All categories

3 years ago

Thermal conductivity of AISI 316 Stainless Steel at 90ºC is 14.54 W/m K. Convert this value to IP system.

Engineering

1 answer:

abruzzese [7]3 years ago

7 0

Answer:

the value of conductivity in IP is $8.406\dfrac{Btu}{ft.hr.F}$

Explanation:

Given that

Thermal conductivity K=14.54 W/m.K

This above given conductivity is in SI unit.

SI unit IP unit Conversion factor

m ft 0.3048

W Btu/hr 0.293

The unit of conductivity in IP is Btu./ft.hr.F.

Now convert into IP divided by 1.73 factor.

$0.57\dfrac{Btu}{ft.hr.F}=1 \dfrac{W}{m.K}$

$0.57\times 14.54\dfrac{Btu}{ft.hr.F}=14.54 \dfrac{W}{m.K}$

$8.406\dfrac{Btu}{ft.hr.F}=14.54 \dfrac{W}{m.K}$

So the value of conductivity in IP is $8.406\dfrac{Btu}{ft.hr.F}$

You might be interested in

Select a research proposal topic that relates to electrical and electronics engineering and write a proposal report taking into

fiasKO [112]

Expand your technical knowledge, form global networks and balance life & work commitments. Our advanced diplomas remain current with technological and industry developments.

8 0

3 years ago

A reservoir is 1 km wide and 10 km long and has an average depth of 100m. Every hour, 0.1% of the reservoir's volume drops throu

Ksju [112]

Answer:

250.7mw

Explanation:

Volume of the reservoir = lwh

Length of reservoir = 10km

Width of reservoir = 1km

Height = 100m

Volume = 10x10³x10³x100

= 10⁹m³

Next we find the volume flow rate

= 0.1/100x10⁹x1/3600

= 277.78m³/s

To get the electrical power output developed by the turbine with 92 percent efficiency

= 0.92x1000x9.81x277.78x100

= 250.7MW

7 0

3 years ago

A consolidation test was performed on a sample of fine-grained soil sample taken from a depth such that the vertical effective s

Scorpion4ik [409]

Answer:

The settlement that is expected is 1.043 meters.

Explanation:

Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil

The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula

$\Delta H=\frac{H_oC_c}{1+e_o}log(\frac{\bar{\sigma_o}+\Delta \bar{\sigma }}{\bar{\sigma_o}})$

where

'H' is the initial depth of the layer

$C_c$ is the Compression index

$e_o$ is the inital void ratio

$\bar{\sigma_o}$ is the initial effective stress at the depth

$\Delta \bar{\sigma_o}$ is the change in the effective stress at the given depth

Applying the given values we get

$\Delta H=\frac{8\times 0.3}{1+0.87}log(\frac{154+28}{154})=1.04$

3 0

3 years ago

PLEASE HURRY!!!

Naily [24]

Answer:

Explanation:

He should get a job in engineering to see what it's like to work in the field.

3 0

2 years ago

Read 2 more answers

An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no

Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus $s_{1} =s_{2}$

From the refrigerant table A11-A13

$P_{1} =0.8MPa$ $\left \{ {{ {{v_{1}=v_{g} @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g} @0.8MPa =246.82 kJ/kg } - also {{s_{1}=s_{g} @0.8MPa =0.91853 kJ/kgK } } \right.$

sat vapor

m= $\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339} = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) = 38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}$