I need help asapppp!!!!!

An 80-L vessel contains 4 kg of refrigerant-134a at a pressure of 160kPa. Determine (a) the temperature, (b) the quality, (c) th

makvit [3.9K]

Answer:

temperature -15.6 C, quality x=0.646, enthalpy h=667.20 KJ, volume of vapor phase Vg= 79.8 L

Explanation:

property table for R-134a

https://www.ohio.edu/mechanical/thermo/property_tables/R134a/R134a_PresSat.html

at 160 KPa , temperature = -15.66 C

quality x=mass of vapour/ total mass of liq-vap mixture

alternaternately: x=(v-vf)/(vg-vf)

v=total volume i.e. volume of container"80L" 80L=0.08 cubic meter

vf=vol of liquid phase vg=vol of vapor phase vf, vg values at 160Kpa

x=(0.08-0.0007437)/(0.1235-0.0007437)=0.646

enthalpy

h=hf+xhfg hf, hfg values at 160Kpa

h=hf+xhfg=31.2+0.646(209.9)=166.80 KJ/Kg

for 4Kg R-134a h=m(166.80 KJ/Kg )=667.20 KJ

volume of vapor phase

vg at 160Kpa=0.1235 cubic meter for quality=1.

in this case quality=0.646 , so it will occupy 64.6% space of the vapor phase at quality=1.

vol. of vapor phase=0.646*0.1235=0.0798 cubic meter=79.8 L

7 0

3 years ago

Block A has a weight of 8 lb. and block B has a weight of 6 lb. They rest on a surface for which the coefficient of kinetic fric

kkurt [141]

Answer:

For block A, a = 9.66 ft/s²

For block B, a = 15 ft/s²

Explanation:

A free body diagram for this force system is attached to this solution

Mass of block A = m₁ = 8 lb

Mass of block B = m₂ = 6 lb

Coefficient of kinetic friction = μ

Normal reaction on the blocks = N

Spring stiffness of the spring btw block A and B = k = 20 lb/ft

Compression of the spring = 0.2 ft

Analysing Block A first

The forces on block A include, the weight, normal reaction, frictional force and the elastic force due to the spring

Sum of forces in the y-direction = 0

So, the weight of the block = Normal reaction of the surface on the block

N = W = 8 lb

Sum of forces in the x-direction = maₓ

(k × x) - (μ × N) = maₓ

m = W/g = 8/32.2 = 0.248 lbm

(20×0.2) - (0.2 × 8) = (0.248) aₓ

aₓ = 9.66 ft/s²

The forces on block B include, the weight, normal reaction, frictional force and the elastic force due to the spring

Sum of forces in the y-direction = 0

So, the weight of the block = Normal reaction of the surface on the block

N = W = 6 lb

Sum of forces in the x-direction = maₓ

(k × x) - (μ × N) = maₓ

m = W/g = 6/32.2 = 0.186 lbm

(20×0.2) - (0.2 × 6) = (0.186) aₓ

aₓ = 15 ft/s²

4 0

3 years ago

3. When performing overhead work on scaffolding, what protective measures must be taken to prevent objects

hjlf

Answer:

Toeboards, debris nets, or canopies

Explanation:

7 0

3 years ago

The rolling process is governed by the frictional force between the rollers and the workpiece. The frictional force at the entra

adell [148]

Answer:

b)false

Explanation:

Rolling is a process in which work piece passes through rolls to produce desired out put of the work piece.Rolling is a metal forming process.

We know that friction force is responsible for motion of work piece between rolls.If friction force is so small at the entrance side then work piece will not enter in the forming zone and forming process will not occurs.So the friction force should be high at the entrance side and low at the exit side.

So given statement is wrong.

3 0

3 years ago

Because assembly language is so close in nature to machine language, it is referred to as a ____________.

My name is Ann [436]

Answer:

symbolic machine code.

Explanation:

The instructions in the language are closely linked to the machine's architecture.

5 0

3 years ago