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pav-90 [236]
3 years ago
14

A simple ideal Rankine cycle with water as the working fluid operates between the pressure limits of 4 MPa in the boiler and 25

kPa in the condenser and a turbine inlet temperature of 700°C. The boiler is sized to provide a steam flow of 50 kg/s. Determine the power produced by the turbine and consumed by the pump. Use steam tables.

Engineering
2 answers:
Maru [420]3 years ago
6 0

Answer:

power produced by turbine = 79656.15 KW

pump work =227.57 KW

Explanation:

here At T1 = 700 C and P1 = 4 Mpa, from steam tabel : h1 =3906.41 KJ/Kg and s1 = 7.62 KJ/Kg.K

now s1 = s2 = 7.62 KJ/Kg.K as 1-2 is isentropic

so at P2 = 20 Kpa and s2 = 7.62 KJ/Kg.K , fom steam table : h2 = 2513.33 KJ/Kg

power produced by turbine = m.( h1 - h2) = 57.18 x ( 3906.41 - 2513.33) = 79656.15 KW

at P3 = 20 Kpa, 3 = vf = 0.001 m^3/kg

so pump work = v3.dP = 57.18 x 0.001 x ( 4000 - 20) =227.57 KW

chubhunter [2.5K]3 years ago
3 0

Answer:

PT = 70 000kw

Pp = 202.4kw

Explanation:

The full details of the explanation is attached.

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A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m ? K), and the wire/sheath interface
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Question

A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m.K), and the wire/sheath interface is characterized by a thermal contact resistance of Rtc = 3E-4m².K/W. The convection heat transfer coefficient at the outer surface of the sheath is 10 W/m²K, and the temperature of the ambient air is 20°C.

If the temperature of the insulation may not exceed 50°C, what is the maximum allowable electrical power that may be dissipated per unit length of the conductor? What is the critical radius of the insulation?

Answer:

a. 4.52W/m

b. 13mm

Explanation:

Given

Diameter of electrical wire = 2mm

Wire Thickness = 2-mm

Thermal Conductivity of Rubberized sheath (k = 0.13 W/m.K)

Thermal contact resistance = 3E-4m².K/W

Convection heat transfer coefficient at the outer surface of the sheath = 10 W/m²K,

Temperature of the ambient air = 20°C.

Maximum Allowable Sheet Temperature = 50°C.

From the thermal circuit (See attachment), we my write

E'q = q' = (Tin,i - T∞)/(R'cond + R'conv)

= (Tin,i - T∞)/(Ln (r in,o / r in,i)/2πk + (1/(2πr in,o h)))

Where r in,i = D/2

= 2mm/2

= 1 mm

= 0.001m

r in,o = r in,i + t = 0.003m

T in, i = Tmax = 50°C

Hence

q' = (50 - 20)/[(Ln (0.003/0.001)/(2π * 0.13) + 1/(2π * 0.003 * 10)]

= 30/[(Ln3/0.26π) + 1/0.06π)]

= 30/[(1.34) + 5.30)]

= 30/6.64

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The critical radius is unaffected by the constant resistance.

Hence

Critical Radius = k/h

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= 0.013m

= 13mm

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