Answer:
401.3 kg/s
Explanation:
The power plant has an efficiency of 36%. This means 64% of the heat form the source (q1) will become waste heat. Of the waste heat, 85% will be taken away by water (qw).
qw = 0.85 * q2
q2 = 0.64 * q1
p = 0.36 * q1
q1 = p /0.36
q2 = 0.64/0.36 * p
qw = 0.85 *0.64/0.36 * p
qw = 0.85 *0.64/0.36 * 600 = 907 MW
In evaporation water becomes vapor absorbing heat without going to the boiling point (similar to how sweating takes heat from the human body)
The latent heat for the vaporization of water is:
SLH = 2.26 MJ/kg
So, to dissipate 907 MW
G = qw * SLH = 907 / 2.26 = 401.3 kg/s
Answer:
what's what is what is it return if you have any clue please let me know Translate in English and send me because most of them are are from English I think so
Answer:
0.4 Dinas*s/cm^2
Explanation:
Tenemos una viscosidad:
V = 0.04 N*s/m^2
Y queremos reescribir esto en Dinas*s/cm^2
Primero transformemos la unidad del denominador, es decir, tenemos que pasar de 1/m^2 a 1/cm^2
Para ello, usamos que:
1m = 100cm
entonces:
(1m/100cm) = 1
Si elevamos ambos lados al cuadrado, obtenemos:
(1m/100cm)^2 = 1
Ahora podemos multiplicar el valor de la viscosidad por esto (que es igual a 1)
V = 0.04 N*s/m^2*((1m/100cm)^2 = 0.00004 N*s/cm^2
Ahora debemos convertir de Newtons a Dinas
Sabemos que:
1 N = 100,000 dinas
1 = (100,000 dinas/1N)
Entonces, de vuelta podemos multiplicar nuestra viscosidad por (100,000 dinas/1N), que es igual a 1 (asi que no cambia el valor, solo sirve para cambiar las unidades)
0.00004 N*s/cm^2 = (100,000 dinas/1N)*(0.00004 N*s/cm^2)
= (100,000 dinas)*(0.00004 s/cm^2)
= 0.4 Dinas*s/cm^2
Answer:
• Gear ratios compare the output (or driven gear)
to the input (or drive gear)
• Gear Ratios can be determined using number
(n) of teeth on the gear or diameter (d) of the
gear
• If the output gear is larger than the
input gear the speed will decrease
• If the output gear is smaller than the
input gear the speed will increase
Answer:
oxygen deficit = 3.851 mg/L
Explanation:
given data
flow rate of the river= 165 × 
gal/d
saturation value of dissolved oxygen = 9.17 mg/L
to find out
oxygen deficit he two flows
solution
we will apply here formula for dissolved oxygen content after dilution is
Do mix = 
..........................1
here Qw is rate of flow of waste water i.e 26 × gal/d
(Do)w is Do of waste waster i.e 1 mg/L
Qr is aret of flow of river i.e 165 × gal/d
(Do)r is do of river water i.e 6 mg/L
so put all value in equation 1 we get
Do mix = 
solve we get
Do mix = 5.319 mg/L
so
oxygen deficit = saturation oxygen - (Do) mix ..............2
oxygen deficit = 9.17 - 5.319
oxygen deficit = 3.851 mg/L
