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aleksley [76]
3 years ago
12

Determine the change in the enthalpy of helium, in kJ/kg, as it undergoes a change of state from 100 kPa and 20 oC to 600 kPa an

d 300 oC
Engineering
1 answer:
MatroZZZ [7]3 years ago
7 0

Answer:

The change in enthalpy of helium is 4073.86kJ/kg

Explanation:

∆H = Cp(T2 - T1)

Cp = 3.5R = 3.5×8.314 = 29.099kJ/kgmolK ÷ 2 (1kgmol of helium = 2kg of helium) = 14.5495kJ/kgK, T2 = 300°C = 300+273K = 573K, T1 = 20°C = 20+273K = 293K

∆H = 14.5495kJ/kgK(573K - 293K) = 14.5495kJ/kgK × 280K = 4073.86kJ/kg

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Paraphin [41]

Answer:

a. 2.08, b. 1110 kJ/min

Explanation:

The power consumption and the cooling rate of an air conditioner are given. The COP or Coefficient of Performance and the rate of heat rejection are to be determined. <u>Assume that the air conditioner operates steadily.</u>

a. The coefficient of performance of the air conditioner (refrigerator) is determined from its definition, which is

COP(r) = Q(L)/W(net in), where Q(L) is the rate of heat removed and W(net in) is the work done to remove said heat

COP(r) = (750 kJ/min/6 kW) x (1 kW/60kJ/min) = 2.08

The COP of this air conditioner is 2.08.

b. The rate of heat discharged to the outside air is determined from the energy balance.

Q(H) = Q(L) + W(net in)

Q(H) = 750 kJ/min + 6 x 60 kJ/min = 1110 kJ/min

The rate of heat transfer to the outside air is 1110 kJ for every minute.

5 0
3 years ago
Select the correct answer.
juin [17]
Orthographic projection, common method of representing three-dimensional objects, usually by three two-dimensional drawings in each of which the object is viewed along parallel lines that are perpendicular to the plane of the drawling.
4 0
2 years ago
Consider air entering a heated duct at P1 = 1 atm and T1 = 288 K. Ignore the effect of friction. Calculate the amount of heat pe
Ne4ueva [31]

Answer:

The solution for the given problem is done below.

Explanation:

M1 = 2.0

\frac{p1}{p*} = 0.3636

\frac{T1}{T*} = 0.5289

\frac{T01}{T0*} = 0.7934

Isentropic Flow Chart:  M1 = 2.0 , \frac{T01}{T1} = 1.8

T1 = \frac{1}{0.7934} (1.8)(288K) = 653.4 K.

In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.

At the inlet,

T02= \frac{T02}{T1}T1 = (1.8)(288K) = 518.4 K.

Q= Cp(T02-T01) = \frac{1.4(287 J / (Kg.K)}{1.4-1}(653.4-518.4)K = 135.7*10^{3} J/Kg.

5 0
3 years ago
Read 2 more answers
How does the clearance volume affect the efficiency of the Otto cycle?
eduard

Answer:

Explanation:

A smaller clearance volume means a higher compression. A higher compression means better thermal efficiency. However a compression ratio too high might be troublesome, as it can cause accidental ignition of the fuel-air mix. This is the reason why Otto cycle engines have lower compressions that Diesel engines. In a Diesel engine the mix ignites by compression instead of a spark.

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W10L1-Show It: Pythagorean Theorem<br> Calculate the total material in the picture.<br> 4<br> 3
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Answer:

35

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