to be used with conductiviti

es of 0.04 W/mK and 0.1 W/mK. If the outside surface temperature of the inner layer (0.04 W/mK) should not go below zero and if the exposed surface temperature should not go below (the expected wet bulb temperature) 220C determine the insulation thicknesses. The heat gain is to be limited to 10 W/m^2. Also estimate the value of convection coefficient at such a situation. Outside is at 250C.
Engineering
1 answer:
choli [55]4 years ago
7 0

Answer:

thickness1=1.4m

thickness2=2.2m

convection coefficient=0.33W/m^2K

Explanation:

you must use this equation to calculate the thickness:

L=K(T2-T1)/Q

L=thickness

T=temperature

Q=heat

L1=0.04*(0--350)/10=1.4m

L2=0.1(220-0)/10=2.2m

Then use this equation to calculate the convective coefficient

H=Q/(T2-T1)

H=10/(250-220)=0.33W/m^2K

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soldier1979 [14.2K]

Answer:

Java program given below

Explanation:

import java.util.*;

import java.io.*;

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{

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class Node

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public void Displaylist(Node q)

 {if (q != null)

       {  

        System.out.println(q.data);

         Displaylist(q.next);

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public void Buildlist()

  {Node q = new Node(0,null);

       head = q;

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                       System.out.println("Please enter a command or a line of text: ");  

          oneLine = indata.readLine();   // always need the following two lines to read data

         head.data = Integer.parseInt(oneLine);

         for (int i=1; i<=head.data; i++)

         {System.out.println("Please enter another command or a new line of text:");

               oneLine = indata.readLine();

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7 0
3 years ago
If the bending moment (M) is 4,176 ft-lb and the beam is an 1 beam, calculate the bending stress (psi) developed at a point with
SpyIntel [72]

Answer:

Bending stress at point 3.96 is \sigma_b = 1.37 psi

Explanation:

Given data:

Bending Moment M is 4.176 ft-lb = 50.12 in- lb

moment of inertia I = 144 inc^4

y = 3.96 in

\sigma_b = \frac{M}{I} \times y

putting all value to get bending stress

\sigma_b = \frac{50.112}{144} \times 3.96  

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3 0
4 years ago
The exhaust steam from a power station turbine is condensed in a condenser operating at 0.0738 bar(abs). The surface of the heat
lozanna [386]

Answer:

Percentage change 5.75 %.

Explanation:Given ;

Given

 Pressure of condenser =0.0738 bar

Surface temperature=20°C

Now from steam table

Properties of steam at 0.0738 bar  

Saturation temperature corresponding to saturation pressure =40°C      

 h_f= 167.5\frac{KJ}{Kg},h_g= 2573.5\frac{KJ}{Kg}

So Δh=2573.5-167.5=2406 KJ/kg

Enthalpy of condensation=2406 KJ/kg

So total heat=Sensible heat of liquid+Enthalpy of condensation

Total\ heat\ =C_p\Delta T+\Delta h

Total heat =4.2(40-20)+2406

Total heat=2,544 KJ/kg

Now film coefficient before inclusion of sensible heat

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h_1=120.3\frac{KJ}{kg-m^2K}

Now film coefficient after inclusion of sensible heat

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h_2=127.2\frac{KJ}{kg-m^2K}

So\ Percentage\ change=\dfrac{h_2-h_1}{h_1}\times 100

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So Percentage change 5.75 %.

3 0
3 years ago
Need help I’m tryna get a good grade
hoa [83]

Answer:

Explanation:

Flowchart

Hope it helped you.

8 0
3 years ago
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