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user100 [1]
3 years ago
14

A steady‐flow gas furnace supplies hot air at a rate of 850 cfm and conditions of 120F and 1.00 atm. The air splits into two bra

nches: a 10" diameter duct and a 12" diameter duct. The velocity in the 12" diameter duct is 800 ft/min.
Determine:
(a) The volumetric flow in the 12"‐dia. duct.
(b) The velocity in the 10"‐dia. duct.
(c) The volumetric flow rate into the furnace if the air enters at 68F and 1.00 atm.
(d) The mass flow rate of air entering the furnace.
Engineering
1 answer:
V125BC [204]3 years ago
8 0

Answer:

if I am not wrong the volumetric flow rate into the finance if the year inter 868 1.00 pm

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A sample of wastewater is diluted 10 times. The diluted solution has an ultimate biochemical oxygen demand (BOD), Lo, of 30 mg/L
zzz [600]

Answer:

474.59 mg/L

Explanation:

Given that

BOD = 30 mg/L

Original BOD  = 30 mg/L × dilution factor

Original BOD  = 30 mg/L  × 10 = 300 mg/L

L_o = \frac{BOD}{1-e^{-5t}}

here L_o is the ultimate BOD ; BOD is the  biochemical oxygen demand ;  t = 0.20 /day

L_o = \frac{300}{1-e^{-5(0.20)}}

L_o = 474.59 \ mg/L

3 0
4 years ago
Does the location of a millimeter change the voltage or current of the circuit?
Cerrena [4.2K]

Answer:

Yes, it does.

Explanation:

8 0
3 years ago
Implement the following Matlab code:
vagabundo [1.1K]
28384 *x soít cos estematema
3 0
3 years ago
There are three options for heating a particular house: a. Gas: $1.33/therm where 1 therm=105,500 kJ b. Electric Resistance: $0.
sergejj [24]

Answer:

Option ‘a’ is the cheapest for this house.

Explanation:

Cheapest method of heating must have least cost per kj of energy. So, convert all the energy in the same unit (say kj) and take select the cheapest method to heat the house.

Given:

Three methods are given to heat a particular house are as follows:

Method (a)

Through Gas, this gives energy of amount $1.33/therm.

Method (b)

Through electric resistance, this gives energy of amount $0.12/KWh.

Method (c)

Through oil, this gives energy of amount $2.30/gallon.

Calculation:

Step1

Change therm to kj in method ‘a’ as follows:

C_{1}=\frac{\$ 1.33}{therm}\times(\frac{1therm}{105500kj})

C_{1}=1.2606\times10^{-5} $/kj.

Step2

Change kWh to kj in method ‘b’ as follows:

C_{2}=\frac{\$ 0.12}{kWh}\times(\frac{1 kWh }{3600kj})

C_{2}=3.334\times10^{-5} $/kj.

Step3

Change kWh to kj in method ‘c’ as follows:

C_{3}=\frac{\$ 2.30}{gallon}\times(\frac{1 gallon }{138500kj})

C_{3}=1.66\times10^{-5} $/kj.

Thus, the method ‘a’ has least cost as compare to method b and c.

So, option ‘a’ is the cheapest for this house.

 

5 0
3 years ago
A body weighs 50 N and hangs from a spring with spring constant of 50 N/m. A dashpot is attached to the body. If the body is rai
lbvjy [14]

Answer:

a) 3.607 m

b) 1.5963 m

Explanation:

See that attached pictures for explanation.

3 0
3 years ago
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