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2 years ago

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8.28 Water is the working fluid in an ideal Rankine cycle with superheat and reheat. Steam enters the first-stage turbine at 140

0 lbf/in.2 and 10008F, expands to a pressure of 350 lbf/in.2, and is reheated to 9008F before entering the secondstage turbine. The condenser pressure is 2 lbf/in.2 The net power output of the cycle is 1 3 109 Btu/h. Determine for the cycle (a) the mass flow rate of steam, in lb/h. (b) the rate of heat transfer, in Btu/h, to the working fluid passing through the steam generator. (c) the rate of heat transfer, in Btu/h, to the working fluid passing through the reheater. (d) the thermal efficiency.

1 answer:

zubka84 [21]2 years ago

3 0

Answer:

Betbtybrbytntrnyrnrynunjhjhnthnnhtnnthnhtnnhnhrnntnthhnhnhtnthn

Explanation:

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Consider a plane composite wall that is composed of two materials of thermal conductivities kA = 0.1 W/m*K and kB = 0.04 W/m*K a

nadya68 [22]

Answer:

q=39.15 W/m²

Explanation:

We know that

Thermal resistance due to conductivity given as

R=L/KA

Thermal resistance due to heat transfer coefficient given as

R=1/hA

Total thermal resistance

$R_{th}=\dfrac{L_A}{AK_A}+\dfrac{L_B}{AK_B}+\dfrac{1}{Ah_1}+\dfrac{1}{Ah_2}+\dfrac{1}{Ah_3}$

Now by putting the values

$R_{th}=\dfrac{0.01}{0.1A}+\dfrac{0.02}{0.04A}+\dfrac{1}{10A}+\dfrac{1}{20A}+\dfrac{1}{0.3A}$

$R_{th}=4.083/A\ K/W$

We know that

Q=ΔT/R

$Q=\dfrac{\Delta T}{R_{th}}$

$Q=A\times \dfrac{200-40}{4.086}$

So heat transfer per unit volume is 39.15 W/m²

q=39.15 W/m²

4 0

2 years ago

A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3

labwork [276]

Answer:

Q = -68.859 kJ

Explanation:

given details

mass $co_2 = 1 kg$

initial pressure P_1 = 104 kPa

Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K

final pressure P_2 = 1068 kPa

Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K

we know that

molecular mass of $co_2 = 44$

R = 8.314/44 = 0.189 kJ/kg K

c_v = 0.657 kJ/kgK

from ideal gas equation

PV =mRT

$V_1 = \frac{m RT_1}{P_1}$

$=\frac{1*0.189*298}{104}$

$V_1 = 0.5415 m3$

$V_2 = \frac{m RT_2}{P_2}$

$=\frac{1*0.189*584}{1068}$

$V_1 = 0.1033 m3$

WORK DONE

$W =P_{avg}*{V_2-V_1}$

w = 586*(0.1033 -0.514)

W =256.76 kJ

INTERNAL ENERGY IS

$\Delta U = m *c_v*{V_2-V_1}$

$\Delta U = 1*0.657*(584-298)$

$\Delta U =187.902 kJ$

HEAT TRANSFER

$Q = \Delta U +W$

= 187.902 +(-256.46)

Q = -68.859 kJ

7 0

3 years ago

A 132mm diameter solid circular section

Ganezh [65]

Answer:

not sure if this helps but

5 0

2 years ago

Radioactive wastes are temporarily stored in a spherical container, the center of which is buried a distance of 10 m below the e

a_sh-v [17]

Answer:

Outside temperature =88.03°C

Explanation:

Conductivity of air-soil from standard table

K=0.60 W/m-k

To find temperature we need to balance energy

Heat generation=Heat dissipation

Now find the value

We know that for sphere

$q=\dfrac{2\pi DK}{1-\dfrac{D}{4H}}(T_1-T_2)$

Given that q=500 W

so

$500=\dfrac{2\pi 2\times .6}{1-\dfrac{2}{4\times 10}}(T_1-25)$

By solving that equation we get

$T_2$ =88.03°C

So outside temperature =88.03°C

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KatRina [158]

Remote?? maybe I’m not really sure

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2 years ago