Ask question

Ask question

All categories

4 years ago

5

Find the total amount of heat in Q lost through a wall 10' by 18' , with R value from q. 1. Inside temperature is 70 degrees F w

hile outside is 45 degrees F

2 answers:

marissa [1.9K]4 years ago

6 0

Answer:

Just think

Explanation:

gogolik [260]4 years ago

5 0

That is really easy just think about it

You might be interested in

A 10-mm steel drill rod was heat-treated and ground. The measured hardness was found to be 290 Brinell. Estimate the endurance s

grandymaker [24]

Answer:

the endurance strength $S_e$ = 421.24 MPa

Explanation:

From the given information; The objective is to estimate the endurance strength, Se, in MPa .

To do that; let's for see the expression that shows the relationship between the ultimate tensile strength and Brinell hardness number .

It is expressed as:

$200 \leq H_B \leq 450$

$S_{ut} = 3.41 H_B$

where;

$H_B$ = Brinell hardness number

$S_{ut}$ = Ultimate tensile strength

From ;

$S_{ut} = 3.41 H_B$ ; replace 290 for $H_B$ ; we have

$S_{ut} = 3.41 (290)$

$S_{ut} =$ 988.9 MPa

We can see that the derived value for the ultimate tensile strength when the Brinell harness number = 290 is less than 1400 MPa ( i.e it is 988.9 MPa)

So; we can say

$S_{ut} < 1400$

The Endurance limit can be represented by the formula:

$S_e ' = 0.5 S_{ut}$

$S_e ' = 0.5 (988.9)$

$S_e '$ = 494.45 MPa

Using Table 6.2 for parameter for Marin Surface modification factor. The value for a and b are derived; which are :

a = 1.58

b = -0.085

The value of the surface factor can be calculate by using the equation

$k_a = aS^b_{ut}$

$K_a = 1.58 (988.9)^{-0.085$

$K_a = 0.8792$

The formula that is used to determine the value of $k_b$ for the rotating shaft of size factor d = 10 mm is as follows:

$k_b = 1.24d^{-0.107}$

$k_b = 1.24(10)^{-0.107}$

$k_b = 0.969$

Finally; the the endurance strength, Se, in MPa if the rod is used in rotating bending is determined by using the expression;

$S_e =k_ak_b S' _e$

$S_e$ = 0.8792×0.969×494.45

$S_e$ = 421.24 MPa

Thus; the endurance strength $S_e$ = 421.24 MPa

8 0

3 years ago

If you know that the change in entropy of a system where heat was added is 12 J/K, and that the temperature of the system is 250

SVEN [57.7K]

Solution:

Given:

Change in entropy of the system, ΔS = 12J/K

Temperature of the system, $T_{o}$ = 250K

Now, we know that the change in entropy of a system is given by the formula:

ΔS = $\frac{\Delta Q}{T_{o}}$

Amount of heat added, ΔQ = $\Delta S\times T_{o}$

ΔQ = 3000J

3 0

3 years ago

What is the dimensionless heat conduction rate for a sphere at surface temperature T1 buried in an infinite medium of temperatur

dexar [7]

Solution :

The dimensionless conduction heat rate, $q_{ss}^*$

$q_{ss}^*=\frac{q\times L_c}{K A_s(T_1-T_2)}$ ...........(1)

where $L_c$ = characteristic length

$=\left(\frac{A}{4\pi} \right)^{1/2}. \sqrt{\frac{D^2}{4}}$

A is surface area

q = heat transfer

$q=Sk(T_1-T_2)$ ..................(2)

where, S = conductor shape factor

Now substituting (2) in (1),

$q_{ss}^* = \frac{Sk(T_1-T_2)L_c}{kA(T_1-T_2)}$

$q_{ss}^* = \frac{S \times L_c}{A}$

$q_{ss}^* = \frac{S \times D/2}{\pi D^2}$

$q_{ss}^* = \frac{S \times D}{2\pi D^2}$ ...................(3)

For a sphere, we know S = 2πD

Putting this in (3),

$q_{ss}^* = \frac{2 \pi D \times D}{2\pi D^2}$

$q_{ss}^* = \frac{2 \pi D^2}{2\pi D^2}$

$q_{ss}^* = 1$

Therefore, the dimensionless heat conduction rate for a sphere is 1.

7 0

3 years ago

A ceramic material can be defined as an organic compound consisting of a metal or semimetal and one or more nonmetals: (a) true

Artyom0805 [142]

Answer:

b)False

Explanation:

A ceramic material is a inorganic compound ,metalloid or non metal which consisting covalent or ionic bond.

Ceramic compounds are hard and brittle.These are also called amorphous material like glass.These compounds have large use in different areas because they can be use as insulator,conductor and also as superconductors.

3 0

3 years ago

Air (ideal gas) is contained in a cylinder/piston assembly at a pressure of 150 kPa and a temperature of 127°C. Assume that the

Stels [109]

Answer:

The process is not possible.

Explanation:

We know for ideal condition, the work done for isothermal process is

$W_{ideal}$ = $P_{1}.V_{1} ln\frac{V_{2}}{V_{1}}$

and for ideal gas, we know PV = mRT

Therefore, $W_{ideal}$ = mRT $ln\frac{V_{2}}{V_{1}}$

= mRT $ln\frac{P_{1}}{P_{2}}$

= 0.287 x 400 $ln\frac{150}{450}$

= -126.12 kJ/kg (negative sign indicates that the process is compressive, so work input to the compressor is 126.12 kJ/kg )

Now we know for adiabatic compression process

P $V^{\gamma }$ = C

We know $\frac{T_{2}}{T_{1}}=(\frac{P_{2}}{P_{1}})^{\frac{\gamma -1}{\gamma }}$

$T_{2}$ = 556 K

For adiabatic work done, $W_{adiabatic}$ = $\frac{P_{1}\times V_{1}-P_{2}\times V_{2}}{\gamma -1}$

= $\frac{mR(T_{1}-T_{2})}{\gamma -1}$

= $\frac{0.287(400-556)}{1.45 -1}$

= -99.49 kJ/kg (negative sign indicates that the process is compressive, so work input to the compressor is 99.49 kJ/kg )

We know that in isothermal process, work input to the compressor is minimum. But in the above adiabatic polytropic process, work input to the compressor is less than the work done in the isothermal process.

Thus the process is not possible.

7 0

4 years ago