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Komok [63]
3 years ago
6

A hot plate with a temperature of 60 C, 50 triangular profile needle wings of length (54 mm), diameter 10 mm (k = 204W / mK) wil

l be added and cooled. Ambient temperature is 20 C and heat transfer coefficient is 20 Since it is W / m2K; a-) Wing efficiency, b-) Total heat transfer rate (W) from the wings, c-) Calculate the effectiveness of a wing.

Engineering
1 answer:
frez [133]3 years ago
3 0

Complete question is;

A hot plate with a temperature of 60 °C will be cooled by adding 50 triangular profile needle blades (k = 204 W/m.K) with a length of 54 mm and diameter 10 mm. According to the ambient temperature 20 °C and the heat transfer coefficient on the surface 20 W/m².K. Calculate,

a-) Wing efficiency

b-) Total heat transfer rate (W) from the wings,

c-) Calculate the effectiveness of a wing.

Answer:

A) Efficiency = 96.05 %

B) Total heat transfer rate = 166.68 W/m

C) Wing Effectiveness = 10.42

Explanation:

Please find attached explanation for all the answers given.

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For a steel alloy it has been determined that a carburizing heat treatment of 11-h duration will raise the carbon concentration
Jobisdone [24]

Answer:

Time =t2=58.4 h

Explanation:

Since temperature is the same hence using condition

x^2/Dt=constant

where t is the time as temperature so D also remains constant

hence

x^2/t=constant

2.3^2/11=5.3^2/t2

time=t^2=58.4 h

4 0
3 years ago
10–25. The 45° strain rosette is mounted on the surface of a shell. The following readings are obtained for each gage: ε a = −20
vazorg [7]

Answer:

The answer is 380.32×10^-6

Refer below for the explanation.

Explanation:

Refer to the picture for brief explanation.

7 0
3 years ago
Water vapor at 5 bar, 320°C enters a turbine operating at steady state with a volumetric flow rate of 0.65 m3/s and expands adia
harina [27]

Answer:

Power = 371.28 kW

Explanation:

Initial pressure, P1 = 5 bar

Final pressure, P2 = 1 bar

Initial temperature, T1 = 320°C

Final temperature, T2 = 160°C

Volume flow rate, V = 0.65m³/s

From steam tables at state 1,

h1 = 3105.6 kJ/kg, s1 = 7.5308 kJ/kgK

v1 = 0.5416 m³/kg

Mass flow rate, m = V/v1

m = 1.2 kg/s

From steam tables, at state 2

h2 = 2796.2 kJ/kg, s2 = 7.6597 kJ/kgK

Power developed, P = m(h1 - h2)

P = 1.2(3105.6-2796.2)

P = 371.28 kW

8 0
3 years ago
A heat pump with an ideal compressor operates between 0.2 MPa and 1 MPa. Refrigerant R134a flows through the system at a rate of
solmaris [256]

Answer:

The mass flow rate of refrigerant is 0.352 kg/s

Explanation:

Considering the cycle of an ideal heat pump, provided in the attachment, we first find enthalpy at state B and D. For that purpose, we use property tables of refrigerant R134a:

<u>At State A</u>:

From table, we see the enthalpy and entropy value of saturated vapor at 0.2 MPa. Therefore:

ha = 244.5 KJ/kg

Sa = 0.93788 KJ/kg.k

<u>At State B</u>:

Since, the process from state A to B is isentropic. Therefore,

Sb = Sa = 0.93788 KJ/Kg

From table, we see the enthalpy value of super heated vapor at 1 MPa and Sb. Therefore:

hb = 256.85 KJ/kg                          (By interpolation)

<u>At State C</u>:

From table, we see the enthalpy and entropy value of saturated liquid at 1 MPa. Therefore:

hc = 107.34 KJ/kg

Now, from the diagram it is very clear that:

Heat Loss = m(hb = hc)

m = (Heat Loss)/(hb - hc)

where,

m = mass flow rate = ?

Heat Loss = (180,000 Btu/hr)(1.05506 KJ/1 Btu)(1 hr/3600 sec)

Heat Loss = 52.753 KW

Therefore,

m = (52.753 KJ/s)/(256.85 KJ/kg - 107.34 KJ/kg)

<u>m = 0.352 kg/s</u>

5 0
3 years ago
A container filled with a sample of an ideal gas at the pressure of 150 Kpa. The gas is compressed isothermally to one-third of
lyudmila [28]

Answer: c) 450 kPa

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}     (At constant temperature and number of moles)

P_1V_1=P_2V_2  

where,

P_1 = initial pressure of gas  = 150 kPa

P_2 = final pressure of gas  = ?

V_1 = initial volume of gas   = v L

V_2 = final volume of gas  = \frac{v}{3}L

150\times v=P_2\times \frac{v}{3}  

P_2=450kPa

Therefore, the new pressure of the gas will be 450 kPa.

7 0
3 years ago
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