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mote1985 [20]
3 years ago
15

Air at T1 = 32°C, p1 = 1 bar, 50% relative humidity enters an insulated chamber operating at steady state with a mass flow rate

of 3 kg/min and mixes with a saturated moist air stream entering at T2 = 7°C, p2 = 1 bar. A single mixed stream exits at T3 = 17°C, p3 = 1 bar. Neglect kinetic and potential energy effects
Determine

(a) the relative humidity and temperature, in °C, of the exiting stream.
Engineering
1 answer:
RUDIKE [14]3 years ago
7 0

Answer:

4.5kg/min

Explanation:

Given parameters

T_1 = 32^0 C,  m_1 = 3 kg/min, T_2 = 7^0 C ,T_3 = 17^0

if we take  

The mass flow rate of the second stream = m_2(kg/min)

The mass flow rate of mixed exit stream = m_3 (kg/min)

Now from mass conservation

m_3 = m_2 + m_1

m_3 = m_2 + 3 (kg/min)

The temperature of the mixed exit stream given as

T_3m_3 = T_2m_2 +T_1m_1\\\\17 ( 3 + m_2) = 7 \times m_2 + 32 \times 3\\\\51 + 17 m_2 = 7 m_2 + 96\\\\10 m_2 = 96 - 51\\\\m_2 = 4.5 kg/min\\\\\\\\

Therefore the mass flow rate of second stream will be 4.5 kg/min.

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Answer:

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3 years ago
Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)
vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

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Read 2 more answers
A rich industrialist was found murdered in his house. The police arrived at the scene at 11:00 PM. The temperature of the corpse
d1i1m1o1n [39]

Answer:

The dude was killed around 6:30PM

Explanation:

Newton's law of cooling states:

    T = T_m + (T_0-T_m)e^{kt}

where,

T_0 = initial temp

T_m = temp of room

T = temp after t hours

k = how fast the temp is changing

t = time (hours)

T_0 = 31     because the body was initlally 31ºC when the police found it

T_m = 22   because that was the room temp

T = 30  because the body temp drop to 30ºC after 1 hour

t = 1 because that's the time it took for the body temp to drop to 30ºC

k=???   we don't know k so we must solve for this

rearrange the equation to solve for k

T = T_m + (T_0-T_m)e^{kt}

T - T_m= (T_0-T_m)e^{kt}

\frac{T - T_m}{(T_0-T_m)}= e^{kt}

ln(\frac{T - T_m}{T_0-T_m})=kt

\frac{ln(\frac{T - T_m}{T_0-T_m})}{t}=k

plug in the numbers to solve for k

k = \frac{ln(\frac{T - T_m}{T_0-T_m})}{t}

k = \frac{ln(\frac{30 - 22}{31-22})}{1}

k=ln(\frac{8}{9})

Now that we know the value for k, we can find the moment the murder occur. A crucial information that the question left out is the temperature of a human body when they're still alive. A living human body is about 37ºC. We can use that as out initial temperature to solve this problem because we can assume that the freshly killed body will be around 37ºC.

T_0 = 37     because the body was 37ºC right after being killed

T_m = 22   because that was the room temp

T = 31  because the body temp when the police found it

k=ln(\frac{8}{9})   we solved this earlier

t = ???   we don't know how long it took from the time of the murder to when the police found the body

Rearrange the equation to solve for t

T = T_m + (T_0-T_m)e^{kt}

T - T_m= (T_0-T_m)e^{kt}

\frac{T - T_m}{(T_0-T_m)}= e^{kt}

ln(\frac{T - T_m}{T_0-T_m})=kt

\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}=t

plug in the values

t=\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}

t=\frac{ln(\frac{31 - 22}{37-22})}{ln(8/9)}

t=\frac{ln(3/5)}{ln(8/9)}

t=\frac{ln(3/5)}{ln(8/9)}

t ≈ 4.337 hours from the time the body was killed to when the police found it.

The police found the body at 11:00PM so subtract 4.337 from that.

11 - 4.33 = 6.66 ≈ 6:30PM

7 0
3 years ago
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