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Sergio039 [100]

3 years ago

12

A reversible refrigerator operates between a low temperature reservoir at TL and a high temperature reservoir at TH . Its coeffi

cient of performance is given by

1 answer:

Anna11 [10]3 years ago

7 0

Answer

TL/TH- TL

Because we know that power coefficient is. = QL/QH-QL

=so using this for performance we have

=>Perf= TL/(TH-TL)

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Which of these are not included as part of a drivetrain

Zielflug [23.3K]

Answer:

transmission, driveshafts, differential and axles

Explanation:

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2 years ago

Refrigerant 134a enters an air conditioner compressor at 4 bar, 208C, and is compressed at steady state to 12 bar, 808C. The vol

SCORPION-xisa [38]

Answer:

heat transfer rate is -15.71 kW

Explanation:

given data

Initial pressure = 4 bar

Final pressure = 12 bar

volumetric flow rate = 4 m³ / min

work input to the compressor = 60 kJ per kg

solution

we use here super hated table for 4 bar and 20 degree temperature and 12 bar and 80 degree is

h1 = 262.96 kJ/kg

v1 = 0.05397 m³/kg

h2 = 310.24 kJ/kg

and here mass balance equation will be

m1 = m2

and mass flow equation is express as

m1 = $\frac{A1\times V1}{v1}$ .......................1

m1 = $\frac{4\times \frac{1}{60}}{0.05397}$

m1 = 1.2353 kg/s

and here energy balance equation is express as

0 = Qcv - Wcv + m × [ ( h1-h2) + $\frac{v1^2-v2^2}{2}$ + g (z1-z2) ] ....................2

so here Qcv will be

Qcv = m × [ $\frac{Wcv}{m} + (h2-h1)$ ] ......................3

put here value and we get

Qcv = 1.2353 × [ {-60}+ (310.24-262.96) ]

Qcv = -15.7130 kW

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6 0

3 years ago

An incompressible fluid flows between two infinite stationary parallel plates. The velocity profile is given by u=umaxðAy2 + By+

nexus9112 [7]

Answer:

the volume flow rate per unit depth is:

$\frac{Q}{b} = \frac{2}{3} u_{max} h$

the ratio is : $\frac{V}{u_{max}}=\frac{2}{3}$

Explanation:

From the question; the equations of the velocities profile in the system are:

$u = u_{max}(Ay^2+By+C)$ ----- equation (1)

The above boundary condition can now be written as :

At y= 0; u =0 ----- (a)

At y = h; u =0 -----(b)

At y = $\frac{h}{2}$ ; u = $u_{max}$ ------(c)

where ;

A,B and C are constant

h = distance between two plates

u = velocity

$u_{max}$ = maximum velocity

y = measured distance upward from the lower plate

Replacing the boundary condition in (a) into equation (1) ; we have:

$u = u_{max}(Ay^2+By+C) \\ \\ 0 = u_{max}(A*0+B*0+C) \\ \\ 0=u_{max}C \\ \\ C= 0$

Replacing the boundary condition (b) in equation (1); we have:

$u = u_{max}(Ay^2+By+C) \\ \\ 0 = u_{max}(A*h^2+B*h+C) \\ \\ 0 = Ah^2 +Bh + C \\ \\ 0 = Ah^2 +Bh + 0 \\ \\ Bh = - Ah^2 \\ \\ B = - Ah \ \ \ \ \ --- (d)$

Replacing the boundary condition (c) in equation (1); we have:

$u = u_{max}(Ay^2+By+C) \\ \\ u_{max}= u_{max}(A*(\frac{h^2}{2})+B*\frac{h}{2}+C) \\ \\ 1 = \frac{Ah^2}{4} +B \frac{h}{2} + 0 \\ \\ 1 = \frac{Ah^2}{4} + \frac{h}{2}(-Ah) \\ \\ 1= \frac{Ah^2}{4} - \frac{Ah^2}{2} \\ \\ 1 = \frac{Ah^2 - Ah^2}{4} \\ \\ A = -\frac{4}{h^2}$

replacing $A = -\frac{4}{h^2}$ for A in (d); we get:

$B = - ( -\frac{4}{h^2})h$ $B = \frac{4}{h}$

replacing the values of A, B and C into the velocity profile expression; we have:

$u = u_{max}(Ay^2+By+C) \\ \\ u = u_{max} (-\frac{4}{h^2}y^2+\frac{4}{h}y)$

To determine the volume flow rate; we have:

$Q = AV \\ \\ Q= \int\limits^h_0 (u.bdy)$

Replacing $u_{max} (-\frac{4}{h^2}y^2+\frac{4}{h}y) \ for \ u$

$\frac{Q}{b} = \int\limits^h_0 u_{max}(-\frac{4}{h^2} y^2+\frac{4}{h}y)dy \\ \\ \frac{Q}{b} = u_{max} \int\limits^h_0 (-\frac{4}{h^2} y^2+\frac{4}{h}y)dy \\ \\ \frac{Q}{b} = u_{max} (-\frac{-4}{h^2}\frac{y^3}{3} +\frac{4}{h}\frac{y^2}{y})^ ^ h}}__0 }} \\ \\ \frac{Q}{b} =u_{max} (-\frac{-4}{h^2}\frac{h^3}{3} +\frac{4}{h}\frac{h^2}{y})^ ^ h}}__0 }} \\ \\ \frac{Q}{b} = u_{max}(\frac{-4h}{3}+\frac{4h}2} ) \\ \\ \frac{Q}{b} = u_{max}(\frac{-8h+12h}{6}) \\ \\ \frac{Q}{b} =u_{max}(\frac{4h}{6})$

$\frac{Q}{b} = u_{max}(\frac{2h}{3}) \\ \\ \frac{Q}{b} = \frac{2}{3} u_{max} h$

Thus; the volume flow rate per unit depth is:

$\frac{Q}{b} = \frac{2}{3} u_{max} h$

Consider the discharge ;

Q = VA

where :

A = bh

Q = Vbh

$\frac{Q}{b}= Vh$

Also; $\frac{Q}{b} = \frac{2}{3} u_{max} h$

Then;

$\frac{2}{3} u_{max} h = Vh \\ \\ \frac{V}{u_{max}}=\frac{2}{3}$

Thus; the ratio is : $\frac{V}{u_{max}}=\frac{2}{3}$

5 0

3 years ago