Ask question

Ask question

All categories

2 years ago

8

A mass of 12 kg saturated refrigerant-134a vapor is contained in a piston-cylinder device at 240 kPa. Now 300 kJ of heat is tran

sferred to the refrigerant at constant pressure while 110-V source supplies current to a resistor within the cylinder for 6 min. Determine the current supplied if the final temperature is 70 degrees C. Also, show the process on a T-v diagram with respect to the saturation lines.

1 answer:

Ket [755]2 years ago

4 0

Answer:

I = 12.706 Amps

Explanation:

Given:

- The mass of saturated R-134a m = 12 kg

- The initial Conditions

P_1 = 240 KPa

Saturated Vapor

- The final Conditions

P_2 = 240 KPa

T_2 = 70° C

- The amount of Heat transferred Q_in = 300 KJ

- The voltage of current source V = 110 V

- The current supplied by the source = I

- The time duration Δt = 6 min

Find:

Determine the current supplied I.

Solution:

- Look-up enthalpies h_1 and h_2 at both states using Tables A-11 and A-13.

P_1 = 240 KPa

Saturated Vapor ----------> h_1 = h_g = 247.32 KJ/kg

P_2 = 240 KPa

T_2 = 70° C ----------> h_2 = 314.53 KJ/kg

- Using First Thermodynamic Law, set up an energy balance:

E_in - E_out = ΔE_system

Q_in + W_electric,in - W_out = Δ U

Q_in + V*I*Δt - W_out = Δ U

Q_in + V*I*Δt = Δ H

Q_in + V*I*Δt = m*( h_2 - h_1 )

- Make the current I the subject of the expression above:

V*I*Δt = m*( h_2 - h_1 ) - Q_in

I = [ m*( h_2 - h_1 ) - Q_in ] / V*Δt

- Plug in the values in the expression derived above and evaluate current of source I:

I = [ 12*( 314.53 - 247.32 ) - 300 ]*1000 / 110*6*60

I = 12.706 Amps

- The required source of current is I = 12.706 Amps.

You might be interested in

Technician A says that TSBs are typically updates to the owner's manual. Technician B says that TSBs are generally

vichka [17]

Answer:

Technician A says that TSBs are typically updates to the owner's manual. Technician B says that TSBs are generally

updated information on model changes that do not affect the technician. Who is correct? the answer is c

4 0

2 years ago

Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid–vapor mixture at a pressure

dezoksy [38]

Answer:

(a). Entropy change of refrigerant is = 0.7077 $\frac{KJ}{K}$

(b). Entropy change of cooled space $dS_{space} = - 0.6844$ $\frac{KJ}{K}$

(c). Total entropy change is dS = 0.0232 $\frac{KJ}{K}$

Explanation:

Given data

Saturation pressure = 140 K pa

Saturation temperature from property table

$T_{sat}$ = - 18.77 °c = - 18.77 + 273 = 254.23 K

(a). Entropy change of refrigerant is given by

$dS_{ref} = \frac{Q}{T_{sat}}$

Since heat absorbed by refrigerant Q = 180 KJ

$dS = \frac{180}{254.23}$

dS = 0.7077 $\frac{KJ}{K}$

(b). Entropy change of cooled space

$dS_{space} = - \frac{Q}{T_{space}}$

$T_{space}$ = - 10 °c = 263 K

$dS_{space} = - \frac{180}{263}$

$dS_{space} = - 0.6844$ $\frac{KJ}{K}$

(c). Total entropy change is given by

$dS = dS_{ref} + dS_{space}$

dS = 0.7077 - 0.6844

dS = 0.0232 $\frac{KJ}{K}$

This is the value of total entropy change.

4 0

2 years ago

A thin-walled tube with a diameter of 12 mm and length of 25 m is used to carry exhaust gas from a smoke stack to the laboratory

nlexa [21]

Answer:

(a) h₁ = 204.45 W/m²k

(b) h₀ = 46.80 W/m².k

(c) T = T = 15.50°C

Explanation:

Given Data;

Diameter = 12mm

Length = 25 m

Entry temperature = 200°C

Flow rate = 0.006 kg/s

velocity = 2.5 m/s.

Step 1: Calculating the mean temperature;

(200 + 15)/2

Mean temperature = 107.5°C = 380.5 K

The properties of air at mean temperature 380.5 K are given as:

v = 24.2689*10⁻⁶m²/s

a = 35.024*10⁻⁶m²/s

μ = 221.6 *10⁻⁷Ns/m²

k = 0.0323 W/m.k

Cp = 1012 J/kg.k

Step 2: Calculating the prantl number using the formula;

Pr = v/a

= 24.2689*10⁻⁶/ 35.024*10⁻⁶

= 0.693

Step3: Calculating the reynolds number using the formula;

Re = 4m/πDμ

= 4 *0.006/π*12*10⁻³ * 221.6 *10⁻⁷

= 0.024/8.355*10⁻⁷

= 28725

Since Re is greater than 2000, the flow is turbulent. Nu becomes;

Nu = 0.023Re^0.8 *Pr^0.3

Nu = 0.023 * 28725^0.8 * 0.693^0.3

= 75.955

(a) calculating the heat transfer coefficient:

Nu = hD/k

h = Nu *k/D

= (75.955 * 0.0323)/12*10^-3

h = 204.45 W/m²k

(b)

Properties of air at 15°C

v = 14.82 *10⁻⁶m²/s

k = 0.0253 W/m.k

a = 20.873 *10⁻⁶m²/s

Pr(outside) = v/a

= 14.82 *10⁻⁶/20.873 *10⁻⁶

= 0.71

Re(outside) = VD/v

= 2.5 * 12*10⁻³/14.82*10⁻⁶

=2024.29

Using Zakauskus correlation,

Nu = 0.26Re^0.6 * Pr^0.37 * (Pr(outside)/Pr)^1/4

= 0.26 * 2024.29^0.6 * 0.71^0.37 * (0.71/0.693)^1/4

= 22.199

Nu = h₀D/k

h₀ = Nu*k/D

= 22.199* 0.0253/12*10⁻³

h₀ = 46.80 W/m².k

(c)

Calculating the overall heat transfer coefficient using the formula;

1/U =1/h₁ +1/h₀

1/U = 1/204.45 + 1/46.80

1/U = 0.026259

U = 1/0.026259

U = 38.08

Calculating the temperature of the exhaust using the formula;

T -T₀/T₁-T₀ = e^-[uπDL/Cpm]

T - 15/200-15 = e^-[38.08*π*12*10⁻³*25/1012*0.006]

T - 15/185 = e^-5.911

T -15 = 185 * 0.002709

T = 15+0.50

T = 15.50°C

6 0

2 years ago

Theorems of Pappus and Guldinus are used to find: a. The surface area and volume of a body of rotation b. The surface area and v

s2008m [1.1K]

Answer:

option (a). The surface area and volume of a body of rotation

is the correct option

Explanation:

Theorems of Pappus and Guldinus are used to find the surface area and volume of a revolving body. It is neither applicable for surface areas and volumes of a symmetric body nor it helps to find the overall mass of any body. Thus, it can help to calculate the surface area and volume of any body rotated in 2-D frame(or any 2-D curve).

It is given or calculated as the product of area, perpendicular distance from the axis and length of the 2-D curve.

6 0

2 years ago

A steady tensile load of 5.00kN is applied to a square bar, 12mm on a side and having a length of 1.65m. compute the stress in t

Shtirlitz [24]

Answer:

The stress in the bar is 34.72 MPa.

The design factor (DF) for each case is:

A) DF=0.17

B) DF=0.09

C) DF=0.125

D) DF=0.12

E) DF=0.039

F) DF=1.26

G) DF=5.5

Explanation:

The design factor is the relation between design stress and failure stress. In the case of ductile materials like metals, the failure stress considered is the yield stress. In the case of plastics or ceramics, the failure stress considered is the breaking stress (ultimate stress). If the design factor is less than 1, the structure or bar will endure the applied stress. By the opposite side, when the DF is higher than 1, the structure will collapse or the bar will break.

we will calculate the design stress in this case:

$\displaystyle \sigma_{dis}=\frac{T_l}{Sup}=\frac{5.00KN}{(12\cdot10^{-3}m)^2}=34.72MPa$

The design factor for metals is:

$DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{y}}$

The design factor for plastic and ceramics is:

$DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{u}}$

We now need to know the yield stress or the ultimate stress for each material. We use the AISI and ASTM charts for steels, materials charts for non-ferrous materials and plastics safety charts for the plastic materials.

For these cases:

A) The yield stress of AISI 120 hot-rolled steel (actually is AISI 1020) is 205 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{205MPa}=0.17$

B) The yield stress of AISI 8650 OQT 1000 steel is 385 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{385MPa}=0.09$

C) The yield stress of ductile iron A536-84 (60-40-18) is 40Kpsi, this is 275.8 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{275.8MPa}=0.125$

D) The yield stress of aluminum allot 6061-T6 is 290 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{290MPa}=0.12$

E) The yield stress of titanium alloy Ti-6Al-4V annealed (certified by manufacturers) is 880 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{880MPa}=0.039$

F) The ultimate stress of rigid PVC plastic (certified by PVC Pipe Association) is 4Kpsi or 27.58 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{27.58 MPa}=1.26$

In this case, the bar will break.

F) You have to consider that phenolic plastics are used as matrix in composite materials and seldom are used alone with no reinforcement. In this question is not explained if this material is reinforced or not, therefore I will use the ultimate stress of most pure phenolic plastics, in this case, 6.31 MPa:

$DF=\displaystyle\frac{34.72MPa}{6.31 MPa}=5.5$

This material will break.

3 0

2 years ago