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2 years ago

Hey friends.... ajao bat Kare ✌️✌️❤️

Engineering

2 answers:

Dvinal [7]2 years ago

4 0

dude thx for the points lol

valina [46]2 years ago

4 0

Hiii the answers b I think

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In a website browser address bar, what does “www” stand for?

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www stands for world wide web

Explanation:

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2 years ago

For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about ____

7nadin3 [17]

Answer:

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For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about _one___

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1 year ago

A reversible power cycle R and an irreversible power cycle I operate between the same hot and cold thermal reservoirs. Cycle I h

anygoal [31]

Answer: Attached below is the missing diagram

answer :

A) 1) Wr > WI, 2) Qc' > Qc

B) 1) QH' > QH, 2) Qc' > Qc

Explanation:

л = w / QH = 1 - Qc / QH and QH = w + Qc

A) each cycle receives same amount of energy by heat transfer

( Given that ; Л1 = 1/3 ЛR )

1) develops greater bet work 

WR develops greater work ( i.e. Wr > WI )

2) discharges greater energy by heat transfer

Qc' > Qc

solution attached below

B) If Each cycle develops the same net work

1) Receives greater net energy by heat transfer from hot reservoir

QH' > QH ( solution is attached below )

2) discharges greater energy by heat transfer to the cold reservoir

Qc' > Qc

solution attached below

4 0

2 years ago

Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a ve

Dmitry [639]

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature $T_1 = -16^0\ C$

Quality $x_1 = 0.2$

Outlet:

Temperature $T_2 = -16^0 C$

Quality $x_2 = 1$

The following data were obtained at saturation properties of R134a at the temperature of -16° C

$v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\ v_g = 0.1247 \ m^3 /kg$

$v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\ v_1 = 0.0255 \ m^3/kg \\ \\ \\ v_2 = v_g = 0.1247 \ m^3/kg$

$m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6 \\ \\ \mathbf{m = 0.0534 \ kg/sec}$

$\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2 \\ \\ \rho_1v_1 = \rho_2v_2 \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}$

3 0

2 years ago

Consider the following ways of handling deadlock: (1) banker’s algorithm, (2) detect

Andrew [12]

Answer:

Explanation:

7 0

2 years ago

Hey friends.... ajao bat Kare ✌️✌️❤️​

Hey friends.... ajao bat Kare ✌️✌️❤️