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arlik [135]
3 years ago
7

A direct contact heat exchanger (where the fluid mixes completely) has three inlets and one outlet. The mass flow rates of the i

nlets are 1kg/s, 1.5kg/s and 2 kg/s. The enthalpy of those inlets are the 100kJ/kg, 120kJ/kg, and 500kJ/kg, respectively. What is the enthalpy at the outlet?
Engineering
1 answer:
lara31 [8.8K]3 years ago
4 0

Answer:

Enthalpy at outlet=284.44 KJ

Explanation:

m_1=1 Kg/s,m_2=1.5 Kg/s,m_3=22 Kg/s

h_1=100 KJ/Kg,h_2=120 KJ/Kg,h_3=500 KJ/Kg

We need to Find enthalpy of outlet.

Lets take the outlet mass m and outlet enthalpy h.

So from mass conservation

m_1+m_2+m_3=m

   m=1+1.5+2 Kg/s

  m=4.5 Kg/s

Now from energy conservation

m_1h_1+m_2h_2+m_3h_3=mh

By putting the values

1\times 100+1.5\times 120+2\times 500=4.5\times h

So h=284.44 KJ

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  1. The conversion of 47,000 Ohms to kilo-ohms is equal to 47 kilo-ohms.
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<h3>What is resistance?</h3>

Resistance can be defined as an opposition to the flow of current in an electric circuit. Also, the standard unit of measurement of the resistance of an electric component is Ohms, which can be converted to kilo-ohms or mega-ohms.

For Ohms to kilo-ohms, we have:

1 Ohms = 0.001 kilo-ohms

47,000 Ohms = X kilo-ohms

Cross-multiplying, we have:

X = 0.001 × 47000

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For Ohms to mega-ohms, we have:

1,000,000 ohms = 1 mega-ohms

47,000 Ohms = X mega-ohms

Cross-multiplying, we have:

X1,000,000 = 47,000

X = 47,000/1,000,000

X = 0.047 kilo-ohms.

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Determine the design moment strength for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compressio
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This question is incomplete, the complete question is;

Determine the design moment strength (ϕMn) for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compression flange is provided at the ends only (i.e., Lb = 18 ft). Report the result in kip-ft.

Use Fy=50 ksi and assume Cb=1.0 (if needed).

Answer: the design moment strength for the W21x73 steel beam is 566.25 f-ft

Explanation:

Given that;

section  W 21 x 73 steel beam;

now from the steel table table for this section;

Zx = Sx = 151 in³

also given that; fy = 50 ksi and Cb = 1.0

QMn = 0.9 × Fy × Zx

so we substitute

QMn = 0.9 × 50 × 151

QMn = 6795 k-inch

we know that;

12inch equals 1 foot

so

QMn = 6795 k-inch / 12

QMn = 566.25 f-ft

Therefore the design moment strength for the W21x73 steel beam is 566.25 f-ft

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