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2 years ago

What is irreversible head loss?

1 answer:

4 0

Irreversible head loss is the loss in head due to the effect of irreversibilities, like friction in piping, and it does not include the losses that occur within the pump or turbine. Mechanical energy loss is the loss of head due to irreversibilities such as friction in pump or turbine.

According to chegg.com

You might be interested in

Write an equation to show how HC2O4− can act as a base with HS− acting as an acid.

Artyom0805 [142]

An acid donates $H^{+}$ ion in aqueous solution. A base accepts $H^{+}$ ion in aqueous solution.

The equation representing the acid base reaction of $HC_{2}O_{4}^{-}$ and $HS^{-}$ :

$HC_{2}O_{4}^{-}(aq) + HS^{-}(aq) ----> H_{2}C_{2}O_{4}(aq)+S^{2-}(aq)$

In the above reaction, as $HC_{2}O_{4}^{-}$ acts as a base it is accepting the hydrogen ion from $HS^{-}$ . Similarly, $HS^{-}$ donates its hydrogen ion to $HC_{2}O_{4}^{-}$ acting as an acid.

7 0

2 years ago

Rhett is solving the quadratic equation 0= x2 – 2x – 3 using the quadratic formula. Which shows the correct substitution of the

bekas [8.4K]

The correct values I believe would be a=1 b=-2 and c=-3.

8 0

2 years ago

The Michaelis‑Menten equation models the hyperbolic relationship between [S] and the initial reaction rate V 0 V0 for an enzyme‑

ipn [44]

Answer:

The Michaelis‑Menten equation is given as

v₀ = Kcat X [E₀] X [S] / (Km + [S])

where,

Kcat is the experimental rate constant of the reaction; [s] is the substrate concentration and

Km is the Michaelis‑Menten constant.

Explanation:

See attached image for a detailed explanation

3 0

2 years ago

A 0.380 kg sample of aluminum (with a specific heat of 910.0 J/(kg x K)) is heated to 378 K and then placed in 2.40 kg of water

lbvjy [14]

Answer:

The equilibrium temperature of the system is 276.494 Kelvin.

Explanation:

Let consider the system formed by the sample of aluminium and water as a control mass, in which the sample is cooled and water is heated until thermal equilibrium is reached. The energy process is represented by First Law of Thermodynamics:

$Q_{water} -Q_{sample} = 0$

$Q_{water} = Q_{sample}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{sample}$ - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

$m_{w}\cdot c_{p,w}\cdot (T-T_{w}) = m_{s}\cdot c_{p,s}\cdot (T_{s}-T)$

Where:

$m_{s}$ , $m_{w}$ - Mass of water and the sample of aluminium, measured in kilograms.

$c_{p,s}$ , $c_{p,w}$ - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

$T_{s}$ , $T_{w}$ - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

$T$ - Temperature which system reaches thermal equilibrium, measured in Kelvin.

The final temperature is now cleared:

$(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}$

$T = \frac{m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}}{m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s}}$

Given that $m_{s} = 0.380\,kg$ , $m_{w} = 2.40\,kg$ , $c_{p,s} = 910\,\frac{J}{kg\cdot K}$ , $c_{p,w} = 4186\,\frac{J}{kg\cdot K}$ , $T_{s} = 378\,K$ and $T_{w} = 273\,K$ , the final temperature of the system is:

$T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}$

$T = 276.494\,K$

The equilibrium temperature of the system is 276.494 Kelvin.

7 0

2 years ago

Calculate the maximum concentration (in m) of silver ions (ag+) in a solution that contains 0.025 m of co32-. the ksp of ag2co3

Helen [10]

Equilibrium equation is

Ag2CO3(s) <---> 2 Ag+(aq) + CO32-(aq) 

From the reaction equation above, the formula for Ksp: 

Ksp = [Ag+]^2 [CO32-] = 8.1 x 10^-12 
You know [CO32-], so you can solve for [Ag+] as: 
(8.1 x 10^-12) = [Ag+]^2 (0.025) 
[Ag+]^2 = 3.24 x 10^-10 
[Ag+] = 1.8 x 10^-5 M

5 0

2 years ago