One of the most important parts about creating great photographs starts with choosing what to photograph

Refrigerant-134a enters the expansion valve of a refrigeration system at 120 psia as a saturated liquid and leaves at 20 psia. D

Shkiper50 [21]

Solution :

$P_1 = 120 \ psia$

$P_2 = 20 \ psia$

Using the data table for refrigerant-134a at P = 120 psia

$h_1=h_f=40.8365 \ Btu/lbm$

$u_1=u_f=40.5485 \ Btu/lbm$

$T_{sat}=87.745^\circ F$

∴ $h_2=h_1=40.8365 \ Btu/lbm$

For pressure, P = 20 psia

$h_{2f} = 11.445 \ Btu/lbm$

$h_{2g} = 102.73 \ Btu/lbm$

$u_{2f} = 11.401 \ Btu/lbm$

$u_{2g} = 94.3 \ Btu/lbm$

$T_2=T_{sat}=-2.43^\circ F$

Change in temperature, $\Delta T = T_2-T_1$

$\Delta T = -2.43-87.745$

$\Delta T=-90.175^\circ F$

Now we find the quality,

$h_2=h_f+x_2(h_g-h_f)$

$40.8365=11.445+x_2(91.282)$

$x_2=0.32198$

The final energy,

$u_2=u_f+x_2.u_{fg}$

$=11.401+0.32198(82.898)$

$=38.09297 \ Btu/lbm$

Change in internal energy

$\Delta u= u_2-u_1$

= 38.09297-40.5485

= -2.4556

5 0

3 years ago

If you are involved in a collision where there is injury, you must report the incident within .......

Misha Larkins [42]

Answer:

24 hours

Explanation:

you must exchange insurance details after a collision if someone is injured. Otherwise you must report the collision to us as soon as possible (and no later than 24 hours). Although you must report such a collision straight away you should always seek medical help in the first instance.

4 0

3 years ago

Determine the mass of a car that weight 3,500 lbs both in slugs and kilograms. The answer should be in kilograms.

Andreas93 [3]

Answer:

The mass of car in slug =108.5 slug.

The mass of car in kilogram =1575 Kg.

Explanation:

Given that

Mass of car = 3500 lbs

We know that

1 lbs =0.031 slug

So

3500 lbs = 0.031 x 3500 slug

So the mass of car in slug =108.5 slug.

We know that

1 lbs =0.45 kilograms

so

3500 lbs = 3500 x 0.45 Kg

So the mass of car in kilogram =1575 Kg.

4 0

3 years ago

Consider a composite wall that includes an 8-mm-thick hardwood siding, 40-mm by 100-mm hardwood studs on 0.65-m centers with gla

mel-nik [20]

Answer:

total resistance = 0.18414 K/W

Explanation:

given data

length L = 8 mm

siding = 40 mm

siding = 100 mm

studs = 0.65-m

paper faced, 28 kg/m³

gypsum layer = 12-mm

to find out

thermal resistance associated with a wall that is 2.5 m high by 6.5 m wide (having 10 studs, each 2.5 m high)

solution

we will apply here resistance formula that is

resistance = $\frac{L}{Ka*A}$ ...................1

here L is length and Ka is thermal conductivity and A id area

thermal conductivity of hard wood siding = 0.94 W/m-K

and thermal conductivity of hard wood stud = 0.16 W/m-K

and thermal conductivity of glass fiber insulation = 0.038 W/m-K

and thermal conductivity of gypsum wall board = 0.17 W/m-K

so resistance for wood is

resistance Rw = $\frac{0.008}{0.094*0.65*2.5}$ = 0.0549 K/W ................2

and resistance for stud is

resistance Rs = $\frac{0.100}{0.16*0.04*2.5}$ = 6.25 K/W ....................3

and resistance for insulation

resistance Ri = $\frac{0.100}{0.038*(0.65 - 0.04)*2.5}$ = 1.7256 K/W .................4

and resistance for wall board

resistance Rg = $\frac{0.012}{0.17*0.65*2.5}$ = 0.4343 K/W .................5

so here stud and insulated are parallel

so resistance = $( Rs^{-1} + Ri^{-1} )^{-1}$

we get resistance = $( 6.25^{-1} + 1.7256^{-1} )^{-1}$ = 1.3522 K/W ..........................6

so total resistance is

total resistance add equation 2 and equation 5 and 6

total resistance = 0.0549 + 0.4343 + 1.3522

total resistance = 1.8414 K/W

and

studs are 10

so total resistance will be

total resistance = $\frac{1.8414}{10}$

total resistance = 0.18414 K/W

7 0

4 years ago

Lab scale tests performed on a cell broth with a viscosity of 5cP gave a specific cake resistance of 1 x1011 cm/g and a negligib

insens350 [35]

Answer:

5.118 m^3/hr

Explanation:

Given data:

viscosity of cell broth = 5cP

cake resistance = 1*1011 cm/g

dry basis per volume of filtrate = 20 g/liter

Diameter = 8m , Length = 12m

vacuum pressure = 80 kpa

cake formation time = 20 s

cycle time = 60 s

<u>Determine the filtration rate in volumes/hr expected fir the rotary vacuum filter</u>

attached below is a detailed solution of the question

Hence The filtration rate in volumes/hr expected for the rotary vacuum filter

V' = ( $\frac{60}{20}$ ) * 1706.0670

= 5118.201 liters ≈ 5.118 m^3/hr

4 0

3 years ago