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11 days ago

Consider a two-stage cascade refrigeration system operating between the pressure limits of 1.2 MPa and 200 kPa with refrigerant-

134a as the working fluid. Heat rejection from the lower cycle to the upper cycle takes place in an adiabatic counterflow heat exchanger where the pressure in the upper and lower cycles are 0.4 and 0.5 MPa, respectively. In both cycles, the refrigerant is a saturated liquid at the condenser exit and a saturated vapor at the compressor inlet, and the isentropic efficiency of the compressor is 80 percent. If the mass flow rate of the refrigerant through the lower cycle is 0.15 kg/s, determine (a) the mass flow rate of the refrigerant through the upper cycle, (b) the rate of heat removal from the refrigerated space, and (c) the COP of this refrigerator.

Engineering

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The molecular weight of a 10g rubber band

Viktor [391]

The response to this query is 1 * 10 g/mole = 10.

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2 months ago

The Machine Shop has received an order to turn three alloy steel cylinders. Starting diameter = 250 mm and length = 625 mm. Feed

alex41 [359]

Response:

The cutting speed is calculated at 365.71 m/min

Clarification:

Given parameters include

diameter D = 250 mm

length L = 625 mm

Feed f = 0.30 mm/rev

cut depth = 2.5 mm

n = 0.25

C = 700

To find

the cutting speed that ensures the tool life coincides with the cutting time for the three parts

The formula for cutting time is given as

Tc =

....................1

where D refers to diameter, L refers to length and f refers to feed while V represents speed $\frac{\pi DL}{1000*f*V}$

Thus, we derive

Tc = $\frac{\pi (250)*625}{1000*0.3*V}$

Tc = $\frac{1636.25}{V}$

Given the tool life is expressed as

T = 3 × Tc............................2

where T denotes tool life and Tc is the cutting duration

Calculating tool life by substituting values into equation 2 yields

T = 3 × $\frac{1636.25}{V}$

According to the Taylor tool formula, cutting speed is expressed as

$VT^{0.25} = 700$

× V × 8.37 = 700

This yields V = 365.71

Thus, the cutting speed calculates to 365.71 m/min

5 0

2 months ago

A spherical hot-air balloon is initially filled with air at 120 kPa and 20°C with an initial diameter of 5 m. Air enters this ba

mote1985 [299]

Answer:

The duration is 17.43 minutes.

Explanation:

Based on the provided information, the initial diameter is 5 m

the velocity is 3 m/s

and the final diameter is 17 m.

To find the solution, we will use the volume change equation expressed as

ΔV = $\frac{4}{3} \pi * (rf)^3 - \frac{4}{3} \pi * (ri)^3$ .............1

where ΔV represents the change in volume, rf is the final radius, and ri is the initial radius.

Calculating ΔV yields

ΔV = $\frac{4}{3} \pi * (8.5)^3 - \frac{4}{3} \pi * (2.5)^3$

ΔV = 2507 m³.

Thus,

Q = velocity × Area

Q = 3 × π ×(0.5)² = 2.356 m³/s.

Next, the change in time can be expressed as

Δt = $\frac{\Delta V}{Q}$

Δt = $\frac{2507}{2.356}$

Δt = 1046 seconds.

Therefore, the total change in time amounts to 17.43 minutes.

8 0

3 months ago

The temperature distribution across a wall 0.3 m thick at a certain instant of time is T(x) a bx cx2 , where T is in degrees Cel

iogann1982 [368]

Answer:

The heat transfer rate into the wall is $\mathbf{q__{in}} \mathbf{ = 200 W/m^2}$

The heat output rate is $\mathbf{q_{out} =182 \ W/m^2}$

The change in stored energy in the wall is $\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }$

The convection coefficient is h = 4.26 W/m².K

Explanation:

Considering the problem:

The temperature profile across the wall is expressed as:

$T(x) = ax+bx+cx^2$

where:

T = temperature in °C

and a, b, & c are constants.

Substituting a = 200° C, b = -200° C/m, and c = 30° C/m² results in:

$T(x) = 200x-200x+30x^2$

This follows the application of Fourier's Law of heat conduction.

$q_x = -k \dfrac{dT}{dx}$

where the heat input rate $q_{in} = q_k$ ; Then x= 0

<pThus:

$q_{in}= -k (\dfrac{d( 200x-200x+30x^2)}{dx})_{x=0}$

$q_{in}= -1 (-200+60x)_{x=0}$

$\mathbf{q__{in}} \mathbf{ = 200 W/m^2}$

Consequently, the heat transfer rate into the wall measures $\mathbf{q__{in}} \mathbf{ = 200 W/m^2}$

The heat output rate is:

$q_{out} = q_{x=L}$ ; where x = 0.3

$q_{out} = -k (\dfrac{dT}{dx})_{x=0.3}$

Replacing T with $200x-200x+30x^2$ and k with 1 W/m.K

$q_{out} = -1 (\dfrac{d(200x-200x+30x^2)}{dx})_{x=0.3}$

$q_{out} = -1 (-200+60x)_{x=0.3}$

$q_{out} = 200-60*0.3$

$\mathbf{q_{out} =182 \ W/m^2}$

Thus, the heat output rate is $\mathbf{q_{out} =182 \ W/m^2}$

Applying the energy balance to find the change in energy (internal energy) stored in the wall.

$\Delta E_{stored} = E_{in}-E_{out} \\ \\ \Delta E_{stored} = q_{in}- q_{out} \\ \\ \Delta E_{stored} = (200 - 182 ) W/m^2 \\ \\$

$\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }$

Thus, the energetic change rate stored in the wall is $\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }$

We know that in a steady state, the heat reaching the end of the plate must be convected to the surrounding fluid.

Thus:

$q_{x=L} = q_{convected}$

$q_{x=L} = h(T(L) - T _ \infty)$

where;

h represents the convective heat transfer coefficient.

Therefore:

$Replacing \ 182 W/m^2 \ for \ q_{x=L}, (200-200x +30x \ for \ T(x) \, 0.3 m \ for \ x \ and \ 100^0 C for \ T$ We find:

182 = h(200-200×0.3 + 30 ×0.3² - 100 )

182 = h (42.7)

h = 4.26 W/m².K

Thus, the convection coefficient equals h = 4.26 W/m².K

6 0

3 months ago