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EleoNora [17]
3 years ago
8

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

as a thermal efficiency equal to one third the thermal efficiency of R.
a. If each cycle receives the same amout of energy by heat transfer from the hot reservor, detrmine which cycle:
1. develops greater net work.
2. discharges greater energy by heat transfer to the cold reservoir

b. If each cycle develops the same net work, determine which cycle:
1. receives greater net energy by heat transfer from the hot reservoir
2. discharges greater energy by heat transfer to the cold reservoir.
Engineering
1 answer:
anygoal [31]3 years ago
4 0

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

<u>A) each cycle receives same amount of energy by heat transfer</u>

<u>(</u> Given that ; Л1 = 1/3 ЛR )

<em>1) develops greater bet work </em>

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

<em>2) discharges greater energy by heat transfer</em>

 Qc' > Qc

solution attached below

<u>B) If Each cycle develops the same net work </u>

<em>1) Receives greater net energy by heat transfer from hot reservoir</em>

QH' > QH   ( solution is attached below )

<em>2) discharges greater energy  by heat transfer to the cold reservoir</em>

Qc' > Qc

solution attached below

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Water is being added to a storage tank at the rate of 500 gal/min. Water also flows out of the bottom through a 2.0-in-inside di
melomori [17]

Answer:

From the answer, the water level is falling (since rate of outflow is more than that of inflow), and the rate at which the water level in the storage tank is falling is

(dh/dt) = - 0.000753

Units of m/s

Explanation:

Let the volume of the system at any time be V.

V = Ah

where A = Cross sectional Area of the storage tank, h = height of water level in the tank

Let the rate of flow of water into the tank be Fᵢ.

Take note that Fᵢ is given in the question as 500 gal/min = 0.0315 m³/s

Let the rate of flow of water out of the storage tank be simply F.

F is given in the form of (cross sectional area of outflow × velocity)

Cross sectional Area of outflow = πr²

r = 2 inches/2 = 1 inch = 0.0254 m

Cross sectional Area of outflow = πr² = π(0.0254)² = 0.00203 m²

velocity of outflow = 60 ft/s = 18.288 m/s

Rate of flow of water from the storage tank = 0.0203 × 18.288 = 0.0371 m³/s

We take an overall volumetric balance for the system

The rate of change of the system's volume = (Rate of flow of water into the storage tank) - (Rate of flow of water out of the storage tank)

(dV/dt) = Fᵢ - F

V = Ah (since A is constant)

dV/dt = (d/dt) (Ah) = A (dh/dt)

dV/dt = A (dh/dt) = Fᵢ - F

Divide through by A

dh/dt = (Fᵢ - F)/A

Fi = 0.0315 m³/s

F = 0.0371 m³/s

A = Cross sectional Area of the storage tank = πD²/4

D = 10 ft = 3.048 m

A = π(3.048)²/4 = 7.30 m²

(dh/dt) = (0.0315 - 0.0370)/7.3 = - 0.000753

(dh/dt) = - 0.000753

4 0
3 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
3 years ago
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