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maw [93]
3 years ago
11

A vertical piston-cylinder assembly contains water. The piston has a mass of 2 kg and a diameter of 7 cm . Assume atmospheric pr

essure to be 101 kPa.
a. Determine the vertical force necessary on the piston to ensure that water inside the cylinder boils at 120 C

b. Determine the vertical force necessary on the piston to ensure that water inside the cylinder boils at 80 C

c. What would the answer in Part A be if the piston mass were neglected?
Engineering
1 answer:
Step2247 [10]3 years ago
4 0

Answer:

for a) F= 744.97 N

for a) F= 167.85 N

for a) F= 764.57 N

Explanation:

the pressure developed by the piston should be higher than the saturated vapor pressure of water for boiling point at T=120 to ensure boiling.

Then from steam tables

T= 120°C → P required=Pr= 198.67 kPa

then the pressure developed by the piston is

P = (m*g + F)/A

where m= mass of the piston ,g= gravity F= force required and A= area of the piston

then

Pr = P = (m*g + F)/A

F = Pr*A-m*g

since A= π/4*D²

F =π/4* Pr*D²-m*g

replacing values

F =π/4* Pr*D²-m*g = π/4*198.67  *10³Pa*(0.07m)² -2kg* 9.8m/s²

F= 744.97 N

b) for T₂=80°C → Pr₂=47.41 kPa

F₂ =π/4* Pr₂*D²-m*g = π/4*47.41*10³Pa*(0.07m)² -2kg* 9.8m/s²

F₂= 167.85 N

c) for m=0 (mass of the piston neglected) ,the force required is

F₃ =π/4*Pr*D² =  π/4*198.67 *10³Pa*(0.07m)²= 764.57 N

F₃ =764.57 N

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The specific volume of a system consisting of refrigerant-134a at 1.0 Mpa is 0.01 m /kg. The quality of the R-134a is: (a) 12.6%
Flura [38]

Answer:

option c is correct

47.2%

Explanation:

given data

consisting of refrigerant = 134 a

volume V = 0.01 m³/kg

pressure P = 1MPa = 1000 kPa

to find out

quality of the R 134a

solution

we will get here value of volume Vf and Vv from pressure table 60 kpa to 3 Mpa for 1 Mpa of R134 a

that is

Vf = 0.0008701 m³/kg

Vv = 0.0203 m³/kg

so we will apply here formula that is

quality = (V - Vf) / (Vv - Vf)    ............1

put here value

quality = (0.01 - 0.0008701 ) / ( 0.0203 - 0.0008701 )

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4 0
3 years ago
(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5
bezimeni [28]

Answer:

a.) 1.453MW/m2,  b.)  2,477,933.33 BTU/hr  c.) 22,733.33 BTU/hr  d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

                        50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, Q = \frac{109*0.5*100}{0.0075} = 726,666.667W

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, Q=\frac{1*0.5*100}{0.0075} = 6,666.67W

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, Q=\frac{109*0.5*100}{0.015} =363,333.33W

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0
3 years ago
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